Answer 1

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

\[ = i^{4 \times 18 + 3} \]

\[ = \left( i^4 \right)^{18} . i^3 \]

\[ = i^3 [ \because i^4 = 1]\]

\[ = - i [ \because i^3 = - i]\]

Let \[z = 0 - i\]

Then, 

\[\left| z \right| = \sqrt{0^2 + \left( - 1 \right)^2} = 1\].

Let θ be the argument of z and α be the acute angle given by 

\[\tan\alpha = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|}\]

Then, 

\[\tan\alpha = \frac{1}{0} = \infty \]

\[ \Rightarrow \alpha = \frac{\pi}{2}\]

Clearly, z lies in fourth quadrant. So, arg(z) = 

\[- \alpha = - \frac{\pi}{2}\].

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

Thus, the polar form of  (i²⁵)³ is \[\cos\left( \frac{\pi}{2} \right) - i\sin\left( \frac{\pi}{2} \right)\].