Question

The polar form of (i²⁵)³ is
(a) $\cos \frac{\mathrm{\pi }}{2}+i\sin \frac{\mathrm{\pi }}{2}$
(b) cos π + i sin π
(c) cos π − i sin π
(d) $\cos \frac{\mathrm{\pi }}{2}-i\sin \frac{\mathrm{\pi }}{2}$

Answer 1

(d) $\cos \frac{\pi }{2}-i\sin \frac{\pi }{2}$
(i²⁵)³ = (i)⁷⁵
= (i)^{4$\times$18+ 3}
= (i)³
= $-$i ($\because$ i⁴ = 1)

$$\begin{gathered} \mathrm{Let}z=0-i \\ \mathrm{Since},\mathrm{the}\mathrm{point}(0,-1)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{negative}\mathrm{direction}\mathrm{of}\mathrm{imaginary}\mathrm{axis}. \\ \mathrm{Therefore},\arg \left(z\right)=\frac{-\mathrm{\pi }}{2} \end{gathered}$$

Modulus, r = $\left|z\right|=\left|1\right|=1$

$\therefore$ Polar form = r (cos $\theta$ + i sin $\theta$)
= cos$\left(\frac{-\mathrm{\pi }}{2}\right)$+i sin$\left(\frac{-\mathrm{\pi }}{2}\right)$
= cos$\frac{\pi }{2}$ $-$ i sin $\frac{\pi }{2}$