Question

Find the multiplicative inverse of each of the complex numbers

$4-3i$

Answer 1

M.I.of $(4-3i)=\frac{1}{4-3i}=\frac{1}{4-3i}\times \frac{4+3i}{4+3i}$

$=\frac{4+3i}{(4{)}^2-(3i{)}^2}=\frac{4+3i}{16-9i^2}$

$=\frac{4+3i}{16+9}=\frac{1}{25}(4+3i)$