Question

Solve using demoivres theorem

Answer 1

It's difficult to make out, but I think the task is to expand

$(\sqrt3-i)^4$

Write the number in polar form first:

$\sqrt3-i=2e^{-i\pi/6}$

By DeMoivre's theorem, you have

$(\sqrt3-i)^4=\left(2e^{-i\pi/6}\right)^4=2^4e^{-i4\pi/6}=16e^{-i2\pi/3}$

and converting back to Cartesian form, this number is equivalent to

$16\left(\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3\right)=16\left(-\dfrac12+-\dfrac{\sqrt3}2\right)=-8(1+i\sqrt3)$