Question

The polar form of (i²⁵)³ is ____________.

Answer 1

(i²⁵)³
= (i²⁴. i)³
Since i²⁴ = 1
= (i)³ = i². i = – i
i.e (i²⁵)³ = – i
Here modulus $r=\sqrt{0^2+1^2}=1$ and argument $\theta ={\tan }^{-1}\left|\frac{-1}{0}\right|=\infty =-\frac{\mathrm{\pi }}{2}$
$$\begin{gathered} \Rightarrow –i\mathrm{has}\mathrm{polar}\mathrm{form}=\cos \left(\frac{-\mathrm{\pi }}{2}\right)+i\sin \left(\frac{-\mathrm{\pi }}{2}\right) \\ \mathrm{Since}\cos \left(-\theta \right)=\cos \theta \mathrm{and}\sin \left(-\theta \right)=-\sin \theta \end{gathered}$$
Polar form of –i is
$$\begin{gathered} \cos \theta -i\sin \theta \mathrm{where}\theta =\frac{\mathrm{\pi }}{2} \\ \mathrm{i}.\mathrm{e}\cos \frac{\mathrm{\pi }}{2}-i\sin \frac{\mathrm{\pi }}{2} \end{gathered}$$