The polar form of
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 $=r(\cos \theta+i \sin \theta)$

$=\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)$

$=\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$

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