Solution:

Given that $Z=\left(i^{25}\right)^{3}$
$\begin{array}{l}
=\dot{i}^{75} \\
=\mathrm{i}^{74} \cdot \mathrm{i} \\
=\left(\mathrm{i}^{2}\right)^{37} \cdot \mathrm{i} \\
=(-1)^{37} \cdot \mathrm{i} \\
=(-1) \cdot \mathrm{i} \\
=-\mathrm{i} \\
=0-\mathrm{i}
\end{array}$
Therefore now,
$\begin{array}{l}
\left.|Z|=\sqrt{(} x^{2}+y^{2}\right) \\
\left.=\sqrt{(} 0^{2}+(-1)^{2}\right) \\
=\sqrt{(0+1)} \\
{ } {=\sqrt{1}} \\
{\theta} {=\tan ^{-1}(|y| /|x|)} \\
{=} {\tan ^{-1}(1 / 0)} \\
{=\tan ^{-1} \infty}
\end{array}$
As $x \geq 0, y<0$ complex number lies in $4^{\text {th }}$ quadrant and the value of $\theta$ is $-90^{0} \leq \theta \leq 0^{0}$
$\begin{array}{l}
\theta=-\pi / 2 \\
Z=1(\cos (-\pi / 2)+i \sin (-\pi / 2)) \\
=1(\cos (\pi / 2)-i \sin (\pi / 2))
\end{array}$
As a result, polar form of $\left(\mathrm{i}^{25}\right)^{3}$ is $1(\cos (\pi / 2)-\mathrm{i} \sin (\pi / 2))$