2+3i divided by 3-3i ( a+bi form)

[latex](2+3i):(3-3i)=\frac{2+3i}{3-3i}\cdot\frac{3+3i}{3+3i}=\frac{(2+3i)(3+3i)}{(3-3i)(3+3i)}=\frac{6+6i+9i-9}{3^2-(3i)^2}=\frac{-3+15i}{9+9}\\\\=\frac{-3+15i}{18}=-\frac{3}{18}+\frac{15i}{18}=-\frac{1}{6}+\frac{5}{6}i[/latex]