Question

Find the multiplicative inverse of the complex number $4-3i$

Answer 1

Let $z=4-3i$
Then, $\overline{z}=4+3i$ and ${|z|}^2=4^2+{(-3)}^2=16+9=25$
Therefore, the multiplicative inverse of $4-3i$ is given by
$z^{-1}=\frac{\overline{z}}{{|z|}^2}=\frac{4+3i}{25}=\frac{4}{25}+\frac{3}{25}i$