Answer
$-8+8i\sqrt {3}$
Work Step by Step
In trigonometric form,
$\sqrt {3}+i=2(\cos 30^{\circ}+i\sin 30^{\circ})$
$2i=2(\cos 90^{\circ}+i\sin90^{\circ})$
$1+i=\sqrt 2(\cos45^{\circ}+i\sin45^{\circ})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(\sqrt {3}+i)^{4}=[2(\cos 30^{\circ}+i\sin 30^{\circ})]^{4}$
$=(2)^{4}[\cos(4\cdot30^{\circ})+i\sin(4\cdot30^{\circ})]$
$=16(\cos120^{\circ}+i\sin120^{\circ})$
$(2i)^{5}=[2(\cos 90^{\circ}+i\sin90^{\circ})]^{5}$
$=(2)^{5}[\cos(5\cdot90^{\circ})+i\sin(5\cdot90^{\circ})]$
$=32(\cos 450^{\circ}+i\sin 450^{\circ})$
$(1+i)^{10}=[\sqrt {2}(\cos 45^{\circ}+i\sin 45^{\circ})]^{10}$
$=(\sqrt {2})^{10}[\cos(10\cdot45^{\circ})+i\sin(10\cdot45^{\circ})]$
$=32(\cos450^{\circ}+i\sin450^{\circ})$
Now, $(\sqrt 3+i)^{4}(2i)^{5}=$
$[16(\cos120^{\circ}+i\sin120^{\circ})][32(\cos 450^{\circ}+i\sin 450^{\circ})]=$
$16\cdot32[\cos(120^{\circ}+450^{\circ})+i\sin(120^{\circ}+450^{\circ})]$
$=512(\cos 570^{\circ}+i\sin570^{\circ})$
$\frac{(\sqrt {3}+i)^{4}(2i)^{5}}{(1+i)^{10}}=\frac{512(\cos 570^{\circ}+i\sin570^{\circ})}{32(\cos450^{\circ}+i\sin450^{\circ})}$
$=\frac{512}{32}[\cos(570^{\circ}-450^{\circ})+i\sin(570^{\circ}-450^{\circ})]$
$=16(\cos120^{\circ}+i\sin120^{\circ})$
which in standard form is
$=16(-\frac{1}{2}+i\cdot\frac{\sqrt 3}{2})$
$=-8+8i\sqrt {3}$
