[tex]2+2i=2\sqrt2e^{i\pi/4}[/tex]
[tex]3-3i=3\sqrt2e^{-i\pi/4}[/tex]
[tex]\implies\dfrac{2+2i}{3-3i}=\dfrac23e^{i\pi/2}[/tex]
By DeMoivre's theorem,
[tex]\left(\dfrac{2+2i}{3-3i}\right)^5=\left(\dfrac23\right)^5e^{i5\pi/2}=\dfrac{32}{243}e^{i\pi/2}=\dfrac{32i}{243}[/tex]
Just to confirm:
[tex]\dfrac{2+2i}{3-3i}\cdot\dfrac{3+3i}{3+3i}=\dfrac{12i}{18}=\dfrac{2i}3\implies\left(\dfrac{2+2i}{3-3i}\right)^5=\dfrac{32i}{243}[/tex]
