Question

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

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. none of these

Answer 1

The correct option is C $\frac{\pi }{2}$

$k+|k+z^2|={\left|z\right|}^2(k\in R^{-})$ ...(1)
Let $z=r(\cos \theta +i\sin \theta )$
Substitute $z$ in equation (1)
$k+|k+{r^2(\cos \theta +i\sin \theta )}^2|=r^2\Rightarrow {(k+r^2\cos 2\theta )}^2+{\left(r^2\sin 2\theta \right)}^2{=(r^2-k)}^2\Rightarrow k^2+r^4({\sin }^{2}2\theta +{\cos }^{2}2\theta )+2k{r}^2\cos 2\theta =k^2+r^4-2k{r}^2\Rightarrow \cos 2\theta =-1$
One of many solution of above equation is $2\theta =\pi$.
$\therefore \theta =\frac{\pi }{2}$
Hence, option 'C' is correct.