Question

If $P$ is a complex number whose modulus is one, then the equation ${\left(\frac{1+iz}{1-iz}\right)}^4=P$ 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,
${\left(\frac{1+iz}{1-iz}\right)}^4=P$
$\Rightarrow {\left(\frac{i-z}{i+z}\right)}^4=P$
$\Rightarrow {\left(\frac{z-i}{z+i}\right)}^4=P$
$\Rightarrow {\left|\frac{z-i}{z+i}\right|}^4=|P|$
$\Rightarrow {\left|\frac{z-i}{z+i}\right|}^4=1$
$[\because |P|=1]$
$\Rightarrow |z-i{|}^4=|z+i{|}^4$
$\Rightarrow |z-i|=|z+i|$
$\therefore z$ lies on perpendicular bisector of $i$ and $-i$
$\therefore z$ lies on $y$ -axis.
$\therefore z$ has all complex roots.