Question

If $P$ is a complex number whose modulus is one, then the equation $\left(\frac{1+i z}{1-i z}\right)^{4}=P$ has

• A. real and equal roots
• B. real and distinct roots
• C. two real and two complex roots
• D. all complex roots

Answer 1

Answer: (D)

We have,
$\left(\frac{1+i z}{1-i z}\right)^{4}=P$
$\Rightarrow \left(\frac{i-z}{i+z}\right)^{4}=P$
$\Rightarrow \left(\frac{z-i}{z+i}\right)^{4}=P$
$\Rightarrow \left|\frac{z-i}{z+i}\right|^{4}=|P|$
$\Rightarrow \left|\frac{z-i}{z+i}\right|^{4}=1$
$[\because|P|=1]$
$\Rightarrow |z-i|^{4}=|z+i|^{4}$
$\Rightarrow |z-i|=|z+i|$
$\therefore z$ lies on perpendicular bisector of $i$ and $-i$
$\therefore z$ lies on $y$ -axis.
$\therefore z$ has all complex roots.