Answer
See the answer below.
Work Step by Step
By replacing $x$ with $2(\cos 60^{\circ}+i\sin60^{\circ})$, we get
$x^{2}-2x+4=[2(\cos 60^{\circ}+i\sin60^{\circ})]^{2}-2[2(\cos 60^{\circ}+i\sin60^{\circ})]+4$
We need to show that
$[2(\cos 60^{\circ}+i\sin60^{\circ})]^{2}-2[2(\cos 60^{\circ}+i\sin60^{\circ})]+4=0$
Using de Moivre's theorem, we have
$[2(\cos 60^{\circ}+i\sin60^{\circ})]^{2}=(2)^{2}[\cos(2\cdot60^{\circ})+i\sin(2\cdot60^{\circ})]$
$=4(\cos120^{\circ}+i\sin120^{\circ})=4(-\frac{1}{2}+i\cdot\frac{\sqrt 3}{2})=-2+2i\sqrt 3$
$2[2(\cos 60^{\circ}+i\sin60^{\circ})]=4(\cos 60^{\circ}+i\sin60^{\circ})$
$=4(\frac{1}{2}+i\cdot\frac{\sqrt 3}{2})=2+2i\sqrt 3$
Therefore, we have
$[2(\cos 60^{\circ}+i\sin60^{\circ})]^{2}-2[2(\cos 60^{\circ}+i\sin60^{\circ})]+4$
$=(-2+2i\sqrt 3)-(2+2i\sqrt 3)+4$
$=-2+2i\sqrt {3}-2-2i\sqrt {3}+4=0$
