I'm here because I can't finish this problem, that comes from a Russian book:
Calculate $z^{40}$ where $z = \dfrac{1+i\sqrt{3}}{1-i}$
Here $i=\sqrt{-1}$. All I know right now is I need to use the Moivre's formula $$\rho^n \left( \cos \varphi + i \sin \varphi \right)^n = \rho^n \left[ \cos (n\varphi) + i \sin (n\varphi) \right]$$
to get the answer of this.
First of all, I simplified $z$ using Algebra, and I got this:
$$z = \dfrac{1-\sqrt{3}}{2} + i \left[ \dfrac{1+\sqrt{3}}{2} \right]$$
Then, with that expression I got the module $|z| = \sqrt{x^2 + y^2}$, and its main argument $\text{arg}(z) = \tan^{-1} \left( \dfrac{y}{x} \right)$.
I didn't have problems with $|z| = \sqrt{2}$, but the trouble begins when I try to get $\text{arg}(z)$. Here is what I've done so far:
$$\alpha = \text{arg}(z) = \tan^{-1} \left[ \dfrac{1+\sqrt{3}}{1-\sqrt{3}} \right]$$
I thought there's little to do with that inverse tangent. So, I tried to use it as is, to get the power using the Moivre's formula.
$$z^{40} = 2^{20} \left[ \cos{40 \alpha} + i \sin{40 \alpha} \right]$$
As you can see, the problem is to reduce a expression like: $\cos{ \left[ 40 \tan^{-1} \left( \dfrac{1+\sqrt{3}}{1-\sqrt{3}} \right) \right] }$.
And the book says the answer is just $-2^{19} \left( 1+i\sqrt{3} \right)$.
I don't know if I'm wrong with the steps I followed, or if I can reduce those kind of expressions. I'll appreciate any help from you people :)
Thanks in advance!