Question

If $|k+z^2|=\left|z^2\right|-k(k\in R^{-}).$ then possible argument of $z$ is

• A. $0$
• B. $\pi$
• C. $\pi /2$
• D. none of these

Answer 1

Answer: (C)

$\pi /2$

According to question,

$\begin{array}{c}k+\left|k+z^2\right|={\left|z\right|}^{2}ifz=niform,\\ then\\ k+\left|k-n^2\right|\\ \Rightarrow k+n^2-k=n^2={\left|z\right|}^2\\ So,inthiscaseargumentis\frac{\pi }{2}\end{array}$

and the correct option is C.