Complex numbers problem with large exponents

Question

I have to simplify the following complex number: $\Big(\dfrac{1+i\sqrt{3}}{1-i}\Big)^{20}$

The solution I've been given is $2^9 (1-i\sqrt{3})$ but I don't know how to get it.

I know how to simplify $\frac{1+i\sqrt{3}}{1-i}$ on its own but I'm not able to simplifiy the same expression with the exponent. Any help would be greatly appreciated.

Answer 1

It may be easier to use polar coordinates in which case $$1+\sqrt{3}i=2\exp{\left(\frac{\pi}{3}\cdot i\right)}$$ $$1-i=\sqrt{2}\exp{\left(-\frac{\pi}{4}\cdot i\right)}$$ Which gives a quotient of $$\frac{1+\sqrt{3}i}{1-i}=\sqrt{2}\exp{\left(\frac{7\pi}{12}\cdot i\right)}$$ and hence raising to the power of $20$ gives \begin{align} \left(\frac{1+\sqrt{3}i}{1-i}\right)^{20} &=\left(\sqrt{2}\exp{\left(\frac{7\pi}{12}\cdot i\right)}\right)^{20}\\ &=2^{10}\exp{\left(\frac{35\pi}{3}\cdot i\right)}\\ &=2^{10}\left(\frac12-\frac{\sqrt{3}}2\cdot i\right)\\ &=2^9\left(1-\sqrt{3}i\right)\\ \end{align}

Answer 2

Hints:

$$\left(\dfrac{1+i\sqrt3}2\right)^{20}=\left(\left(\dfrac{1+i\sqrt3}2\right)^3\right)^6\left(\dfrac{1+i\sqrt3}2\right)^2=(-1)^6\left(\dfrac{-1+i\sqrt3}2\right)$$

$$\left(\dfrac{1-i}{\sqrt2}\right)^{20}=\left(\left(\dfrac{1-i}{\sqrt2}\right)^{2}\right)^{10}=(-i)^{10}=-1$$