TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the complex number
(‌

1+iz

1−iz

)4=p
The above complex number is solved as
(‌

1+iz

1−iz

)4=p
(‌

i−z

i+z

)4=p
|‌

z−i

z+i

|4=|p|
|‌

z−i

z+i

|4=1
Thus,
|z−i|4=|z+i|4
|z−i|=|z+i|
Therefore, z lies on perpendicular bisector of i and −i
So z lies on y axis.
Therefore, z has all complex roots.