Is my method to solve this problem on argument of complex numbers correct?

Question

$k+ |k+z^2|= |z|^2 ~ k \in \mathbb{R^-}$, 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?

Answer 1

Now to obtain the same thing as LHS on RHS we get $2\theta = \pi $

I don't see offhand how that step is justified.

thus, the argument is $\theta = \frac \pi2$

This does not give the complete solution set. Consider for example $k=-1$, $z=-i$, $\arg z = - \dfrac{\pi}{2}$.

For an alternative hint:  by the triangle inequality $\,\require{cancel} k + |k+z^2| \le \cancel{k} + \cancel{|k|} + |z|^2 = |z|^2\,$ with equality iff $z^2 = \lambda k$ for some real $\,\lambda \ge 0\,$.

Answer 2

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 ...