Q. If $z$ is a complex number such that Re $(z) = Im\ (z)$, then
A) Re $\left( z^{2} \right) = 0$
B) Im $\left( z^{2} \right) = 0$
C) Re $\left( z^{2} \right) = Im\ (z^{2})$
D) Re $\left( z^{2} \right) = - Im\ (z^{2})$
Explanation
(a)
Let $z = x + iy$
$\Rightarrow \ \ zz = (x + iy)(x + iy)$
$= x^{2} - y^{2} + 2ixy$
$= 0 + 2ixy\ \ \ \ \ \ \ \ \lbrack\because\ \ Re\ (z) = Im\ (z)\ \ \Rightarrow \ \ x = y\rbrack$
$\Rightarrow \ \ Re\left( z^{2} \right) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.