If $z$ is a non-real complex number , find the minimum value of $$\frac{\operatorname{Im}(z^5)}{(\operatorname{Im}(z))^5}$$
where $\operatorname{Im}(z)$ is the imaginary part of a given complex number $z$.
My attempt:
Let $z =x+iy$ so we have $$\frac{\operatorname{Im}(x^5+5ix^4-10x^3 y^2 -10ix^2y^3 +5 xy^4+iy^5)}{y^5}$$
which is $$\frac{5x^4 -10x^2y^3+y^5}{y^5}$$
Beyond this, I'm not sure what I'm supposed to do.
There is an error in your calculation, it should be $$ \frac{\operatorname{Im}(z^5)}{(\operatorname{Im}(z))^5} = \frac{5x^4y -10x^2y^3+y^5}{y^5} $$ which is homogeneous in $(x, y)$, as one might expect.
Setting $q = (x/y)^2$ we get $$ \frac{\operatorname{Im}(z^5)}{(\operatorname{Im}(z))^5} = 5 q^2 - 10 q + 1 = 5(q-1)^2 - 4 \ge -4 $$ so that the minimal value is $-4$.
Equality holds if $x/y = \pm 1$, that is for all complex numbers of the form $z = a (1\pm i)$ with $a \in \Bbb R$, $a \ne 0$.
$f(z)=\dfrac{\operatorname{Im}(z^5)}{\operatorname{Im}(z)^5}=\dfrac{\operatorname{Im}(r^5e^{i5\theta})}{\operatorname{Im}(re^{i\theta})^5}=\dfrac{r^5\sin(5\theta)}{r^5\sin(\theta)^5}=\dfrac{\sin(5\theta)}{\sin(\theta)^5}$
Since $z\notin\mathbb R$ then $\theta\neq 0\pmod{\pi}$ and the denominator is not vanishing, note also that $f$ is $\pi$-periodic.
$f'(\theta)=\dfrac{5\cos(5\theta)\sin(\theta)-5\sin(5\theta)\cos(\theta)}{\sin(\theta)^6}=-5\ \dfrac{\sin(4\theta)}{\sin(\theta)^6}$
The denominator is of constant sign thank to the even power, so only the sign of numerator is important, and it is easy to see that on $(0,\pi)$ the function variations are like this : $\searrow\,\nearrow\,\searrow\,\nearrow$ with critical points in $\dfrac{n\pi}4$.
We have local minima $f(\frac{\pi}4)=\frac{-1/\sqrt{2}}{(1/\sqrt{2})^5}=-\sqrt{2}^{\ 4}=-4$ and $f(\frac{3\pi}4)=-4$ as well.
Therefore $-4$ is a global minimum.
