Question:
If $z=(1-i)$, find $z^{-1}$
Solution:
We have, $z=(1-i)$
$\Rightarrow \overline{\mathrm{z}}=1+\mathrm{i}$
$\Rightarrow|\mathrm{z}|^{2}=(1)^{2}+(-1)^{2}=2$
$\therefore$ The multiplicative inverse of $(1-\mathrm{i})$,
$z^{-1}=\frac{\bar{z}}{|z|^{2}}=\frac{1+i}{2}$
$z^{-1}=\frac{1}{2}+\frac{1}{2} i$