Is my method to solve this problem on argument of complex numbers correct? 
$k+ |k+z^2|= |z|^2 ~ k \in \mathbb{R^-}$, then the possible argument of $z$ is ? 

I converted $z$ to polar form and got:
$\underbrace{r^2 - k}_{\text{positive}} = |\underbrace{k}_{\text{negative}} + r^2 e^{i2\theta}|$ where $r= |z|
$
Note that $|z|= |-z|$, 
Now to obtain the same thing as LHS on RHS we get $2\theta = \pi $
$\implies r^2 - k= |k - r^2|$
thus, the argument is $\theta = \frac \pi2$. 
The answer is correct but I am unsure about my method. Are there any discrepancies? 

Let's start with the $$r^2-k=|k+r^2e^{2i\theta}|. $$
We have:
$$(r^2-k)^2=|(k+r^2\cos(2\theta)+ir^2\sin(2\theta)|^2 $$
or
$$r^2-2kr^2+r^4=((k+r^2\cos(2\theta))^2+r^4\sin^2(2\theta) $$
after simplifying, we reach to
$$-2kr^2=2kr^2\cos(2\theta). $$
Thus $$\cos(2\theta)=-1 $$
and ...
