Respuesta :

LammettHash
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]

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