InterviewSolution
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InterviewSolution
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Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`. The value of `theta` for which z is purely real areA. `npi-(pi)/(4), n in I`B. `pin +(pi)/(4) , n in I`C. `npi, n in I`D. None of these |
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Answer» Correct Answer - A `z=(1-i sin theta)/(1+ i cos theta) = ((1- isin theta)( 1- icos theta))/((1+ i cos theat)(1-i cos theta))` `=((1-sin theta cos theta) - icos theta + sintheta)/((1+ cos^(2)theta))` If z is purely real , then `cos theta + sin theta=0` or `tan theta = -1` `rArr = npi - (pi)/(4), n in I` If z is purely imaginary, `1- sin theta cos theta = 0 or sin theta coe theta = 1`, which is not possible `|z|= |(1- isin theta)/(1+ i cos theta)| = (sqrt(1+ sin^(2) theta))/(sqrt(1+ cos^(2) theta))` If `|z|=1`,then `cos^(2) theta = sin theta^(2) theta` `rArr tan^(2) theta = 1` `rArr theta = npipm(pi)/(4), ninI` We have, `agr(z) = tan^(-1) ((-(cos theta + sin theta))/((1-sin theta cos theta)))` Now `arg(z) = pi//4` `rArr (-(costheta + sintheta))/((1-sin theta cos theta)) = 1` `rArr cos^(2) theta+ sin^(2) + 2 sin theta cos theta = 1 + sin^(2) theta cos^(2) theta - 2 sin theta cos theta` `rArr 1+ 4 sin theta cos theta = 1 + sin ^(2) theta cos^(2) theta` `rArr sin^(2) theta cos^(2) theta - 4 sin theta cos theta =0` `rArr sin theta cos theta (sin theta cos theta -4) = 0` `rArr sin theta cos theta = 0 " "(because sin theta cos theta = 4" is not possible ")` `rArr = (2n + 1) pi or theta ( 4n - 1) pi//2, n in I " "(because - cos theta - sin theta gt 0)` |
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