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)`
