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If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' is
  • a)
    x2 + y2 + 2x cot 2α = 1
  • b)
    (cot 2α) (x2 + y2) = 1 + x
  • c)
    x2 + y2+ 2y tan 2α = 1
  • d)
    x2 + y2 + 2x sin 2α = 1
Correct answer is option 'A'. Can you explain this answer?

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Pavan Patil
May 20, 2020
Related If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer?

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Jananda Vardhan
May 08, 2020
Correct answer is option a

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If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer? for JEE 2022 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2022 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If α + i β = tan⁻1 z, z = x + yi and α is constant, then locus of \'z\' isa)x2 + y2 + 2x cot 2α = 1b)(cot 2α) (x2 + y2) = 1 + xc)x2 + y2+ 2y tan 2α = 1d)x2 + y2 + 2x sin 2α = 1Correct answer is option 'A'. Can you explain this answer?.
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