The polar form of $(\text{i}^{25})^3$ is:
View full solution →- $\cos\frac{\pi}{2}+\text{i}\sin\frac{\pi}{2}$
- $\cos\pi+\text{i}\sin\pi$
- $\cos\pi-\text{i}\sin\pi$
- $\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$