The polar form of

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

$=\left(i^{24} \cdot i\right)^{3}$

Since $i^{24}=1$

$=(i)^{3}=i^{2} \cdot i=-i$

i.e $\left(i^{25}\right)^{3}=-i$

Here modulus $r=\sqrt{0^{2}+1^{2}}=1$ and argument $\theta=\tan ^{-1}\left|\frac{-1}{0}\right|=\infty=-\frac{\pi}{2}$

$\Rightarrow-i$ has polar form $=\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)$

Since $\cos (-\theta)=\cos \theta$ and $\sin (-\theta)=-\sin \theta$

Polar form of $-i$ is

$\cos \theta-i \sin \theta \quad$ where $\theta=\frac{\pi}{2}$

i. e $\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$