It's difficult to make out, but I think the task is to expand
[tex](\sqrt3-i)^4[/tex]
Write the number in polar form first:
[tex]\sqrt3-i=2e^{-i\pi/6}[/tex]
By DeMoivre's theorem, you have
[tex](\sqrt3-i)^4=\left(2e^{-i\pi/6}\right)^4=2^4e^{-i4\pi/6}=16e^{-i2\pi/3}[/tex]
and converting back to Cartesian form, this number is equivalent to
[tex]16\left(\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3\right)=16\left(-\dfrac12+-\dfrac{\sqrt3}2\right)=-8(1+i\sqrt3)[/tex]
