Find the multiplicative inverse of the complex number 4 – 3i

Let $z=4-3 i$

Then, $\bar{z}=4+3 i$ and $|z|^{2}=4^{2}+(-3)^{2}=16+9=25$

Therefore, the multiplicative inverse of $4-3 i$ is given by

$z^{-1}=\frac{\bar{z}}{|z|^{2}}=\frac{4+3 i}{25}=\frac{4}{25}+\frac{3}{25} i$