Mahesh

Here students can locate TS Inter 2nd Year Zoology Notes 3rd Lesson Human Anatomy and Physiology – III to prepare for their exam.

TS Inter 2nd Year Zoology Notes 3rd Lesson Human Anatomy and Physiology – III

→ Human body has two important systems to give it the right posture and movement of body parts
(a) Muscular system
(b) Skeletal system.

→ Human body has hundreds of muscles, about 640 in total.

→ Skeletal muscles constitute the bulk of our body.

→ Skeletal muscles are controlled by the somatic nervous system and the cardiac and visceral muscles are controlled by the autonomous system.

→ Skeletal system has two major components, the axial skeleton (skull, vertebral column, ribs and sternum) and appendicular system (the limb skeletons, girdles etc.).

→ We use limbs for changing body postures and locomotion too.

→ In animals locomotion is performed generally to search for food, shelter, , mate suitable breeding grounds, and favourable climatic conditions or to escape from enemies/ predators.

→ In adult human beings skeletal system is made up of 206 bones and a few cartilages. It is grouped into two principal divisions. The axial skeleton and the appendicular skeleton.

→ Axial skeleton comprises of 80 bones.

→ Appendicular skeleton comprises of 126 bones.

→ Nervous system evolved from the basic non-polarised nerve cells forming a diffuse nerve net as seen in the diploblastic organisms to a highly organized integrating system with the brain.

→ Man came to know more about brain with the help of “Functional Magnetic Resonance Imaging” (FMRI) technique.

→ Nervous tissue has connecting cells called neurons and supporting cells called glial cells.

→ The system mainly controls the body activities by receiving stimuli, processing them and reacting to them by sending motor signals.

→ Hippocampus of the brain is responsible for the formation and recall of memory.

→ Research is going on why people develop dementia, involving memory loss as seen in Alzheimers.

→ Co- ordination is the process through which two or more organs interact and complement the functions of one another.

→ The neural system provides an organized network of point to point connections for a quick co-ordination.

→ Human neural system is divided into two parts

• Central Neural System (CNS)
• Peripheral Neural System (PNS).

→ CNS includes the brain and the spinal cord.

→ Archibald Hill:
A.V. Hill christened Archibald Vivian, CH OBE FRS (26 September 1886 – 3 June 1977), was an English physiologist one of the founders of the diverse disciplines of biophysics and operations research. He shared the 1922 Nobel Prize in Physiology or Medicine for his elucidation of the production of heart and mechanical work in muscles.

→ Otto Fritz Meyerhof:
Biography; Meyerhof was born in Hildesheim. He spent most of his childhood in Berlin, where he started his study of medicine.
In 1912, he moved to the University of Kiel, where he became professor in 1918. In 1922, he was awarded the Nobel Prize in Medicine, with Archibald Vivian Hill, for his work on muscle metabolism, including glycolysis.

→ Camillo Golgi:
Camillo Golgi : (1843-1926) was an Italian physician, pathologist, scientist; and Nobel laureate. Camillo Golgi was born in July 1843 in the village of Corteno, Lombardy, then part of the Austrian Empire. The village is now named Corteno Golgi in his honour. His father was a physician and district medical officer. Golgi studied at the University of Pavia, where he worked in the experimental pathology laboratory under Giulio Bizzozero, who elucidated the properties of bone marrow. He graduated in 1865. He spent much of his career studying the central nervous system. Tissue staining techniques in the later half of the 19th century were inadequate for studying nervous tissue. While working as chief medical officer in a psychiatric hospital he experimented with metal impregnation of nervous tissue, using mainly silver (silver staining). He discovered a method of staining nervous tissue which would stain a limited number of cells at random, in their entirety. This enabled him to view the paths of nerve cells in the brain for the first time. He called his discovery the “black reaction” (in Italian, reazine nera), which later received his name (Golgi’s method) or Golgi stain.

→ Santiago Ramon y Cajal:
Santiago Ramon y Cajal For MemRS (Spanish 1852-1934) was a Spanish pathologist, histologist, neuroscientist, and Nobel laureate. His pioneering investigations of the microscopic structure of the brain were original: he is considered by many to be the father of modern neuroscience. He was skilled at drawing, and hundreds of his illustrations of brain cells are still used for educational purposes today.

Here students can locate TS Inter 2nd Year Zoology Notes 2nd Lesson Human Anatomy and Physiology – II to prepare for their exam.

TS Inter 2nd Year Zoology Notes 2nd Lesson Human Anatomy and Physiology – II

→ All multicellular animals require a circulatory system to carry nutrients and oxygen to the body parts, collect wastes including carbondioxide from the tissues and carry them to the appropriate excretory organs.

→ Blood and lymph are the only fluid tissues of human body.

→ Blood vascular system is involved in the defence system of the body.

→ Blood contains several types of proteins such as albumins and globulins.

→ Blood carries various hormones to body, hence deserves to be reckoned as the third integrating system. ,

→ Circulatory system has a pumping station, the heart.

→ Different groups of animals have evolved various methods for transport of substances between body parts.

→ Blood is the most commonly used circulatory fluid by higher organisms.

→ In higher animals in addition to blood, another body fluid, lymph also helps in the transport of certain substances.

→ Blood is referred to as “red river of life”.

→ Excretion is the elimination of nitrogenous and other waste materials from the body.

→ Kidneys are the chief excretory organs.

→ Aquatic animals excrete Ammonia (ammonotelic).

→ In others, ammonia is converted into urea which is less toxic via the ornithine cycle in the liver (ureotelic).

→ In insects, reptiles, birds and land snails the end product of nitrogen metabolism is uric acid (uricotelic).

→ Some animals excrete amino acids also (aminotelic).

→ Kidneys also maintain acid – base balance by excreting H⁺ ions.

→ The regulation of internal fluids (water balance) is an example of homeostasis.

→ A variety of excretory organs are present in the animal kingdom.

→ Some of the excretory structures in invertebrates are mentioned here. They are Protonephridium, Metanephridia, Malpighian tubules, Antennary

→ glands or green glands, Coxal glands and Kidneys.

→ William Harvey:
William Harvey : (1 April 1578 – 3 June 1657) was an English physician. He was the first to describe completely and in detail the systemic circulation and properties of blood being pumped to the body by the heart, though earlier writers had provided precursors of the theory. After his death the William Harvey Hospital was constructed in the town of Ashford, several miles from his birthplace of Folkestone.

→ Sir William Bowman:
Sir William Bowman, 1st Baronet (20 July 1816 – 29 March 1892) was an English surgeon, histologist and anatomist. He is best known for his research using microscopes to study various human organs, though during his lifetime he pursued a successful career as an opthalmologist.
His earliest notable work was on the structure of striated muscle.

He was elected a Fellow of the Royal Society in 1841. At the young age of 25, he identified what then became known as the Bowman’s capsule, a key component of the nephron. He presented his findings in 1842 in his paper “On the Structure and Use of the Malpighian Bodies of the Kidney” to the Royal Society and was awarded the Royal MedaL

→ Sir William Bowman, 1st Baronet

• Born: 20 July 1816 Nantwich, Cheshire, England
• Died: 29 March 1892 (aged 75) Dorking, Surrey
• Nationality: English
• Fields: Surgeon, Histologist Anatomist

Here students can locate TS Inter 2nd Year Zoology Notes 1st Lesson Human Anatomy and Physiology – I to prepare for their exam.

TS Inter 2nd Year Zoology Notes 1st Lesson Human Anatomy and Physiology – I

→ Digestive system : Fuel provider and Life sustainer.

→ It consists of alimentary canal and associated glands.

→ Human beings are mostly omnivores in food habits.

→ Complex items of food have to be broken down into simple, absorbable substances.

→ Food mainly consists of carbohydrates, fats, proteins, minerals and vitamins.

→ Alimentary canal is generally long in herbivores and comparitively shorter in carnivores.

→ The major site of digestion in the human gut is the small intestine.

→ Food is one of the basic requirements of all living organisms as it provides energy and organic material for growth and repair of tissues.

→ The breakdown of biomacro molecules into simple substances is required for their absorption into cells.

→ The process of conversion of the complex food substances into their simple absorbable forms is called digestion and carried out by our digestive system. Digestion involves both mechanical and biochemical processes.

→ Respiratory system is a catabolic process of release of energy mostly by oxidation of foods.

→ Oxygen obtained from surrounding environment is utilised in production of metabolic water.

→ Birds have developed a technique of continuous exchange of gases even during expiration due to presence of air sacs and parabronchii.

→ The rib cage and the diaphragm help mammals breath in air more effectively.

→ Homeostasis of oxygen and carbondioxide are under the control of the respiratory centre.

→ Inhalation and exhalation are under the control of the medulla oblongata.

→ There is a pneumotaxic centre in the pons and it controls the rate and depth of breathing.

→ The muscles of the elephant, seal like in some other aquatic mammals contains myoglobin which has more affinity for oxygen.

→ Processes leading to and including the chemical break down of food materials to provide energy for life is called Respiration.

→ Respiration is a vital feature of life.

Here students can locate TS Inter 2nd Year Botany Notes Unit 6 Plants, Microbes and Human Welfare to prepare for their exam.

TS Inter 2nd Year Botany Notes Unit 6 Plants, Microbes and Human Welfare

→ Plant breeding is a technique of manipulation of plant species in order to create desire plant types that are better suited for cultivation, give better yields and are disease resistant

→ Green revolution was dependent mainly on plant breeding techniques.

→ Methods of breeding for disease resistance are

• Conventional breeding method
• Mutation breeding.

→ Insect resistance in host crop plants may be due to morphological, biochemical or physiological characteristics.

→ Biofortification is the process of developing crops with higher levels of vitamins, minerals, proteins and healthier fats to improve public health.

→ Single Cell Protein (SCP) is an alternate protein source for animal and human nutrition from certain Microorganisms like Spirulina.

→ Techniques of tissue culture and somatic hybridisation offer vast potential for manipulation of plants in vitro to produce new varieties.

→ Microbes are the major components of biological systems on the earth.

→ Various types of microbes are protozoa, bacteria, fungi, viruses, viroids and prions.

→ Microbes in household products are common experience of everyday life.

→ The common products obtained by the use of microbes are curd, dough, toddy, cheese eta

→ Microbes are used to produce industrial products like lactic acid, acetic acid and alcohol which are used in a variety of processes in the industry.

→ Antibiotics are chemical substances which are produced by some microbes and can kill or retard the growth of the disease causing microbes.

→ Microbes play a major role in sewage treatment

→ Biogas produced by microbes is used as the source of energy in rural areas.

→ Microbes can also be used to kill harmful pests. This process is referred to as biocontrol. There is a need for the aggressive use of biofertilisers in place of chemical fertilisers.

→ Microbes, play an important role in the welfare of human society and have been put to adverse use.

→ Dr. Swaminathan at the 100th Indian Science Congress, Kolkata.

• Born: 7 August 1925 Kumbakonam, Tamil Nadu
• Residence: Chennai, Tamil Nadu
• Nationality: India
• Fields: Agricultural science
• Institutions: MS Swaminathan Research Foundation
• Alma mater:
  • Maharajas College
  • Tamil Nadu Agricultural University
  • University of Cambridge
  • University of Wisconsin-Madison
• Known for: High-yielding varieties of wheat in India
• Influences: Dr. Noman Borlaug
• Notable awards:
  • Padma Shri (1967)
  • Padma Bhushan (1972)
  • Padma Vibhushan (1989)
  • World Food PrIze (1987)

→ Mankombu Sambasivan Swaminathan:
(born 7 August 1925) is an Indian geneticist and international administrator, renowned for his leading role in India’s “Green Revolution”, a program under which high- yield varieties of wheat and” rice seedlings were planted in the fields of poor farmers.
Swaminathan is known as the “Father of the Green Revolution in India”, for his leadership and success in introducing and further developing high-yielding varieties of wheat in India. He is the founder and Chairman of the MS Swaminathan Research Foundation.
From 1972 to 1979 he was director general of the Indian Council of Agricultural Research, and he was minister of Agriculture from 1979 to 1980. He served as director general of the International Rice Research Institute (1982 – 88) and became president of the International Union for the Conservation of Nature and Natural Resources in 1988.

Here students can locate TS Inter 2nd Year Botany Notes Unit 5 Biotechnology to prepare for their exam.

TS Inter 2nd Year Botany Notes Unit 5 Biotechnology

→ Biotechnology is the science of utilising the properties and uses of microorganisms or to exploit cells and the cell constituents at industrial level for generating useful products essential to life and human welfare.

→ The two core techniques that enabled birth of modern biotechnology are Genetic engineering and tissue culture.

→ Tools of Recombinant DNA Technology are restriction enzymes, polymerase enzymes, ligases, vectors and the host organism.

→ Restriction endonuclease recognises a specific palindromic nucleotide sequence in the DNA.

→ The DNA used as a carrier for transferring a fragment of foreign DNA into a suitable host is called vector. ^

→ Large scale production involves the use of bioreaction.

→ Downstream processing and quality control testing vary from product to product Biotechnology has given humans several useful products by using microbes, plants, animals and their metabolic machinery.

→ The applications of biotechnology include therapeutics, diagnostics, genetically modified crops for agriculture processed food, bioremediation, waste treatment and energy production.

→ Genetically Modified Organisms (GMO) are plants, animals, bacteria and fungi whose genes have been altered by manipulation.

→ Biotechnology in agriculture are the production of pest resistant plants, eg :Bt cotton. Biotechnology in medicine have made immense impact in the area of health care. Gene therapy is the insertion of genes into an individuals cells and tissues to treat heriditary diseases.

→ Molecular diagnosis helps to solve the problems of early diagnosis and treatment of diseases.

→ Herbert Boyer

• Fields: biology scientist
• Notable awards: National Medal of Science (1990)

Herbert Boyer : Born in 1936. He graduated at University of Pittsburgh in 1963, followed by three years of post – graduate studies at Yale in 1966. He took assistant professorship at the University of California. Boyer had discovered a restriction enzyme of E.coli, that cut DNA strands at specific DNA sequence producing “Sticky ends” that could stick to otherpieces of DNA.

→ Stanley Cohen

• Born: November 17, 1922 Brooklyn, New York
• Nationality: American
• Fields: Biochemiry
• Institutions: Washington University in St. Louis
• Alma mater: University of Michigan
• Thesis: The Nitrogenous Metabolism of the Earthworm (1949)
• Doctoral advisor: Howard B. Lewis
• Known for: Nerve growth factor
• Notable awards: Nobel Prize In Physiology or Medidne (1986), Franklin Medal (1987)

Stanley Cohen : a Stanford University professor, had been working on ways to isolate specific genes in antibiotic carrying plasmids and clone them individually through introducing them to E.coli bacteria.
Boyer and Cohen met in 1972 while presenting papers in Hawaii at a conference or bacterial plasmids.
Following the conference, the two colleagues met and discussed with each other. Both agreed to collaborate and in a matter of months succeeded in splicing a piece of foreign DNA into a plasmid carrier, which then inserted genetic information into a bacterium. Then the bacterium reproduced, copied the foreign DNA into its offspring. This break through was the basis upon which the discipline of biotechnology was founded.

Here students can locate TS Inter 2nd Year Botany Notes Unit 4 Molecular Biology to prepare for their exam.

TS Inter 2nd Year Botany Notes Unit 4 Molecular Biology

→ Deoxyribon ucleic acid (DNA) and ribonucleic acid (RNA) are the two types ofn ucleic acids found in living systems.

→ Nucleic acids are long polymers of nucleotides.

→ In 1953, Watson and Crick, based on the X-ray diffraction proposed a very simple but i famous Double Helix model for the structure of DNA.

→ RNA was the first to evolve and DNA was derived from RNA.

→ Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA → RNA → Protein.

→ DNA replicates semi conservatively, the process is guided by the complementary H-bonding.

→ The proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase.

→ The process ofcopying genetic information from one strand ofDNA into RNA is termed as transcription.

→ The gene is defined as the functional unit of inheritance.

→ The DNA sequence coding for tRNA or rRNA molecule also define a gene.

→ Cistron is defined as a segment of DNA coding for a polypeptide.

→ In bacteria, the transcribed mRNA is functional hence can be directly translated

→ In eukaryotes, the gene is split

→ The coding sequences exons, are interrupted by non-coding sequences called introns.

→ Introns are removed and exons are joined to produce functional RNA by a process called splicing.

→ In capping an unusual nucleotide is added to the 5′ end ofhn RNA.

→ In tailing, adenylate residues are added at 3′- end in a template independent manner.

→ Genetic code is that which direct the sequence of amino acids during synthesis of proteins.

→ The codon is triplet 61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons.

→ The relationships between genes and DNA are best understood by mutation studies.

→ Regulation of transcription is the primary step for regulation of gene expression.

→ In bacteria more than one gene is arranged together and regulated in units called operons.

→ Lac operon is the prototype operon in bacteria, which codes fongenes responsible for metabolism of lactose.

→ The operon is regulated by the amount of lactose in the medium where the bacteria are grown. Therefore, this regulation can also be viewed as regulation of enzyme I synthesis by its substrate.

→ James Dewey Watson

• Born: James Dewey Watson April 6, 1928 (Chicago, Illinois, U.S.)
• Nationality: American
• Fields: Genetics
• Institutions:
  • Indiana University
  • Cold Spring Harbor Laboratory
  • Harvard University
  • University of Cambridge
  • National Institute of Health
• Alma mater:
  • University of Chicago
  • Indiana University
• Thesis:
  • The Biological Properties of XRay
  • Inactivated Bacteriophage (1951)
• Doctoral advisor: Salvador Luna
• Doctoral students: Mario Capecchi
• Other notable students: Ewan Bimey
• Known for: DNA structure
• Molecular biology
• Notable awards: Nobel Prize for Physiology or Medione(1962), Copley Medal (1993)
• spouse: Elizabeth Watson (née Lewis)

→ Francis Crick:

• Born: Franas Harry Compton Crick 8 June 1916 Weston Favell, Northam ptonshire, England, UK
• Died: 28 July 2004 (age 88) San Diego, Cahfornia, US. Cdori oenoer
• Residence: UK, US.
• Nationality: British
• Fields: Physics, Molecular biology
• Institutions:
  • University of Cambridge
  • University College London
  • Salk Institute for Biological Studies
• Alma mater: University College London (BSC) Gonville and Calus College, Cambridge (PhD)
• Thesis: Polypeptides and Proteins : X-ray studies (1954)
• Doctoral advisor: Max Perutz
• Known for: DNA structure consciousness
• Notable awards: Nobel Prize (1962)

→ In 1962 James Watson and Francis Crick jointly received the Nobel Prize in Physiology or Medicine for their determination of the structure of deoxyribonucleic acid (DNA)

• They both were also honoured by the John Collins Warren Prize of the Massachusetts General Hospital in 1959, the Lasker Award in 1960, the Research Corporation Prize in 1962.
• James Dewey Watson was born in Chicago on 6 April 1928. In 1951, 23 year old James Watson a Chicago – born American, arrived at the Cavendish Laboratory in Cambridge. Watson had two degrees in Zoology, a bachelor’s degree from the University of Chicago and a doctorate from the University of Indiana where he became interested in genetics his doctorate degree in 1950 was on the study of effect of hard x -rays on bacteriophage multiplication.
• Francis Harry Compton Crick was born on 8 June 1916 at Northampton, England. He studied physics at University College London and obtained a B.Sc in 1937. He completed Ph.D in 1954 on a thesis entitled ‘X-ray Diffraction Polypeptides and Proteins”.
• Watson and Crick had common interest in solving the DNA structure. Their first serious effort was unsatisfactory. Their second effort based upon more experimental evidence and better appreciation of nucleic acid literature resulted in the proposal of the complementary double helical configuration in 1953.

Here students can locate TS Inter 2nd Year Botany Notes Unit 3 Genetics to prepare for their exam.

TS Inter 2nd Year Botany Notes Unit 3 Genetics

→ The biological science that deals with the study ofheriditary and variation is known as Genetics.

→ Gregor Johann Mendel proposed the ‘laws of inheritance’ those led down foundation of the Genetics through his famous hybridization experiments on Phisum sativum.

→ Law of segregation states, that “the two alleles of a gene present together in het-erozygous state, do not fuse, remain distinct and segregate during meiosis.

→ Law of independent assortment states that “Genes for different characters are inherited independently of one other”.

→ The dominant characters are expressed when factors are either in homozygous or in heterozygous condition. (law of dominance)

→ The recessive characters are expressed only in homozygous condition. Different com- biriations of gametes are theoretically represented in a square tabular form known as “Punnett square.”

→ A cross involving a F₁ individual and any one of its parents is called Back cross.

→ A cross between a F₁ individual and its double recessive parents is called Test cross.

→ Monohybrid cross: Phenotypic ratio is 3 :1 and genotypic ratio is 1; 2:1.

→ Dihybrid cross: Phenotypic ratio is 9:3:3:1 and genotypic ratio is 1:2:2:4:1:2 : 1:2:1.

→ Mendel’s laws were extended in the form of “Chromosomal theory of inheritance.”

→ Later it was found that Mendel’s law of independent assortment did not hold true for the genes that are located on the same chromosomes. These genes are called linked genes.

→ Closely located genes assorted together and distantly located genes, due to recombi-nation, assorted independently.

→ Mutations involves changes in chromosomes and/or genes. They help to increase variability which might be useful in crop improvement

→ Gregor Johann Mendel:

• Born: Johann Mendel, July 22, 1822 Heinzendorf bei Odrau, Austrian Empire (Now Hync (ice, czech Republic))
• Died: January 6,1884 (aged 61) Brno (Brunn), Austria – Hungary (now Czech Republic)
• Nationality: Empire of Austria – Hungary’
• Fields: Genetics
• Institutions: Abbey of St. Thomas in Brno
• Alma mater: University of Olomouc University of Vienna
• Known for: Creating the science of genetics

Gregor Johann Mendel laid foundation for the Science of Genetics and Study of Hereditary. Mendel demonstrated that the inheritance of certain traits in pea plants follows particular patterns, now referred to as the laws of Mendelian inheritance. The profound significance of Mendel’s work was not recognized until the turn of the 20th century, when the independent rediscovery of these laws initiated the modern science of genetics. In 1 843 Mendel began his training as a priest. Brno Johann Mendel, he took the name Gregor upon entering religious life. In 1851 he was sent to the University of Vienna to study. At Vienna, his professor of physics was Christian Doppler. Mendel returned to his abbey in 1853 as a teacher, principally of physics, and by 1867, he had replaced Napp as abbot of the monastery. During this time he began the experiments for which he is best known. He published the results of his studies under the title “Experiments on Plant Hybrids”

Here students can locate TS Inter 2nd Year Botany Notes Unit 2 Microbiology to prepare for their exam.

TS Inter 2nd Year Botany Notes Unit 2 Microbiology

→ Bacteria is omnipresent It occurs in all climatic condition even in extreme conditions.

→ Bacteria are important microbes.

→ Even though bacteria are microscopic, they perform certain acts of living things i.e., they take food, grow and reproduce.

→ Cell wall provides shape and protection to bacteria.

→ Bacteria occur in several shapes – cylindrical (Bacillus), spherical (Cocci), spiral (Spirillum).

→ Flagella are present in the motile bacteria depending upon the number and distribution of flagella bacteria are classified as Monotrichous, Lophotrichous, Amphitrichous and Peritrichous.

→ Bacteria normally reproduce by binary fission.

→ Bacterial plasmids can be manipulated in the laboratory and are used as vectors in rDNA technology.

→ Genetic exchange among bacteria takes place by conjugation, transformation and transduction.

→ Bacteria can be considered as both friends and foes of man.

→ The study of viruses is known as virology.

→ Viruses are very small infectious, obligate, intracellular parasites.

→ Viruses are acellular and lack metabolism of their own.

→ Virus has a central core of nucleic acid (DNA/RNA single stranded/double stranded). It is covered by protein coat capsid

→ Viroids have only nucleic acid Prions have only proteins.

→ Viruses are grouped according to their shared properties and are classified by ICTV system of classification.

→ TMV is rod shaped virus.

→ T – even bacteriophage is tadpole shaped

→ Virus replication is of two types namely Lytic cycle (virulent phages) and Lysogenic cycle (temperate phages).

→ Lytic cycle can be seen in T-even phage. This replication can be divided into 5 phases namely attachment, penetration, biosynthesis, maturation and release.

→ Lysogenic cycle can be seen in lambda bacteriophage in which the phage DNA becomes an integral part of bacterial DNA and replicates along with it Viruses causes several human, plant and animat diseases.

→ Some of the viruses can cause cancer and they are called oncogenic viruses.

→ Portrait of Antonie van Leeuwenhoek (1632 – 1723) by Jan Verkolje

• Born: October 24, 1632 Delft, Dutch Republic
• Died: August 26, 1723 (age 90) Delft, Dutch Republic
• Residence: Netherlands
• Nationality: Dutch
• Fields: Microscopist and Bilogist

He is commonly known as “the Father of Microbiology” and considered to be the first microbiologist. He is best known for his work on the improvement of the microscope and for his contributions towards the establishment of microbiology. Using his handcrafted microscopes, he was the first to observe and describe single – celled organisms, which he originally referred to as animalcules, and which are now referred to as microorganisms. He was also the first to record microscopic observations of muscle fibers, bacteria, spermatozoa, and blood flow in capillaries (small blood vessels). Van Leeuwenhoek did not author any books, although he did write many letters. In 1674 Leeuwenhoek communicated his discoveries to the Royal Society of London, sending detailed drawings which were the first representations of Bacteria and Protozoa.

Here students can locate TS Inter 2nd Year Botany Notes Unit 1 Plants Physiology to prepare for their exam.

TS Inter 2nd Year Botany Notes Unit 1 Plants Physiology

→ Transport over longer distances through the vascular system is called translocation.

→ The movement of molecules from high concentration to lower concentration till it reaches equilibrium is called diffusion.

→ Membrane proteins help the movement of molecules across the membrane by diffusion is called facilitated diffusion.

→ Porins are proteins that form huge pores in the outer membrane of the plastids, mitochondria and some bacteria; allowing molecules upto the size of small proteins to pass through.

→ Water potential is a relative term, which refers to the chemical potential of pure water to that of chemical potential of a solution.

→ Water potential of pure water is zero.

→ Solute potential (Ψ_s) and pressure potential (Ψ_p) are the two main components that determine water potential

→ Osmosis is the term used to refer the diffusion of water across semi-permeable membrane.

→ The shrinkage of protoplast of cell due to loss of water and turgor is called plasmolysis.

→ The absorption of water by seeds and dry wood is done by imbibition.

→ Water, minerals and food cannot be moved within the body of a plant by diffusion alone therefore, they are transported by a mass flow system.

→ The movement of substance in bulk from one point to another as a result of pressure differences between the two points is called mass flow or pressure flow.

→ Water absorbed by root hairs moves deeper into the root by two distinct pathways. They are:

• Apoplast
• Symplast

→ Transpiration pull is the most acceptable factor to explain the transport of water or Ascent of sap.

→ Loss of water from the aerial pqrts of the plant body in the form of vapour is called transpiration.

→ Removal of excess, water through the tips of leaves of plant is done by guttation.

→ Phloem is responsible for transport of food (primary sucrose) from the source to the sink.

→ The translocation in phloem is bidirectional as source-sink relationship is variable due to seasonal changes. . .

→ The translocation in phloem is explained by the pressure-flow hypothesis.

→ Plants obtain their inorganic nutrients from air, water and soil

→ The technique of growing plants in a specified nutrient solution is known as hydroponics.

→ Not all the mineral elements that they absorb are required by plants.

→ Based upon the criteria for essentiality few elements have been found to be absolutely essential for plant growth and metabolism.

→ The elements required in large quantities are called macronutrients while the elements required in less quantities or in trace are termed as micronutrients.

→ Deficiency of essential elements may lead to symptoms called deficiency symptoms.

→ Plants absorb elements through roots by either passive or active processes.

→ Mineral salts are translocated through thexylem along with the ascending stream of water, which is pulled up through the plant by transpiration pull

→ Nitrogen is the essential mineral element required in the largest quantity by plants.

→ Plants cannot use atmospheric nitrogen directly.

→ Leguminous plants and some non-leguminous plants (alnus) can fix atmospheric nitrogen into the soil in the form of nitrates with the help of Rhizobium and Frankia bacteria.

→ The pink colour in the root nodules is due to the presence of leguminous haemoglobin or leg-haemoglobin.

→ The enzyme nitrogenase plays very important role in biological N₂ fixation and it is very sensitive to oxygen.

→ Enzymes are biological catalysts. All enzymes are proteins.

→ Enzymes through their active site, catalyse reactions at a high rate.

→ Inorganic catalysts work efficiently at high temperature whereas enzymes get damaged at high temperatures.

→ Catalysed reactions proceeds at rates vastly higher than that of uncatalysed ones.

→ A multistep chemical reaction, when each of the steps is catalysed by the same enzyme complex or different enzymes it is called metabolic pathway.

→ In our skeletal muscle, under anaerobic condition lactic acid is formed Under normal aerobic conditions, pyruvic acid is formed

→ Enzymes i.e., proteins with three dimensional structures including an active site, convert a substrate (S) into a product (P).

→ The difference between average energy content of S and transition state is called activation energy.

→ The activity of enzyme is affected by different factors.

→ Ribozymes are nucleic acids with catalytic power.

→ Enzymes exhibit substrate specificity, require optimum temperature and pH for maximum activity.

→ Enzymes are divided into six classes.

→ Each enzyme can be identified by 4 numbers that denote – major class, sub-class, sub – subclass and its serial number.

→ Green plants are autotrophs. They form their food by means of a process Photosynthesis.

→ Photosynthesis is a Physico-chemical process by which they use light energy to drive the synthesis of carbohydrates.

→ Photosynthesis takes place in the green leaves of the plant due to presence ofchloroplast

→ Chlorophyll pigments are chlorophyll a, chlorophyll b, xanthophylls and carotenoids.

→ Graph showing light absorption by photosynthetic pigment as a function ofwavelength of light is Absorption spectrum.

→ Graph showing rate of photosynthesis as a function of wavelength is Action spectrum.

→ Photosynthesis includes two phases viz., light reaction and dark reaction.

→ Light phase occurs in grana ofchloroplast. In light phase NADPH and ATP are generated.

→ There are two types of photosystems PS I & PS II.

→ PS I has a 700 nm absorbing chlorophyll a P 700 molecule at its reaction centre.

→ PS II has a P 680 reaction centre that absorbs red light at 680 nm.

→ After absorbing light electron get exited and transferred through PS II and PS I and finally to NAD forming NADH.

→ The break down of proton gradient due to movement through the part of the ATPase enzyme releases enough energy for synthesis of ATP.

→ Splitting of water molecules is associated with PS II resulting in the release of O₂ protons and transfer of electrons to PS II.

→ Dark phase occurs in the stroma of chloroplast During dark phase, CO₂ is reduced to carbohydrates by utilising ATP and NADPH produced in light phase.

→ Based on the first stable compound produced during CO₂ reduction, two pathways are identified They are Calvin cycle and Hatch – Slack pathway.

→ Calvin cycle includes 3 phases viz carboxylation, reduction and regeneration. The first stable compound is PGA which is 3 carbon compound Hence Calvin cycle is also called C₃ cycle.

→ Hatch-Slack pathway operates in two photosynthetic cells – Mesophyll and bundle sheath cells. In this the first stable compound is OAA which is 4 carbon compound Hence this pathway is also called C₄ cycle.

→ In C₃ plants, RuBisCO also catalysed a wasteful oxygenation reaction called photorespiration.

→ The breaking of the C-C bonds of the complex compounds through oxidation within the cells leading to release of considerable amount of energy is called respiration.

→ The compounds that are oxidised during respiration are known as respiratory substrates.

→ A cellular process in which glucose is broken down into two molecules of pyruvic acid is termed as glycolysis. It occurs in cytoplasm.

→ The fate of the pyruvate depends on the availability of oxygen and the organism.

→ Under anaerobic condition, either lactic acid fermentation or alcohol fermentation occurs.

→ Fermentation occurs in many prokaryotes, unicellular eukaryotes and in germinating seeds.

→ Aerobic respiration occurs in the presence of oxygen.

→ In aerobic respiration, pyruvic acid is transported into the mitochondria where it is converted into acetyl CoA with the release of CO₂

→ Acetyl CoA then enters the tricarboxylic acid pathway or Krebs cycle in the matrix of mitochondria.

→ In Krebs cycle NADH + H⁺ and FADH₃ are formed

→ After Krebs cycle, Electron transport system occurs in on the inner membrane of mitochondria.

→ As the electrons are moved through the electron transport system, energy released that are trapped to synthesis ATP.

→ This synthesis of ATP is called oxidative phosphorylation as O₂ is the ultimate electron acceptor in this process and it gets reduced to water.

→ The respiratory pathway is involved both anabolism and catabolism. So it is termed as amphibolic pathway.

→ In aerobic respiration 2 ATP molecules are formed Respiratory quotient is defined as the ratio of CO₂ evolved to O₂ taken during respiration.

→ RQ indicates the kind of substate undergoing oxidation in the respiration

→ Growth is regarded as one of the most fundamental and conspicuous characteristics of a living being.

→ Growth is an irreversible permanent increase in size of an organ on its parts or even of an individual cells.

→ Growth conspicuously involves increase protoplasmic material

→ In plants, meristems are the sites of growth.

→ Root and Shoot meristems sometimes along with intercalary meristem, contribute to the growth of plant axes.

→ Growth is indeterminate in higher plants.

→ The growth could be arithmetic or geometrical

→ There are three phases of growth – the lag, the log and the senescent phase.

→ When a cell loses the capacity to divide, it leads to differentiation.

→ Differentiation results in development of structure that is commensurate with the function of the cells finally have to perform.

→ General principles of differentiation for cells, tissues and organs are similar.

→ A differentiated cell may be dedifferentiate and then redifferentiate.

→ Since differentiation in plants is open, the development could also be flexible i.e., the development is the sum of growth and differentiation.

→ Plant exhibit plasticity in development

→ Plant growth and development are under the control of both intrinsic and extrinsic factors.

→ Intrinsic factors are the chemical substances called plant growth regulators (PGR) whereas the extrinsic factors are light, temperature, nutrition, gravity, oxygen etc.

→ PGRbelong to five groups. They are auxins, gibberellins, cytokinins, abscisicacid, andethylene.

→ The influence of the relative duration of day and night on the flowering response of plants is called photoperiodism.

→ According to critical day length, plants are grouped into three

• Short day plants
• Long day plants
• Day – neutral plants.

→ The method ofinducing early flowering in plants by pretreatment at low temperature is called vernalization.

→ JAGADISH CHANDRA BOSE: Jagadish Chandra Bose in Royal Institution, London

• Born: 30 November 1858 Bikrampur, Bengal Presidency, British India
• Died: 23 November 1937 (age 79) Giridih, Bengal Presidency, British India
• Residence: Kolkata, Bengal Presidency, British India
• Nationality: British Indian
• Field: Physics, Biophysics, Biology, Botany, Archaeology, Bengali Literature, Bengali Science Fiction
• Institution:
  • University of Calcutta
  • University of Cambridge
  • University of London
• Alma mater: St. Xavier’s College, Calcutta University of Cambridge
• Academic advisors John Strutt (Rayleigh)
• Notable students: Satyendra Nath Bose, Meghnad Saha
• Known for:
  • Millimetre waves Radio
  • Crescograph Plant soence
• Notable awards:
  • Companion of the Order of the Indian
  • Empire (CIE) (1903)
  • Companion of the Order of the Star of India (CSI) (1911)
  • Knight Bachelor (1917)

→ Acharya Sir Jagadish Chandra Bose:
He pioneered the investigation of radio and microwave optics, made very significant contributions to plant science, and laid the foundations of experimental science in the Indian subcontinent. IEEE named him one of the fathers of radio science. He is also considered as the father of Bengali science fiction. He also invented the crescograph. He used his own invention, the crescograph, to measure plant response to various stimuli, and thereby scientifically proved parallelism between animal and plant tissues. Bose’s demonstration of an apparent power of feeling in plants, exemplified by the quivering of injured plants. His books include Response in the Living and Non-Living (1902) and The Nervous Mechanism. He explained various bio-electrical responses shown by plants through his pulsating theory. He was the first Indian to become a fellow of Royal Society.

Here students can locate TS Inter 2nd Year Physics Notes 10th Lesson Alternating Current to prepare for their exam.

TS Inter 2nd Year Physics Notes 10th Lesson Alternating Current

→ Alternating Current (AC): If the supplied voltage varies like a sine function then it is called “alternating voltage” and the current driven in the circuit is called “alternating current”.

→ Preference of AC:

• We prefer AC supply to DC supply because it is easy to produce.
• It can be easily transformed from high voltage to low voltage or from low voltage to high voltage. 3) It is easy to transmit to longer distances with less line losses.

→ AC Voltage applied to a Resistor: When a AC Voltage V = V_m sin cot is applied to a pure resistor then current, I = \(\frac{V_m}{R}\) sin ωt = \(\frac{\mathrm{V}}{\mathrm{R}}\) and I_m = \(\frac{V_m}{R}\)
In a pure resistor applied voltage and current through resistor are in phase, i.e., when voltage is maximum then current is also maximum. Similarly when voltage is minimum the current is also minimum.

→ Average power consumed in a resistor P = \(\frac{1}{2}\)I_m² R(or) P = I²R = VI (∵I = I_m/√2)

→ AC Voltage through an Inductor: in a pure inductor (i.e., where resistance R = 0) applied voltage V = V_m sin ω₀t.
Current through inductor I = I_m sin (ωt – π/2) In a pure inductor current lags behind
voltage by a phase angle Φ = \(\frac{\pi}{2}\)
Reactance of Inductor X_L = ω_L
I_m = \(\frac{V_m}{X_L}\)

→ AC Voltage through a pure Capacitor:
When AC voltage is applied to a pure capacitor its reactance X_c = \(\frac{1}{\omega \mathrm{C}}\). (called capacitive reactance)
Voltage across capacitor V = V_m sin ωt
Current in the capacitor I = I_m sin (ωt + \(\frac{\pi}{2}\)) In a capacitor current (I) leads the applied voltage by an angle \(\frac{\pi}{2}\).
Power dissipation in a pure capacitor P_c = 0

→ Reactance: The resistance of active components like Inductance (L) and Capacitance (C) changes with frequency (co) of current supplied.
The resistance that changes with frequency is called reactance (X).
Reactance of Inductance = X_L = ωL
Reactance of capacitance = X_C = \(\frac{1}{\omega \mathrm{C}}\)

→ Impedance (Z): The total resistance of a circuit with reactive components like inductance or capacitance or both along with resistance ‘R’ is given by Z = R + X_C + X_L
The total resistance of a circuit with reactive components is called Impedance.
Impedence (Z) of R-L circuit Z = R + ωL
Impedance of C – R circuit Z = \(\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}\)
Impedance of L – C circuit Z = \(\sqrt{\left(\omega L-\frac{1}{\omega \mathrm{C}}\right)^2}\)

→ Phasor diagram: Phasor diagram represents the current in a circuit which contains resistance (R) and reactive components like inductance (L) and capacitance ‘C’.

→ In a circuit let I is the phasor represents the current it is always parallel to VR i.e., vol¬tage along resistance.
Voltage across capacitor V_c lags behind I by an angle \(\frac{\pi}{2}\).
Voltage across inductor VR leads current phasor I by an angle \(\frac{\pi}{2}\)
Phasor relation is V = V_R + V_L + V_c

→ Impedance diagram: Graphical represen-tation impedance of Z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2}\) in the form of a right angle triangle is called “impedance diagram”.

→ Impedance diagram

Where
X – direction represents Resistance R.
Y – direction represents total reactance (X_C – X_L) and Hypotenuse represents impedance.

→ Series LCR Circuit: In series LCR circuit an inductance (L), capacitance ‘C’ and resistance ‘R’ are connected to an AC source. AC voltage through LCR ciruit. Let a voltage V = V_m sin cot is applied to series LCR circuit.

Total voltage in the circuit V = V_L + V_R + V_C
Impedance of circuit,
Z = \(\sqrt{R^2+\left(X_C-X_L\right)^2}=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)
Maximum current I_m = \(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{Z}}\)
Phase angle Φ = tan\(\left(\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right)\)
Resonating frequency, ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}\)

→ Resonance: Resonance is a physical phenomenon at which a system tends to oscillate freely. This particular frequency is called natural frequency. At resonance amplitude of oscillations is large.

→ Series LCR circuit-Resonant frequency:
Resonant frequency of series LCR circuit is ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}\)

→ Sharpness of resonance: In case of series LCR circuit resonant frequency ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}\)
However the amplitude of oscillation is high in between the frequencies ω₂ = ω₀ + Δω and ω₁ = ω₀ – Δω. Where Δω is a small frequency change from ω₀?
ω₂ – ω₁ = 2Δω is called band width of resonance, ω₀/2Δω is called sharpness of resonance.
Note: Tuning circuits with sharp resonance are considered as very good frequency selectors.

→ Power of AC circuit and Power factor:
Power of AC circuit P = I²Z cosΦ. It indicates that power depends not only on current I and Impedance Z of circuit but also cosine of phase angle between I and Z.
The term cos Φ is called power factor.

→ Wattless current: In a circuit with pure inductance or pure capacitance the phase angle between voltage and currents are Φ = \(\frac{\pi}{2}\), so cos Φ = 0. hence no power is dissipated through then even though current passes through them. This current is referred as wattless current.

→ Transformer: A transformer works on the principle of electromagnetic induction”. A transformer will convert high voltage AC current into low voltage AC current or Low voltage AC current into high voltage AC current by keeping VI = constant.

→ Turns ratio: The ratio of number of turns in primary coil (N_p) to number of turns in secondary coil (N_s) is called “transformer turns ratio”.
Voltage at secondary, V_s = \(\)V_p and
Current at secondary, I_s = \(\)I_p

Note:

• Transformers with N_s/ N_p > 1 are called step up transformers.
• Transformers with N_s / N_p < 1 are called step down transformers.
• A transformer will work for a.c currents only.
• The Equation V_s = \(\left(\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\right)\)V_p is applicable for ideal transformers. But in real case a small amount of energy (less than 5%) is wasted due to
  • Flux leakage
  • Resistance of windings
  • Eddy currents and
  • Hysteresis.

→ In AC circuits: Voltage at any instant V = V_m sin cot current at any instant I = I_m sin cot (where I_m = \(\frac{V_m}{R}\))

→ Root mean square values (R.M.S Values):
R.M.S. value of current I = I_m / √2 = 0.707 I_m
R.M.S. value of voltage V = V_m /√2 = 0.707 V_m
Average power
P̅ = \(\frac{1}{2}\)I_m²R = \(\frac{1}{2}\)Iv²/R = \(\frac{1}{2}\)I_mV_m

→ Relation between V and I: V_m = I_m R and V
= I R and P = VI = I²R = \(\frac{\mathrm{V}^2}{\mathrm{R}}\).

→ In pure Inductors : V = V_m sin ωt ; I = I_m (sin ωt – π/2)
In inductance current lags behind voltage by 90° or \(\frac{\pi}{2}\) radians.
Inductive reactance X_L = ωL = 2πυL
Maximum current through inductor I_m = \(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{X}_{\mathrm{L}}}\)

→ In pure Capacitors : Voltage across capacitor V = \(\frac{q}{c}\) = V_m sin ωt.
Current in capacitor I = I_m sin (ωt + π/2).
In a pure capacitor current I leads voltage V by a phase angle 90° or \(\frac{\pi}{2}\) radians.
Reactance of capacitor X_c = \(\frac{1}{\omega \mathrm{C}}\); Maximum current I_m = \(\frac{V_m}{(1 / \omega c)}=\frac{V_m}{X_c}\)
Instantaneous power supplied P_c = \(\frac{\mathrm{i}_{\mathrm{m}} \cdot \mathrm{V}_{\mathrm{m}}}{2}\) sin 2ωt.
Power supplied to capacitor over one complete cycle is zero.

→ In a resistor: Applied Voltage V = V_m sin ωt
Current I = \(\frac{I_m}{R}=\frac{V_m}{R}\) sin ωt
Average power dissipated
P̅ = \(\frac{1}{2}\)I2mR = \(\frac{V^2}{R}\) = VI = \(\frac{\mathrm{V}_{\mathrm{m}} \mathrm{I}_{\mathrm{m}}}{2}\).

→ In series LCR circuit,
Total potential in circuit V = V_L + V_R + V_c.
= V_m/\(\frac{V_m}{Z}\)
Impedance of circuit Z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2}\)
Phase difference Φ = tan^-1\(\left[\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right]\)

Resonant frequency ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}\)
At Resonance ωL = \(\frac{1}{\omega \mathrm{C}}\) and Impedance Z = R.
Band width of circuit 2Δω = ω₂ – ω₁
(where ω₂ > ω₁)
Sharpness of circuit Q = \(\frac{\omega_0}{2 \Delta \omega}=\frac{\omega_0 L}{R}\)
Power in AC circuit P = I²Zcos Φ = VIcos Φ
where is phase between voltage V and current I.

→ In LC CircuIt : Electrical energy stored In charged capacitor U_E = \(\) where q_m = I_m/ω₀
Resonant frequency ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}\)
In transformer;
Turns ratio = \(\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)
Secondary voltage V_s = \(\left[\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\right]\)V_p
Secondary current I_s = \(\left[\frac{\mathrm{N}_{\mathrm{p}}}{\mathrm{N}_{\mathrm{s}}}\right]\)I_p
Back emf in primary V_p = -N_p \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\)
Induced emf or voltage in secondary V_s = N_s \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Complex Numbers Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Complex Numbers Important Questions Short Answer Type

Question 1.
Determine the locus of z, z ≠ 2i, such that Re (\(\frac{z-4}{z-2 i}\)) = 0.
Solution:
Let z = (x + iy)
Now,

x² + y² – 4x – 2y = 0
x² + 4 – 4x – 4 + y² – 2y + 1 – 1 = 0
(x – 2)² + (y – 1)² = 5
∴ The locus of z is (x – 2)² + (y – 1)² = 5
The locus represents a circle with centre (2, 1) and radius √5 units.

Question 2.
Show that \(\frac{2-i}{(1-2 i)^2}\) and \(\frac{-2-11}{25}\) conjugate to each other. [May ’97]
Solution:
Let the given complex number are
z₁ = \(\frac{2-\mathrm{i}}{(1-2 \mathrm{i})^2}\) and z₂ = \(\frac{-2-11 i}{25}\)

∴ z₁ and z₂ are conjuagte to each other.

Question 3.
If x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then, show that x² + y² = 4x – 3.
Solution.

L.H.S:

From (1) and (2);
L.H.S = R.H.S
x² + y² = 4x – 3.

Question 4.
If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\) then, show that 4x² – 1 = 0.
[March ’06, May ‘82, TS- Mar. ‘15, May ’16; AP- May, Mar. 2016, AP & TS – Mar. 2019]
Solution:

Comparing real parts on both sides
x = \(\frac{1}{2}\)
x² = \(\frac{1}{4}\)
4x² = 1
∴ 4x – 1 = 0

Question 5.
If u + iv = \(\frac{2+i}{z+3}\) and z = x + iy, find u, v. [March ’81]
Solution:

Question 6.
If z = 2 – i√7 then show that 3z³ – 4z² + z + 88 = 0. [March 2000]
Solution:
Given that,
z = 2 – i√7
Now,
z² = (2 – i√7)²
= 4 + i² 7 – 4√7i
= 4 – 7 – 4√7i
= – 3 – 4√7i

L.H.S:
3z³ – 4z² + z + 88 = 3z² . z – 4z² + z + 88
= 3 (- 3 – 4i√7) (2 – i√7) – 4(- 3 – 4√7) + (2 – i√7) + 88
= – 18 + 9√7i – 24√7i + 84i² + 12 + 16√7i + 2 -i√7 + 88
= – 18 – 84 + 12 + 2 + 88 = 0 = R.H.S
∴ 3z³ – 4z² + z + 88 = 0.

Question 7.
If (x – iy)^1/3 = a – ib then show that \(\) = 4(a² – b²). (Ap-Mar 2018)
Solution:
Given that
(x – iy)^1/3 = a – ib
Cubing on both sides,
[(x – iy)^1/3]³ = (a – ib)³
x – iy = a³ – i³b³ – 3iab (a – ib)
x – iy = a³ – i² . ib³ – 3 ia²b + 3i²ab²
x – iy = a³ – 3ab² + ib³ – 3ia²b
x – iy = a (a² – 3b²) + ib (b² – 3a²)
Comparing real and imaginary parts on both sides
x = a(a² – 3b²)
\(\frac{x}{a}\) = a² – 3b² ……………(1)

– y = b(b² – 3a²)
\(-\frac{y}{b}\) = – (3a² – b²)
\(\frac{y}{b}\) = 3a² – b² …………..(2)
Now. (1) + (2)
\(\frac{x}{a}\) + \(\frac{y}{b}\) = a² – 3b² + 3a² – b²
∴ \(\frac{x}{a}\) + \(\frac{y}{b}\) = 4 (a² – b²)

Question 8.
Write \(\left(\frac{a+i b}{a-i b}\right)^2-\left(\frac{a-i b}{a+i b}\right)^2\) in the form x + iy.
Solution:
Given that,

Question 9.
If x and y are real numbers such that \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}\) = i then determine the values of x and y.
Solution:
\(\frac{(1+\mathrm{i}) \mathrm{x}-2 \mathrm{i}}{3+\mathrm{i}}+\frac{(2-3 \mathrm{i}) \mathrm{y}+\mathrm{i}}{3-\mathrm{i}}\) = i

i = \(\frac{[(1+\mathrm{i}) \mathrm{x}-2 \mathrm{i}](3-\mathrm{i})+[(2-3 \mathrm{i}) \mathrm{y}+\mathrm{i}](3+\mathrm{i})}{(3+\mathrm{i})(3-\mathrm{i})}\)

i = \(\frac{(2-3 i) 3 y+3 i+(2-3 i) i y+i^2}{\left(9-i^2\right)}\)
10i = (3x + 3ix – 6i – xi – xi² – 2) + 6y – 9iy + 3i + 2iy – 3i²y – 1
(3x + x – 2 + 6y + 3y – 1) + i(3x – 6 – x – 9y + 3 + 2y) = 10i
(4x + 9y – 3) + i(2x – 7y – 3) = 10i + 0
Comparing real and imaginary parts on both sides
4x + 9y – 3 = 0 …………..(1)
2x – 7y – 3 = 10
2x – 7y – 3 = 10 ………….(2)
From (1) and (2);

Question 10.
If \(\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3\) = x + iy find x and y.
Solution:
Given that,

x + iy = i³ + i³
2i³ = x + iy
2i²(i) = x + iy
2(- 1)i = x + iy
0 + i(- 2) = x + iy
Comparing real and imaginary parts on both sides
x = 0, y = – 2.

Question 11.
Find the real values of θ in order that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is a
a) real number
b) purely imaginary number.
Solution.

a) If z is a real number
⇒ Imaginary part of z = 0
⇒ \(\frac{8 \sin \theta}{1+4 \sin ^2 \theta}\) = 0
8 sin θ = 0
sin θ = 0
∴ θ = nπ, n ∈ Z

b) If z ¡s purely imaginary
⇒ Real part of z = 0
⇒ \(\frac{3-4 \sin ^2 \theta}{1+4 \sin ^2 \theta}\) = 0
3 – 4 sin² θ = 0
sin² θ = \(\frac{3}{4}\)
sin² θ = \(\left(\frac{\sqrt{3}}{2}\right)^2\)
sin² θ = sin² \(\frac{\pi}{3}\)
∴ θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z.

Question 12.
Find the real values of x and y if \(\frac{x-1}{3+1}+\frac{y-1}{3-1}\) = i.
Solution:
Given that,
\(\frac{x-1}{3+i}+\frac{y-1}{3-i}\) = i
\(\frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)}\) = i
\(\frac{3 x-x i-3+i+3 y+y i-3-i}{9-i^2}\) = i
3x + 3y – 6 + i(- x + y) = 10i
3x + 3y – 6 + i(- x + y) = 0 + i(10)
Comparing the real and imaginary parts on both sides
3x + 3y – 6 = 0 ………….(1)
y – x = 10
x – y + 10 = 0 …………..(2)
From (1) and (2)

Question 13.
If the amplitude of \(\left[\frac{z-2}{z-6 i}\right]=\frac{\pi}{2}\), find its locus. [May ‘06]
Solution:
Let, z = x + iy
Now,

Given that,
the amplitude of \(\left[\frac{z-2}{z-6 i}\right]=\frac{\pi}{2}\) then,
tan^-1 \(\left[\frac{\frac{6 x+2 y-12}{x^2+(y-6)^2}}{\frac{x^2+y^2-2 x-6 y}{x^2+(y-6)^2}}\right]=\frac{\pi}{2}\)
\(\frac{6 x+2 y-12}{x^2+y^2-2 x-6 y}\) = tan \(\frac{\pi}{2}\) = ∞ = \(\frac{1}{0}\)
x² + y² – 2x – 6y = 0
∴ Locus of z is x² + y² – 2x – 6y = 0.

Question 14.
If the point P denotes the complex number z = x + iy in the Argand plane and if \(\frac{z-i}{z-1}\) is a purely imaginary number, find the locus of P. [March 06, May 08 (old)] [AP-Mar.2017]
Solution:
Given, z = x + iy
⇒ P = (x, y)

Given that,
\(\frac{z-i}{z-1}\) is a purely imaginary then real part = 0
\(\frac{x^2+y^2-x-y}{(x-1)^2+y^2}\) = 0
x² + y² – x – y = 0
∴ Locus of P is x² + y² – x – y = 0.

Question 15.
If z = x + iy and if the point P in the Argand plane represents z, then describe geometrically the locus of z satisfying the equation |z – 2 – 3i| = 5. [Board Paper]
Solution:
Given, z = x + iy
⇒ P (x, y)
|z – 2 – 3i| = 5
|x + iy – 2 – 3i| = 5
|(x – 2) + i (y – 3)| = 5
\(\sqrt{(x-2)^2+(y-3)^2}\) = 5
Squaring on both sides
(x – 2)² + (y – 3)² = 25
x² – 4x + 4 + y² + 9 – 6y – 25 = 0
x² + y² – 4x – 6y – 12 = 0
∴ Locus of P is x² + y² – 4x – 6y – 12 = 0
Locus represents a circle with centre (2, 3) and radius 5 units.

Question 16.
Show that the points in the Argand plane represented by the complex numbers -2 + 7i, \(\frac{-3}{2}\) + \(\frac{1}{2}\) i, 4 – 3i, \(\frac{7}{2}\) (1 + i) are the vertices of a rhombus. [AP – Mar. 2015] [TS – Mar. 2016; March’05, ‘05, ‘04, 2000]
Solution:
Let, the given points are
A = – 2 + 7i, B = \(\frac{-3}{2}+\frac{1}{2} \mathrm{i}\)
C = 4 – 3i, D = \(\frac{7}{2}\) (1 + i)
A = – 2 + 7i = (- 2, 7)
B = \(\frac{-3}{2}+\frac{1}{2} \mathrm{i}=\left(\frac{-3}{2}, \frac{1}{2}\right)\)
C = 4 – 3i = (4, – 3)
D = \(\frac{7}{2}\) (1 + i) = \(\left(\frac{7}{2}, \frac{7}{2}\right)\)

∴ AB = BC = CD = DA and AC ≠ BD
∴ The given points form a rhombus.

Question 17.
The points P, Q denote the complex numbers z₁, z₂ in the Argand diagram. O is the origin. If z₁\(\overline{\mathbf{z}}_2\) + \(\overline{\mathbf{z}}_1\)z₂ = 0, then show that ∠POQ = 90° [March 07]
Solution:
Given points P, Q denote the complex numbers z₁, z₂.
Let z₁ = x₁ + iy₁
⇒ P (x₁, y₁)
\(\overline{\mathrm{Z}}_1\) = x₁ – iy₁
⇒ Q (x₂, y₂)
z₂ = x₂ + iy₂

Since, z1\(\overline{\mathrm{Z}}_2\) + \(\overline{\mathrm{Z}}_1\)z₂ = 0
(x₁ + iy₁) (x₂ – iy₂) + (x₁ – iy₁) (x₂ + iy₂) = 0
x₁x₂ – ix₁y₂ + ix₂y₁ + y₁y₂ + x₁x₂ + ix₁y₂ – ix₂y₁ + y₁y₂ = 0
2(x₁x₂ + y₁y₂) = 0
x₁x₂ + y₁y₂ = 0
x₁x₂ = – y₁y₂
\(\frac{\mathrm{y}_1 \mathrm{y}_2}{\mathrm{x}_1 \mathrm{x}_2}\) = – 1
\(\left(\frac{\mathrm{y}_1}{\mathrm{x}_1}\right)\left(\frac{\mathrm{y}_2}{\mathrm{x}_2}\right)\) = – 1
Slope of \(\overline{\mathrm{OP}}\) × Slope of \(\overline{\mathrm{OQ}}\) = – 1
∴ ∠POQ = 90°

Some More Maths 2A Complex Numbers Important Questions

Question 1.
Find the multiplicative inverse of (3, 4).
Solution:
Let z = (3, 4)
The multiplicative inverse of z is
\(\mathrm{z}^{-1}=\frac{1}{\mathrm{z}}=\left(\frac{\mathrm{a}}{\mathrm{a}^2+\mathrm{b}^2}, \frac{-\mathrm{b}}{\mathrm{a}^2+\mathrm{b}^2}\right)\)

\(z^{-1}=\left(\frac{3}{3^2+4^2}, \frac{-4}{3^2+4^2}\right)\)

= \(\left(\frac{3}{9+16}, \frac{-4}{9+16}\right)=\left(\frac{3}{25}, \frac{-4}{25}\right)\)

Question 2.
Write \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\) in the form a + ib.
Solution:

Question 3.
Write the complex number (1 + 2i)³ in the form a + ib.
Solution:
Let z = (1 + 2i)³
= 1³ + (2i)³ + 3(1) (2i)² + 3(1)² (2i)
= 1 + 8i³ + 12i² + 6i
=1 + 8i²(i) + 6i – 12
= 1 – 81 + 6i – 12
= – 11 – 2i
= – 11 + i(- 2)
It is in the form a + ib.

Question 4.
Find the multiplicative inverse of √5 + 3i.
Solution:
Let z = √5 + 3i
The multiplicative inverse of z is
\(z^{-1}=\frac{1}{z}=\frac{1}{\sqrt{5}+3 i}\)
= \(\frac{1}{\sqrt{5}+3 \mathrm{i}} \times \frac{\sqrt{5}-3 \mathrm{i}}{\sqrt{5}-3 \mathrm{i}}\)
= \(\frac{\sqrt{5}-3 \mathrm{i}}{5-9 \mathrm{i}^2}=\frac{\sqrt{5}-3 \mathrm{i}}{5+9}=\frac{\sqrt{5}-3 \mathrm{i}}{14}\)

Question 5.
Write the conjugate of (2 + 5i) (- 4 + 6i).
Solution:
Let z = (25i) (- 4 + 6i)
z = – 8+ 12i – 20i + 30i²
z = – 8 – 8i – 30
z = – 38 – 8i
The conjugate of z is \(\bar{z}=(\overline{-38-8 i})\) = -38 + 81

Question 6.
Find the square roots of 3 + 4i.
Solution:
Let \(\sqrt{3+4 i}\) = ± (x + iy)

Question 7.
Find the square roots of 7 + 24i.
Solution:
Let \(\sqrt{7+24 i}\) = ± (x + iy)

Question 8.
Find the square roots of – 8 – 6i.
Solution:
Let \(\sqrt{-8-6 i}\) = ± (x – iy)

Question 9.
Show that z₁ = \(\frac{2+11 i}{25}\), z₂ = \(\frac{-2+i}{(1-2 i)^2}\) are conjugate to each other.
Solution:
Given that,

Since, this complex number is conjugate of \(\frac{2+11 i}{25}\).
∴ z₁ and z₂ are conjugate to each other.

Question 10.
If z = 3 – 5i then show that z³ – 10z² + 58z – 136 = 0.
Solution:
Given that, z = 3 – 5
Now, z² = (3 – 5i)² = 9 + 25i² – 30i
= 9 – 25 – 30i = – 16 – 30i
L.H.S:
z³ – 10z² + 58z – 136 = z²z – 10z² + 58z – 136
= (- 16 – 30i) (3 – 5i) – 10 (- 16 – 30i)+ 58 (3 – 5i) – 136
= – 48 + 80i – 90i + 150i² + 160 + 300i + 174 – 290i – 136
= – 48 – 150 + 160 + 174 – 136 = 0 = R.H.S.
∴ z³ – 10z² + 58z – 136 = 0.

Question 11.
If z₁ = (3, 5) and z₂ = (2, 6) find z₁ z₂.
Solution:
Given that,
z₁ = (3, 5), z₂ = (2, 6)
z₁ z₂ = (3, 5) . (2, 6)
= (3.2 – 5.6, 3.6 + 2.5)
= (6-30, 18 + 10)
= (- 24, 28)

Question 12.
Write the additive Inverse of (- 6, 5) + (10, – 4).
Solution:
Let,
z = (- 6, 5) + (10, – 4)
= (- 6 + 10, 5 – 4)
= (4, 1)
∴ The additive inverse of z is
– z = – (4, 1) = (- 4, – 1).

Question 13.
Write the additive inverse of (2, 1) (- 4, 6).
Solution:
Let,
z = (2, 1) (- 4, 6)
= (2 . (- 4) – 1 . 6, 2 . 6 + 1 . (- 4))
= (- 8 – 6, 12 – 4)
= (- 14, 8)
∴ The additive inverse of z is
– z = – (- 14, 8) = (14, – 8).

Question 14.
If z₁ = (6, 3), z₂ = (2, – 1) find \(\frac{z_1}{z_2}\).
Solution:
Given that,
z₁ = (6, 3), z₂ = (2, – 1)
\(\frac{z_1}{z_2}=\frac{(6,3)}{(2,-1)}=\left(\frac{6 \cdot 2+3 \cdot-1}{2^2+(-1)^2}, \frac{3 \cdot 2-6 \cdot-1}{2^2+(-1)^2}\right)\)
= \(\left(\frac{12-3}{4+1}, \frac{6+6}{4+1}\right)\)
= \(\left(\frac{9}{5}, \frac{12}{5}\right)\)

Question 15.
VrItethe multiplicative inverse of (7, 24).
Solution:
Let z = (7, 24)
The multiphcative inverse of z ¡s
\(z^{-1}=\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right)\)
= \(\left(\frac{7}{7^2+24^2}, \frac{-24}{7^2+24^2}\right)\)
= \(\left(\frac{7}{49+576}, \frac{-24}{49+576}\right)\)
= \(\left(\frac{7}{625}, \frac{-24}{625}\right)\).

Question 16.
Write the complex number (2 – 3i) (3 + 4i) in the form A + iB. [AP – Mar. 2017]
Solution:
Let z = (2 – 3i) (3 + 4i)
= 6 + 8i – 9i – 12i²
= 6 – i + 12
= 18 – i = 18 + (- 1)i
It is in the form A + iB.

Question 17.
Write the complex number \(\frac{a-i b}{a+i b}\) in the form A + iB.
Solution:

Question 18.
Write the complex number \(\frac{4+3 i}{(2+3 i)(4-3 i)}\) in the form A + iB.
Solution:

Question 19.
Write the complex number (- √3 + √-2) (2√3 – i) in the form A + iB.
Solution:
Let, z = (- √3 + √-2) (2√3 – i)
= (- √3 + √2i) (2√3 – i)
= – 6 + √3i + 2√6i – \(\sqrt{2} \mathrm{i}^2\)
= – 6 + (√3 + 2√6)i + √2
= (√2 – 6) + i(√3 + 2√6)
It is in the form A + iB.

Question 20.
Write the complex number (- 5i) (i/8) in the form A + iB.
Solution:
Let, z = (- 5i) (\(\frac{i}{8}\))
= \(\frac{-5 \mathrm{i}^2}{8}\)
= \(\frac{5}{8}\)
= \(\frac{5}{8}\) + i(0)
It is in the form A + iB.

Question 21.
Write the complex number (- i) 2i in the form A + iB.
Solution:
Let, z = (- i) (2i) = – 2i²
= 2 = 2 + i(0)
It is in the form A + iB.

Question 22.
Write the complex number i⁹ in the form A + iB.
Solution:
Let, z = i⁹ = i⁸ i = (i²)⁴ i
= (- 1)⁴ i = i = 0 + i(1)
It is in the form A + iB.

Question 23.
Write the complex number 3 (7 + 7i) + i(7 + 7i) in the form A + IB.
Solution:
Let, z = 3 (7 + 7i) + i (7 + 7i)
= 21 + 21i + 7i + 7i
= 21 + 28i – 7
= 14 + 28i
= 14 + i(28)
It is in the form A + iB.

Question 24.
Find the complex conjugate of (15 + 3i) – (4 – 20i).
Solution:
Let, z = (15 + 3i) – (4 – 20i)
= 15 + 3i – 4 + 20i
= 11 + 23i
The conjugate of z is \(\overline{\mathrm{z}}=(\overline{11+23 \mathrm{i}})\) = 11 – 23i.

Question 25.
Find the complex conjugate of \(\frac{5 i}{7+i}\).
Solution:

Question 26.
Find the multiplicative inverse of the complex number – i.
Solution:
Let, z = – i
The multiplicative inverse of z is
\(z^{-1}=\frac{1}{z}=\frac{1}{-i}=\frac{-1}{i}=\frac{i^2}{i}\) = i.

Question 27.
Find the multiplicative inverse of the complex number i^-35.
Solution:
Let, z = i^-35
The multiplicative inverse of z^-1 is
\(z^{-1}=\frac{1}{z}=\frac{1}{i^{-35}}\) = i³⁵
= i³⁴ . i
= (i²)¹⁷ i
= (- 1)¹⁷ i = – i.

Question 28.
Show that the equation of any circle in the complex plane is of the form \(\mathbf{z} \overline{\mathbf{z}}+\mathbf{b} \overline{\mathbf{z}}+\overline{\mathbf{b}} \mathbf{z}+\mathbf{c}\) = 0.
Solution:
Let the equation of a circle in cartesian coordinates ¡s
x² + y² + 2gx + 2fy + c = 0 …………….(1)
Let, z = x + iy then \(\overline{\mathrm{z}}\) = x – iy
Now, z\(\overline{\mathrm{z}}\) = (x + iy) (x – iy)
= x² + y²
z\(\overline{\mathrm{z}}\) = x² + y²
z + \(\overline{\mathrm{z}}\) = x + iy + x – iy = 2x
∴ z + \(\overline{\mathrm{z}}\) = 2x
z – \(\overline{\mathrm{z}}\) = x + iy – x + iy = 2iy
\(\frac{z-\bar{z}}{i}\) = 2y
\(\frac{z-\bar{z}}{i} \times \frac{i}{i}\) = 2y
\(\frac{\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}})}{-1}\) = 2y
2y = – i (z – \(\overline{\mathrm{z}}\))
Substituting these results in (1) we obtain
z\(\overline{\mathrm{z}}\) + g(z + \(\overline{\mathrm{z}}\))+ f[- i(z – \(\overline{\mathrm{z}}\))] + c = 0
z\(\overline{\mathrm{z}}\) + gz + g\(\overline{\mathrm{z}}\) – ifz + if\(\overline{\mathrm{z}}\) + c = 0
Let g + if = b, then g – if = \(\overline{\mathrm{b}}\)
∴ z\(\overline{\mathrm{z}}\) + \(\overline{\mathrm{b}}\)z + b\(\overline{\mathrm{z}}\) + c = 0.

Question 29.
Exprese the complex iiumber 1 – i in modulus amplitude form. [AP- Mar, 2015]
Solution:

Let. 1 – i = r(cos θ + i sin θ)
then r cos θ = 1, r sin θ = – 1
r = \(\sqrt{x^2+y^2}=\sqrt{(1)^2+(-1)^2}\)
= \(\sqrt{1+1}=\sqrt{2}\)
Hence,
√2 cos θ = 1
cos θ = \(\frac{1}{\sqrt{2}}\)

√2 sin θ = – 1
sin θ = \(\frac{-1}{\sqrt{2}}\)

∴ θ lies in Q₄
∴ θ = \(\frac{-\pi}{4}\)
∴ 1 – i = √2 (cos (\(\frac{-\pi}{4}\)) + i sin (\(\frac{-\pi}{4}\)))

Question 30.
Exprcs the complex number 1 + i√3 in modulus amplitude form. [Mar. 14, TS – Mar.2017]
Solution:

Let, 1 + i√3 = r(cos θ + i sin θ)
then r cos θ = 1, r sin θ = √3
∴ r = \(\sqrt{x^2+y^2}=\sqrt{(1)^2+(\sqrt{3})^2}\)
= \(\sqrt{1+3}=\sqrt{4}\) = 2
Hence,
2 cos θ = 1
cos θ = \(\frac{1}{2}\)

2 sin θ = √3
sin θ = \(\sqrt{3}{2}\)

∴ θ lies in Q₁
∴ θ = \(\frac{\pi}{3}\)
∴ 1 + i√3 = 2 (cos (\(\frac{\pi}{3}\)) + i sin (\(\frac{\pi}{3}\)))

Question 31.
If z ≠ 0, find Arg z + Arg \(\overline{\mathbf{z}}\).
Solution:
If z = x + iy then
Arg z = tan^-1 (\(\frac{y}{x}\))

If \(\overline{\mathbf{z}}\) = x – iy then
Arg \(\overline{\mathbf{z}}\) = tan^-1 (\(\frac{-y}{x}\))
= – tan^-1 (\(\frac{y}{x}\))
Now,
Arg z + Arg \(\overline{\mathbf{z}}\) = tan^-1 (\(\frac{y}{x}\)) – tan^-1 (\(\frac{y}{x}\)) = 0

Question 32.
If \(\frac{\mathbf{z}_2}{\mathbf{z}_1}\), 1 ≠ o, is an imaginary number then find the value of \(\left|\frac{2 z_1+z_2}{z_1-z_2}\right|\).
Solution:
Given, \(\frac{\mathbf{z}_2}{\mathbf{z}_1}\), z₁ is an imaginary numbers.
⇒ \(\frac{\mathbf{z}_2}{\mathbf{z}_1}\) = ki (say)

Question 33.
If z = x + iy and if the point P in the Argand plane represents z, then describe geometrically the locus of P satisfying the equations
(i) |2z – 3| = 7
(ii) |z|² = 4Re(z + 2)
(iii) |z + i|² – |z – i|² = 2
(iv) |z + 4i| + |z – 4i| = 10
Solution:
(i) Given,
z = x + iy
⇒ P = (x, y)
|2z – 3| = 7
|2(x + iy) – 3| = 7
|2x + 2iy – 3| = 7
|(2x – 3) + i2y| = 7
\(\sqrt{(2 x-3)^2+(2 y)^2}\) = 7
Squaring on both sides,
(2x – 3)² + (2y)² = 49
4x² + 9 – 12x + 4y² – 49 = 0
4x² + 4y² – 12x – 40 = 0
x² + y² – 3x – 10 = 0
∴ Locus of P is x² + y² – 3x – 10 = 0
The locus represents a circle with centre at (\(\frac{3}{2}\), o) and radius \(\frac{7}{2}\) units.

(ii) Given, z = x + iy
⇒ P = (x, y)
|z|² = 4 Re(z + 2)
|x + iy|² = 4 Re(x + iy + 2)
\(\left(\sqrt{x^2+y^2}\right)^2\) = 4 Re(x + 2 + iy)
x² + y² = 4 (x + 2)
x² + y² = 4x + 8
x² + y² – 4x – 8 = 0
Locus of P is x² + y² – 4x – 8 = 0
The locus represents a circle with centre at (2, 0) and radius 2√3 units.

(iii) Given, z = x + iy
⇒ P = (x, y)
|z + i|² – |z – i|² = 2
|x + iy + i|² – |x + iy – i|² = 2
|x + i(y + 1)|² – |x + i(y – 1)²| = 2
\(\left(\sqrt{(x)^2+(y+1)^2}\right)^2-\left(\sqrt{x^2+(y-1)^2}\right)^2\) = 2
x² + (y + 1)² – (x² + (y – 1)²) = 2
x² + (y + 1)² – x² – (y – 1)² = 2
y² + 2y + 1 – y² + 2y – 1 – 2 = 0
4y – 2 = 0
2y – 1 = 0
∴ Locus of P is 2y – 1 = 0
The locus represents a straight line parallel to X – axis.

(iv) Given, z = x + iy
⇒ P = (x, y)
|z + 4i| + |z – 4i| = 10
|x + iy + 4i| + |x + iy – 4i| = 10
|x + i(y + 4)| + |x + i(y – 4)| = 10
\(\sqrt{x^2+(y+4)^2}+\sqrt{x^2+(y-4)^2}\) = 10
\(\sqrt{x^2+(y+4)^2}\) = 10 – \(\sqrt{x^2+(y-4)^2}\)
Squaring on both sides,
x² + (y + 4)² = 100 + (x² + (y – 4)²) – 20 \(\sqrt{x^2+(y-4)^2}\)
x² + y² + 16 + 8y = 100 + x² + y² + 16 – 8y – 20 \(\sqrt{x^2+y^2+16-8 y}\)
8y + 8y – 100 = – 20 \(\sqrt{x^2+y^2+16-8 y}\)
16y – 100 = – 20 \(\sqrt{x^2+y^2+16-8 y}\)
4y – 25 = – 5 \(\sqrt{x^2+y^2+16-8 y}\)
Again squaring on both sides
16y² + 625 – 200y = 25(x² + y² + 16 – 8y)
16y² + 625 – 200y = 25x² + 25y² 400 – 200y
25x² + 25y² + 400 – 16y – 625 = 0
25x² + 9y² – 225 = 0
25x² + 9y² = 225
\(\frac{x^2}{9}+\frac{y^2}{25}\)
∴ Locus of P is \(\frac{x^2}{9}+\frac{y^2}{25}\)= 1.

The locus represents an ellipse with centre (0, 0) with eccentricity \(\frac{4}{5}\) and major axis is parallel to Y – axis.

Question 34.
If z₁, z₂ are two non-zero complex numbers satisfying |z₁ + z₂| = |z₁ + z₂| then show that Arg z₁ – Arg z₂ = 0.
Solution:
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ then
|z₁ + z₂| = |x₁ + iy₁ + x₂ + iy₂|
= |(x₁ + x₂) + i(y₁ + y₂)|
= \(\sqrt{\left(\mathrm{x}_1+\mathrm{x}_2\right)^2+\left(\mathrm{y}_1+\mathrm{y}_2\right)^2}\)
and |z₁| = |x₁ + iy₁|
= \(\sqrt{x_1^2+y_1^2}\)
|z₂| = |x₂ + iy₂|
= \(\sqrt{\mathrm{x}_2^2+\mathrm{y}_2^2}\)

Given that,
|z₁ + z₂| = |z₁| + |z₂|
\(\sqrt{\left(x_i+x_2\right)^2+\left(y_1+y_2\right)^2}\) = \(\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}\)

Squaring on both sides,
(x₁ + x₂)² + (y₁ + y₂)² = x₁² + y₁² + x₂² + y₂² + 2 \(\sqrt{\left(x_1^2+y_1^2\right)} \sqrt{x_2^2+y_2^2}\)

x₁² + x₂² + 2x₁x₂ + y₁² + y₂² + 2y₁y₂ = x₁² + y₁² + x₂² + y₂² + 2 \(\sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2}\)

2x₁x₂ + 2y₁y₂ = 2 \(\sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2}\)
x₁x₂ + y₁y₂ = \(\sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2}\)
Again squaring on both sides,
(x₁x₂ + y₁y₂)² = (x₁² + y₁²) (x₂² + y₂²)
x₁²x₂² + y₁²y₂² + 2x₁x₂y₁y₂ = x₁²x₂² + x₁²y₂² + x₂² y₂² + y₁² y₂²
x₁² y₂² + x₂² y₁² – 2 x₁ x₂y₁ y₂ = 0
(x₁y₂ – x₂ y₁)² = 0
x₁y₂ – x₂ y₁ = 0
x₁y₂ = x₂ y₁
\(\frac{\mathrm{y}_2}{\mathrm{x}_2}=\frac{\mathrm{y}_1}{\mathrm{x}_1}\)
\(\frac{\mathrm{y}_1}{\mathrm{x}_1}=\frac{\mathrm{y}_2}{\mathrm{x}_2}\)
\(\tan ^{-1}\left(\frac{\mathrm{y}_1}{\mathrm{x}_1}\right)=\tan ^{-1}\left(\frac{\mathrm{y}_2}{\mathrm{x}_2}\right)\)
Arg z₁ = Arg z₂
Arg z – Arg z₂ = 0.

Question 35.
Describe geometrically the subsets of ‘C’ suchthat {z ∈ C/|z – 1 + i|= i}
Solution:
Let z = x + iy
⇒ P = (x, y)
Let,
C = {z ∈ C/|x + iy – 1 + i| = 1}
= {(x, y) ∈ R²/|x + iy – 1 +i| = i}
= {(x, y) ∈ R²/|(x – 1) + i(y + 1)i = i}
={(x, y)∈ R²/(x – 1)² + (y + 1)² = i}
Hence C is a circle with centre (1, – 1) and radius 1 unit.

Question 36.
If z = x + iy and if the point P in the Argand plane represents z, then describe geometrically the locus of z satisfying the equations
(i) 2|z – 2| = |z – 1|
(ii) Im z² = 4
(iii) Arg \(\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)
Solution:
(i) Given, z = x + iy
2|z – 2| = |z – 1|
2|x + iy – 2| = |x + iy – 1|
2|(x – 2) + iy| = |(x – 1) + iy|
2\(\sqrt{(x-2)^2+y^2}=\sqrt{(x-1)^2+y^2}\)
Squaring on both sides,
4(x – 2)² + 4y² = (x – 1)² + y²
4(x² – 4x + 4) + 4y² = (x² – 2x + 1) + y²
4x² – 16x + 16 + 4y² – x² + 2x – 1 – y² = 0
3x² + 3y² – 14x + 15 = 0
∴ Locus of P is 3x² + 3y² – 14x + 15 = 0.
Locus represents a circle with centre (\(\frac{7}{3}\) , 0) and radius \(\frac{2}{3}\) units.

(ii) Given, z = x + iy
⇒ P = (x, y)
Given, Imz² = 4
Im (x + iy)² = 4
Im (x² – y² + 2ixy) = 4
2xy = 4
xy = 2
∴ The locus of P is xy = 2
The locus represents a rectangular hyperbola.

(iii) Given, z = x + iy P = (x, y)
Now, \(\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}=\frac{(x-1)+i y}{(x+1)+i y}\)

2y = x² + y² – 1
x² + y² – 2y – 1 = 0
∴ Locus of P is x² + y² – 2y – 1 = 0
∴ The locus represents a circle with centre (0, 1) and radius √2 units.

Question 37.
If \(\frac{\mathbf{z}_3-\mathbf{z}_1}{\mathbf{z}_2-\mathbf{z}_1}\) is a real number, show that the points represented by the complex numbers z₁, z₂, z₃ are collinear.
Solution:
Given that,
\(\frac{\mathbf{z}_3-\mathbf{z}_1}{\mathbf{z}_2-\mathbf{z}_1}\) is a real number.
⇒ \(\frac{\mathbf{z}_3-\mathbf{z}_1}{\mathbf{z}_2-\mathbf{z}_1}\) = k (say)
z₃ – z₁ = k (z₂ – z₁)
z₃ – z₁ = kz₂ – kz₁
z₃ – kz₂ = z₁ – kz₁
z₃ – kz₂ = z₁ (1 – k)
z₁ = \(\frac{\mathrm{z}_3-\mathrm{kz}_2}{1-\mathrm{k}}\)

z₁ represents a point on the Line joining, the points represented by z₂, z₃ in the ratio 1 : k externally.
∴ z₁, z₂, z₃ are collinear.

Question 38.
Show that the four points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are vertices of a square.
Solution:

Let, the given points are A (2, 1), B (4, 3), C(2, 5), D(0, 3)
AB = \(\sqrt{(2-4)^2+(1-3)^2}\)
= \(\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}\)

BC = \(\sqrt{(4-2)^2+(3-5)^2}\)
= \(\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}\)

CD = \(\sqrt{(2-0)^2+(5-3)^2}\)
= \(\sqrt{(2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}\)

DA = \(\sqrt{(2-0)^2+(1-3)^2}\)
= \(\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}\)

AC = \(\sqrt{(2-2)^2+(1-5)^2}\)
= \(\sqrt{0+(-4)^2}=\sqrt{0+16}=4\)

BD = \(\sqrt{(4-0)^2+(3-3)^2}\)
= \(\sqrt{(4)^2+0}=\sqrt{16}=4\)

∴ AB = BC = CD = DA and AC = BD.
∴ The given points form a square.

Question 39.
Show that the points in the Argand diagram represented by the complex numbers z₁, z₂, z₃are colilnear if and only if there exist three real numbers p, q, r not all zero, satisfying pz₁ + qz₂ + rz₃ = 0 and p + q + r = 0.
Solution:

z₁, z₂, z₃ are collinear then z₃ lies on the line joining z₁ and z₂.
⇒ z₃ = \(\frac{\mathrm{mz}_2+\mathrm{nz}_1}{\mathrm{~m}+\mathrm{n}}\)
⇒ z₃ (m + n) = mz₂ + nz₁
⇒ nz₁ + mz₂ – (m + n) z₃ = 0
which is of the form pz₁ + qz₂ + rz₃ = 0
where
p = n, q = m, r = – m – n
Now, p + q + r = n + m – m – n = 0
∴ p + q + r = 0.

Question 40.
Represent the complex number 2 + 3i in Argand plane. [TS – Mar.2015]
Solution:
In Argand plane the complex number 2 + 3i is represented as the point (2, 3).

Here students can locate TS Inter 2nd Year Physics Notes 11th Lesson Electromagnetic Waves to prepare for their exam.

TS Inter 2nd Year Physics Notes 11th Lesson Electromagnetic Waves

→ Electromagnetic waves: Electromagnetic waves consists of time varying electric and magnetic field. A time varying electric field will produce a magnetic field and vice versa. In this way in electromagnetic waves energy oscillates between electric and magnetic fields.

→ Ampere-Maxwell Law: The total current passing through any surface of which the closed loop as the perimeter is the sum of conduction current and the displacement current.
i.e ∮\(\overline{\mathrm{B}} \cdot \mathrm{d} \bar{l}\) = μ₀ + μ₀ε₀ \(\frac{\mathrm{d} \phi_E}{\mathrm{dt}}\)

→ Displacement current: According to Maxwell a time changing electric flux through a surface will also generate a current. This current produced due to rate of change in electric flux is called displacement current i_d.
Its value Is ε₀ times greater than \(\frac{\mathrm{d} \phi_E}{\mathrm{dt}}\)
Displacement current, i_d = ε₀\(\frac{\mathrm{d} \phi_E}{\mathrm{dt}}\)
where \(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{Q}}{\epsilon_0}\right)=\frac{1 \mathrm{dQ}}{\epsilon_0} \frac{\mathrm{dt}}{\mathrm{dt}}\)
But \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = Rate of change of charge displacement current i_d

→ Conduction current (i_c): Current carried by conductors due to flow of charges is called “conduction current” (i_c).

Note:

• From Maxwell’s theory total current ‘i’ is the sum of conductiqn current i_c and displacement current i_d.
  i = i_c + i_d
• There are many regions in space which contains only displacement current due to time varying electric fields.
• From Maxwell’s theory the source of magnetic field is not just due to conduction current produced by flow of charges. Magnetic field can also be produced due to time rate of change of electric field.

→ Electromagnetic waves (Maxwell’s concepts):

• According to Maxwell’s theory, accelerated charges radiate electromagnetic waves.
• These electric and magnetic fields are mutually perpendicular and also perpendicular to direction of propagation.
• Frequency of electromagnetic wave is equal to frequency of oscillator.
• Energy associated with the propagation of wave is obtained from oscillating source.
• In an electromagnetic wave let electric field vibrations are along X-axis and magnetic field vibrations are along Y -axis then direction of propagation of electromagnetic wave is along Z-axis.
• Electric field component E_x = E₀ sin (kz – cot)
  Magnetic field component B_y = B₀ sin (kz – cot)
  where k = \(\frac{2 \pi}{\lambda}\) and speed of wave v = \(\frac{\omega}{k}\).

→ From Maxwell’s equations relation between
E₀ and B₀ is \(\frac{\mathrm{E}_0}{\mathrm{~B}_0}\) = c or B₀ = \(\frac{E_0}{c}\)

→ In vacuum velocity of electromagnetic wave c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

→ In a medium velocity of electromagnetic v =\(\frac{1}{\sqrt{\mu \epsilon}}\)

→ Hertz experiments on electromagnetic waves showed that electromagnetic waves of wavelength 10 million times more than light waves could be diffracted, reflected and polarised.

→ Electromagnetic waves carry energy and momentum like other waves.

→ Amount of pressure of visible light is in the order of 7 × 10^-6 N / m².

→ Electromagnetic spectrum: Electromagnetic spectrum extends over a wide wavelength region of 107 m to 10-14 m. It consists of longer wavelength radio waves. Television and F.M Radio waves, Microwaves, Infrared, Visible and Ultra violet light, X-Rays and high energetic y – Rays.

→ Radio waves: Radio waves are produced by the accelerated motion of charges in conducting wires. They are generally in the range of 500 kHz to 1000 MHz.

• Amplitude Modulation is in the range of 530 kHz to 1710 kHz.
• Short waves are lip to 54 MHz.
• TV signals range is 54 MHz to 890 MHz.

→ Microwaves: Frequency of microwaves is in the region of gegahertz. They are produced by klystrons and magnetrons.
In Microwave ovens the frequency of magnetron is matched to resonant frequency of water molecules. So that energy of microwaves is rapidly and efficiently transferred to food material containing water molecules and their temperature rises quickly.

→ Infrared waves: Infrared rays are produced by hot bodies and molecules. Their wave-length is in the range of 1 mm to 700 nm.
Infrared rays plays an important role in keeping warm atmosphere of earth through green house effect.

→ Visible light The part of electromagnetic spectrum that can be detected by human eye is called visible spectrum. Wavelength range of visible spectrum is 700 nm to 400 nm. Note: Snakes can defect. infrared rays. Many insects can detect ultraviolet rays.

→ Ultraviolet rays: Wavelength range of ultra-violet rays is 400 nm to 0.6 nm. They are more energetic. Exposure to UV rays will cause tanning to skin. They cannot travel through glass. In atmosphere Ozone layer in upper atmosphere filters UV rays.

→ X-rays: Wavelength range of X-rays 10^-10 nm to 10^-4 nm. Commonly used method to produce X-rays is to bombard metal target with high energy electrons. X-rays are widely used as diagnostic tools in medicine.

→ Gamma rays: Wavelength range of gamma rays is 10-10 nm to 10~14 nm. They are very high energetic radiation. These rays are produced in nuclear reactions and also by radio active nuclie.
In medicine they are used to destroy cancer cells.

→ Electric flux Φ_E = \(\frac{\mathrm{Q}}{\epsilon_0}\)

→ Displacement current i = e₀\(\left(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\right)\)

→ Ampere – Maxwell’s Law
∮\(\overline{\mathrm{B}} \cdot \mathrm{d} \bar{l}\) = μ₀ + μ₀ε₀ \(\frac{\mathrm{d} \phi_E}{\mathrm{dt}}\)

→ Maxwell’s equations
(a) \(\oint \overline{\mathrm{E}} \cdot \mathrm{d} \overline{\mathrm{A}}=\frac{\mathrm{Q}}{\epsilon_0}\) (Gauss law for Electricity)
(b) \(\oint \overline{\mathrm{B}} \cdot \mathrm{d} \overline{\mathrm{A}}\) = 0 (Gauss Law for Magnetism)
(c) \(\oint \overline{\mathrm{E}} \cdot \mathrm{d} \bar{l}=\frac{-\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\) (Faraday’s Law)
(d) ∮\(\overline{\mathrm{B}} \cdot \mathrm{d} \bar{l}\) = μ₀ + μ₀ε₀ \(\frac{\mathrm{d} \phi_E}{\mathrm{dt}}\)
(Ampere-Maxwell’s Law)

→ In Electromagnetic waves
(a) Electric field E_x = E₀ sin (kz – ωt)
(b) Magnetic field B_y = B₀ sin (kz – ωt)
(c) Speed of wave v = \(\frac{\omega}{k}\), ω = ck where k = \(\frac{2 \pi}{\lambda}\)
(d) Velocity of electromagnetic wave in vacuum c = \(\frac{1}{\sqrt{\mu_0 \in_0}}\).
(e) Velocity of electromagnetic wave in medium c = \(\frac{1}{\sqrt{\mu \epsilon}}\).

→ In electromagnetic waves \(\frac{\mathrm{E}_0}{\mathrm{~B}_0}\) = C or B₀ = E₀c
where E₀ and B₀ are magnitudes of electric field and magnetic field in vacuum.