Question
Write in $(\text{i}^{25})^3$ polar form.

Answer

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

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

$=(\text{ i}^4)^{ 18}.\text{ i}^3$

$=\text{i}^3\ \big[\because \text{i}^4=1\big]$

$=\text{-i}\ \big[\because \text{i}^3=-\text{i}\big]$

Let $\text{z} = 0 - \text{i}$

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

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

Then,

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

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

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

$\therefore$ the polar from of z is $\text{|z|}(\cos\theta+\text{i }\sin\theta)=\cos(-\frac{\pi}{2})+\text{i }\sin(-\frac{\pi}{2})$

Thus, the polar from of $(\text{i}^{25})^3$ is $\cos(\frac{\pi}{2})-\text{i }\sin(\frac{\pi}{2})$

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