find multiplicative inverse of $\left(z^{-}\right)$
$z=3+4i$ then $z^{-1}=?$

Consider the given complex number,

$z=3+4i$

Now,

$\frac{1}{z}=\frac{1}{3+4i}$

$z^{-1}=\frac{1}{3+4i}\times \frac{3-4i}{3-4i}$

$=\frac{3-4i}{3^2-16i^2}$

$\because i^2=-1$

$\therefore z^{-1}=\frac{3-4i}{3^2-16i^2}=\frac{3-4i}{3^2-16\left(-1\right)}$

$z^{-1}=\frac{3-4i}{25}=\frac{3}{25}-\frac{4i}{25}$

Hence, this is the answer.