Solve the following

$\left(i^{25}\right)^{3}=i^{75}$

$=i^{4 \times 18+3}$

$=\left(i^{4}\right)^{18} \cdot i^{3}$

$=i^{3} \quad\left[\because i^{4}=1\right]$

$=-i \quad\left[\because i^{3}=-i\right]$

Let $z=0-i$

Then, $|z|=\sqrt{0^{2}+(-1)^{2}}=1$

Let $\theta$ be the argument of $z$ and $\alpha$ be the acute angle given by $\tan \alpha=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}$.

Then, 

$\tan \alpha=\frac{1}{0}=\infty$

$\Rightarrow \alpha=\frac{\pi}{2}$

Clearly, $z$ lies in fourth quadrant. So, $\arg (z)=-\alpha=-\frac{\pi}{2}$.

$\therefore$ the polar form of $z$ is $|z|(\cos \theta+i \sin \theta)=\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)$.

Thus, the polar form of $\left(i^{25}\right)^{3}$ is $\cos \left(\frac{\pi}{2}\right)-i \sin \left(\frac{\pi}{2}\right)$.