lf $z$ is a non-real complex number, then the minimum value of $\frac{\mathrm{Im}{z}^5}{(\mathrm{Im}z{)}^5}$ is 

Let , $z=r{e}^{i\theta }$

$\begin{array}{l}\Rightarrow \frac{\mathrm{lm}{z}^5}{(\mathrm{lm}z{)}^5}=\frac{r^5(\sin 5\theta )}{r^5(\sin \theta {)}^5}=\frac{\sin 5\theta }{{\sin }^5\theta }\\ \Rightarrow \frac{dz}{d\theta }=\frac{{\sin }^5\theta \cdot 5\cos 5\theta -5\sin 5\theta {\sin }^4\theta \cos \theta }{{\left({\sin }^5\theta \right)}^2}\end{array}$

Put , $\frac{dz}{d\theta }=0$

$\begin{array}{l}\Rightarrow 5{\sin }^4\theta (\sin \theta \cos 5\theta -\cos \theta \sin 5\theta )=0\\ \Rightarrow \sin \theta =0\text{ or }\sin (-4\theta )=0\end{array}$

$\Rightarrow \theta =n\pi$ or $\theta =\frac{n\pi }{4}$where $n\in Z$ 

As$z$ is non-real complex number. 

 $\therefore$   Only $\theta =\frac{n\pi }{4}$ is possible .