Question

Simplify and hence find the modulus and argument $\frac{(3+i)(3-i)}{2+i}$.

Answer 1

$\frac{(3+i)(3-i)}{2+i}$

$=\frac{(3{)}^2-(i{)}^2}{2+i}=\frac{9+1}{2+i}$

$=\frac{10}{2+i}\times \frac{2-i}{2-i}$

$=\frac{10(2-i)}{5}$

$=2(2-i)$

$z=4-2i$

$|z|=\sqrt{4^2+22}=2\sqrt{5}$

$arg=+n^{-1}(-\frac{1}{2})$.