Question

The complex number $\frac{{\left(1-i\right)}^3}{1-i^3}$ in polar form is ____________.

Answer 1

$$\begin{gathered} \frac{{\left(1-i\right)}^3}{1-i^3}=\frac{{\left(1-i\right)}^3}{1+i}\left(\because i^2=-1\right) \\ =\frac{{\left(1-i\right)}^3}{1+i}\times \frac{1-i}{1-i} \\ =\frac{{\left(1-i\right)}^4}{1-i^2} \\ =\frac{1}{2}\left\{{\left(1-i\right)}^2{\left(1-i\right)}^2\right\} \\ =\frac{1}{2}\left\{\left(1+i^2-2i\right)\left(1+i^2-2i\right)\right\} \\ =\frac{1}{2}\left\{\left(1-1-2i\right)\left(1-1-2i\right)\right\} \\ =\frac{1}{2}\left\{+2i\times 2i\right\} \\ =\frac{1}{2}\times 4i^2 \\ =-2=-2\left(\cos 0+i\sin 0\right) \\ \because \cos 0=1\mathrm{and}\sin 0=0 \\ \mathrm{Hence}\frac{{\left(1-i\right)}^3}{1-i^3}=2\left(\mathrm{cos\pi }+i\mathrm{sin\pi }\right) \end{gathered}$$