Question:
The polar form of $\left(i^{25}\right)^{3}$ is
(a) $\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}$
(b) $\cos \pi+i \sin \pi$
(c) $\cos \pi-i \sin \pi$
(d) $\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$
Solution:
(d) $\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$
$\left(i^{25}\right)^{3}=(i)^{75}$
$=(i)^{4 \times 18+3}$
$=(i)^{3}$
$=-i \quad\left(\because i^{4}=1\right)$
Let $z=0-i$
Since, the point $(0,-1)$ lies on the negative direction of imaginary axis.
Therefore, $\arg (z)=\frac{-\pi}{2}$
Modulus, $r=|z|=|1|=1$
$\therefore$ Polar form $=r(\cos \theta+i \sin \theta)$
$=\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)$
$=\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$