$z^3 = i * \frac{|z^5|}{z* \bar z}$ Did I do it right? I just want to check whether I understand the basic algebra of complex numbers.
I have to find solution to: $z^3 = i * \frac{|z^5|}{z* \bar z}$.
So I transform that expression into: $z^3 = i * \frac{|z^5|}{|z^2|} \iff z^3 = i * |z^3|$.
Then I take trigonometric form of $z^3$: $|z^3|[\cos(3\alpha) + \sin(3\alpha) i] = i * |z^3| \iff \cos(3\alpha) + \sin(3\alpha) i = i \iff 3\alpha = \frac{\pi}{2} \iff \alpha = \frac{\pi}{6}$
So ultimately I get $|z|(\frac{\sqrt{3}}{2}+\frac{1}{2}i)$. Is that the final solution? Did I get it right?
 A: Watch our for more solutions (there are lots of them)!
Note that, if
$$z^3 = i |z^3|$$
then $\alpha = z/|z|$ satisfies $\alpha^3 = i$. Hence $\alpha$ is a cube root of $i$, i.e. $\alpha$ could be any of
$$ \alpha_k = e^{(\pi/2 + 2\pi i k)/3}$$
for $k= 0, 1,2$. Therefore $z$ is of the form $z = |z|\alpha_k$ for some $k=0, 1,2$. Since this equation does not force any restriction on the size of $|z|$ we see that there are infinitely many solutions each of the form $r \alpha_k$ for some $k=0,1,2$ and $r \in \mathbb{R}$.

Edit: The single solution you have found is correct but there are more (so it's not `bad' but just missing all of the solutions). In your calculation you have $\cos(3\alpha) + \sin(3\alpha) i = i  \iff 3\alpha = \frac{\pi}{2}$. This is not correct as it misses out the other two possibilities $3\alpha = \frac{5\pi}{2}$ and $3\alpha = \frac{7\pi}{2}$ - you should really check all of the possible solutions to $3\alpha = \frac{\pi}{2} + 2\pi  k$ for $k \in \mathbb{N}$ and see that there are $3$.
