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?

### Answers

 Pavan Patil May 20, 2020

 Jananda Vardhan May 08, 2020
Correct answer is option a

This discussion on If α + i β = tan1 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? is done on EduRev Study Group by JEE Students. The Questions and Answers of If α + i β = tan1 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? are solved by group of students and teacher of JEE, which is also the largest student community of JEE. If the answer is not available please wait for a while and a community member will probably answer this soon. You can study other questions, MCQs, videos and tests for JEE on EduRev and even discuss your questions like If α + i β = tan1 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? over here on EduRev! Apart from being the largest JEE community, EduRev has the largest solved Question bank for JEE.

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This discussion on If α + i β = tan1 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? is done on EduRev Study Group by JEE Students. The Questions and Answers of If α + i β = tan1 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? are solved by group of students and teacher of JEE, which is also the largest student community of JEE. If the answer is not available please wait for a while and a community member will probably answer this soon. You can study other questions, MCQs, videos and tests for JEE on EduRev and even discuss your questions like If α + i β = tan1 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? over here on EduRev! Apart from being the largest JEE community, EduRev has the largest solved Question bank for JEE.