Question

Write (i²⁵)³ in polar form.

Answer 1

$$\begin{gathered} {\left(i^{25}\right)}^3=i^{75} \\ =i^{4\times 18+3} \\ ={\left(i^4\right)}^{18}.i^3 \\ =i^3[\because i^4=1] \\ =-i[\because i^3=-i] \end{gathered}$$

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 $\tan \alpha =\frac{\left|\mathrm{Im}\left(z\right)\right|}{\left|\mathrm{Re}\left(z\right)\right|}$.
Then,
$$\begin{gathered} \tan \alpha =\frac{1}{0}=\infty \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{2} \end{gathered}$$

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

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

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