Question

If $k+\mid k+z^2\mid =\mid z{\mid }^2\left(kϵR^{-}\right)$, 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 $\frac{\pi }{2}$

$|k+z^2|=|k|+|z^2|$
Hence $k$ and $|z^2|$ are collinear.
Now $kϵR^{-}$
Since, $k$ and $z^2$ is collinear, then $z^2$ has to be purely real.
Thus, $arg\left(z^2\right)=(2n-1)\pi$
Now $z=r{e}^{i\theta }$
$z^2=r^2{e}^{2i\theta }=m{e}^{2i\theta }$
Hence $arg\left(z^2\right)=2arg\left(z\right)$
Therefore, $arg\left(z\right)=\frac{1}{2}arg\left(z^2\right)$
$=\frac{(2n-1)\pi }{2}$
Putting $n=1$, we get
$arg\left(z\right)=\frac{\pi }{2}$
Hence, option 'C' is correct.