\frac{(\sqrt3+j\sqrt2)^2}{\sqrt2-\sqrt3j}=\frac{3-2+2\sqrt6j}{\sqrt2-\sqrt3j}=\frac{1+2\sqrt6j}{\sqrt2-\sqrt3j}\cdot \frac{\sqrt2+\sqrt3j}{\sqrt2+\sqrt3j}=\\
=\frac{(...)(...)}{2+3}=\frac{(1+2\sqrt6j)(\sqrt2+\sqrt3j)}{5}=\frac{\sqrt2+\sqrt3j+2\sqrt{12}j-2\sqrt{18}}{5}=\\
=\frac{\sqrt2-6\sqrt2+j(\sqrt3+4\sqrt3)}{5}=-\sqrt2+\sqrt3j
