Question

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

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

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

$|k+z^2|=|z^2|-k\Rightarrow |k+z^2|=|z^2|+|k|(\because k\in R^{-})$
$\Rightarrow k,z^2$ and $0+i0$ are collinear
$\Rightarrow \arg \left(z^2\right)=\arg \left(k\right)$
$\because k\in R^{-}$ So, $\arg \left(k\right)=\pi$
$\Rightarrow \arg \left(z^2\right)=\pi$
$\Rightarrow 2\arg \left(z\right)=\pi$
$\Rightarrow \arg \left(z\right)=\frac{\pi }{2}$