Question:
The complex number $\frac{(1-i)^{3}}{1-i^{3}}$ in polar form is _________________
Solution:
$\frac{(1-i)^{3}}{1-i^{3}}=\frac{(1-i)^{3}}{1+i} \quad\left(\because i^{2}=-1\right)$
$=\frac{(1-i)^{3}}{1+i} \times \frac{1-i}{1-i}$
$=\frac{(1-i)^{4}}{1-i^{2}}$
$=\frac{1}{2}\left\{(1-i)^{2}(1-i)^{2}\right\}$
$=\frac{1}{2}\left\{\left(1+i^{2}-2 i\right)\left(1+i^{2}-2 i\right)\right\}$
$=\frac{1}{2}\{(1-1-2 i)(1-1-2 i)\}$
$=\frac{1}{2}\{+2 i \times 2 i\}$
$=\frac{1}{2} \times 4 i^{2}$
$=-2=-2(\cos 0+i \sin 0)$
$\because \cos 0=1$ and $\sin 0=0$
Hence $\frac{(1-i)^{3}}{1-i^{3}}=2(\cos \pi+i \sin \pi)$