How do you simplify $\frac{1+2i}{2-3i}$ ?

$\frac{1+2i}{2-3i}=\frac{-4+7i}{13}=-\frac{4}{13}+\frac{7}{13}i$
Explantion:
1. Find the complex conjugate of denominator
denominator: $z=2-3i$
denominator complex coniugate $\overline{z}=2+3i$
1. Multiply both nemerator and denominator for the complex coniugate
$\frac{1+2i}{2-3i}\times \frac{2+3i}{2+3i}=\frac{2+3i+4i+6i^2}{2^2-(3i{)}^2}=$
Rememberig that $i^2=-1$
$=\frac{2+6\times (-1)+7i}{4-9i^2}=\frac{2-6+7i}{4-9\times (-1)}=\frac{-4+7i}{4+9}=$
$=\frac{-4+7i}{13}=-\frac{4}{13}+\frac{7}{13}i$