Answer
$-4-4i$
Work Step by Step
In trigonometric form,
$1+i=\sqrt {2}(\cos 45^{\circ}+i\sin 45^{\circ})$
$2i=2(\cos 90^{\circ}+i\sin90^{\circ})$
$-2+2i=2\sqrt 2(\cos135^{\circ}+i\sin135^{\circ})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(1+i)^{4}=[\sqrt {2}(\cos 45^{\circ}+i\sin 45^{\circ})]^{4}$
$=(\sqrt {2})^{4}[\cos(4\cdot45^{\circ})+i\sin(4\cdot45^{\circ})]$
$=4(\cos180^{\circ}+i\sin180^{\circ})$
$(2i)^{2}=[2(\cos 90^{\circ}+i\sin90^{\circ})]^{2}$
$=(2)^{2}[\cos(2\cdot90^{\circ})+i\sin(2\cdot90^{\circ})]$
$=4(\cos 180^{\circ}+i\sin 180^{\circ})$
Now, $(1+i)^{4}(2i)^{2}=$
$[4(\cos 180^{\circ}+i\sin 180^{\circ})][4(\cos 180^{\circ}+i\sin 180^{\circ})]=$
$4\cdot4[\cos(180^{\circ}+180^{\circ})+i\sin(180^{\circ}+180^{\circ})]$
$=16(\cos 360^{\circ}+i\sin360^{\circ})$
$\frac{(1+i)^{4}(2i)^{2}}{-2+2i}=\frac{16(\cos 360^{\circ}+i\sin360^{\circ})}{2\sqrt 2(\cos135^{\circ}+i\sin135^{\circ})}$
$=\frac{16}{2\sqrt 2}[\cos(360^{\circ}-135^{\circ})+i\sin(360^{\circ}-135^{\circ})]$
$=4\sqrt {2}(\cos225^{\circ}+i\sin225^{\circ})$
which in standard form is
$=4\sqrt {2}(-\frac{1}{\sqrt {2}}+i\cdot-\frac{1}{\sqrt {2}})$
$=-4-4i$