Problems with trigonometry getting the power of this complex expression 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!
 A: Oh my. We have:
$$1+i\sqrt{3} = 2\exp\left(i\arctan\sqrt{3}\right)=2\exp\frac{\pi i}{3}$$
$$\frac{1}{1-i} = \frac{1}{2}(1+i) = \frac{1}{\sqrt{2}}\exp\frac{\pi i}{4},$$
hence:
$$z=\frac{1+i\sqrt 3}{1-i} = \sqrt{2}\exp\frac{7\pi i}{12},$$
so:
$$ z^{40} = 2^{20}\exp\frac{70\pi i}{3}=2^{20}\exp\frac{4\pi i}{3}=-2^{20}\exp\frac{\pi i}{3}=-2^{19}(1+i\sqrt{3}).$$
As an alternative way, if you set $a=1+i\sqrt{3}$ and $b=\frac{1}{1-i}$ you have:
$$ a^3 = -2^3,\qquad b^4 = -2^{-2}, $$
hence:
$$ z^{40} = a^{40} b^{40} = a(a^3)^{13} (b^4)^{10} = a\cdot(-2^{39})\cdot(2^{-20}) = -2^{19}(1+i\sqrt{3}).$$ 
A: $$
z=\sqrt 2\frac{\frac{1+\sqrt 3i}{2}}{\frac{\sqrt 2-\sqrt 2i}{2}}=\sqrt 2 e^{i(\pi/3-(-\pi/4))}
$$
and therefore:
$$
z^{40}=2^{20}e^{i2\pi/3}=-2^{19}(1+i\sqrt3)
$$
A: To tackle the arctangent head-on, we make use of an identity derived from the tangent angle addition formula:
$$\tan\left(\frac\pi4+x\right)=\frac{\tan\frac\pi4+\tan x}{1-\tan\frac\pi4 \tan x}=\frac{1+\tan x}{1-\tan x}$$ since $\tan \dfrac\pi4=1$. Recalling that $\tan\dfrac\pi3=\sqrt{3}$, then we immediately have 
$$\alpha=\tan^{-1}\left(\frac{1+\sqrt{3}}{1-\sqrt{3}}\right)=\tan^{-1}\left(\frac{1+\tan\frac\pi3}{1-\tan\frac\pi3}\right)=\tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{\pi}{3}\right)\right]=-\frac{5\pi}{12}$$ since $\tan^{-1}$ takes values in $(-\pi/2,\pi/2]$.
From here, things are clear: 
$$\cos(40\alpha)+i\sin(40\alpha)
=\cos\left(-\frac{2\pi}{3}-16\pi \right)+i\sin\left(-\frac{2\pi}{3}-16\pi \right)=-\frac{1+i\sqrt{3}}{2}$$
Multiplying through by the modulus $2^20$ gives the desired result.
