Discussion Forum



1)

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:

Option D

Explanation:

We have

 $\left(\frac{1+iz}{1-iz}\right)^{4}$  =P

$\Rightarrow$      $\left(\frac{i-z}{i+z}\right)^{4}  =P$

$\Rightarrow$    $\left(\frac{z-i}{z+i}\right)^{4}  =P$

$\Rightarrow$      $|\frac{z-i}{z+i}|^{4}  =|P|$

$\Rightarrow$   $|\frac{z-i}{z+i}|^{4}  =1$

$\Rightarrow$     $|z-i|^{4}= |z+i|^{4}$

$\Rightarrow$     |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