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

1. Sep 6, 2010

### abruski

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$z = a+ib$$

1. $$z = r(cos \phi + isin \phi)$$

$$r = \sqrt{a^2+b^2}$$

$$cos \phi = \frac{a}{r}$$

$$sin \phi = \frac{b}{r}$$

$$\frac{z_1}{z_2} = \frac{r_1}{r_2}\left( cos(\phi_1-\phi_2) + isin(\phi_1 - \phi_2)\right)$$

$$z^{n} = r^{n}(cos n \phi + isin n \phi)$$

3. 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: $$2^{74}(-\sqrt{3}+i)$$

$$\frac{2 \sqrt{3}+2i}{1-i}\cdot\frac{1+i}{1+i}=\sqrt{3}-1+(1+\sqrt{3})i$$