Calculate complex number to nth power with de Moivre's formula

[abruski]

Homework Statement

\left( \frac{2 \sqrt{3}+2i}{1-i} \right)^{50}

Homework Equations

z = a+ib

1. 
   z = r(cos \phi + isin \phi)
   
   r = \sqrt{a^2+b^2}
   
   cos \phi = \frac{a}{r}
   
   sin \phi = \frac{b}{r}
   
   \frac{z_1}{z_2} = \frac{r_1}{r_2}\left( cos(\phi_1-\phi_2) + isin(\phi_1 - \phi_2)\right)
   
   z^{n} = r^{n}(cos n \phi + isin n \phi)

The Attempt at a Solution

First I assign the numerator to z_1 = 2 \sqrt{3} + 2i and convert it to trigonometric form:

a=2\sqrt{3} , b=2
r = \sqrt{(2\sqrt{3})^2+2^2} = \sqrt{12+4} = 4
cos \phi_1 = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}
sin \phi_1 = \frac{2}{4} = \frac{1}{2}

From the sin \phi_1 and cos \phi_1 can be concluded that that \phi_1 = \frac{\pi}{6}. Using the conversion formula we get the trigonometric form of z_1: z_1 = 4(cos \frac{\pi}{6} + isin \frac{\pi}{6})

Second I assign the denominator to z_2 = 1-i

a = 1, b = -1, r = \sqrt{2}
cos \phi_2 = \frac{1}{\sqrt{2}}
sin \phi_2 = -\frac{1}{\sqrt{2}}

So \phi_2 = \frac{7\pi}{4} and now we can derive the trigonometric form of z_2 = \sqrt{2}(cos \frac{7\pi}{4} + isin \frac{7\pi}{4})

What I get is:\left( \frac{4(cos \frac{\pi}{6} + isin \frac{\pi}{6})}{\sqrt{2}(cos \frac{7\pi}{4} + isin \frac{7\pi}{4})} \right)^{50}

Using the division formula I get the trigonometric version of the main equation:
\left( \frac{4}{\sqrt{2}}(cos(\frac{\pi}{6}-\frac{7\pi}{4}) + isin (\frac{\pi}{6}-\frac{7\pi}{4}) \right)^{50}

How do I continue? I can simplify it further and I get cos (\frac{19\pi}{12}) and the same for sin but with the - (minus) sign in front.

I can apply the de Moivre's formula but I can't simplify it to the answer given below.

Can some one give me any guidance (step by step).

Maybe I have an error somewhere or I just don't know how to arrive to the right answer which is: 2^{74}(-\sqrt{3}+i)


Thank you in advance.

 

[hunt_mat]

The first thing I would do is this:
<br /> \frac{2 \sqrt{3}+2i}{1-i}\cdot\frac{1+i}{1+i}=\sqrt{3}-1+(1+\sqrt{3})i<br />
The the application of de Movire's theorem is relatively straight forward.

Mat