Find real part of $\frac{1}{1-e^{i\pi/7}}$ How can you find
$$\operatorname{Re}\left(\frac{1}{1-e^{i\pi/7}}\right).$$
I put it into wolframalpha and got $\frac{1}{2}$, but I have no idea where to begin. I though maybe we could use the fact that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2},$$ where $\bar{z}$ is the conjugate of $z$. Unfortunately, the magnitude doesn't seem to be a nice number. I feel like this might be a trigonometry question in disguise, but converting $e^{i\pi/7}=\cos\left(\frac{\pi}{7}\right)+i\sin\left(\frac{\pi}{7}\right)$ hasn't been very fruitful.

Well, firstly, since $z\bar{z} = |z|^2$ we know $\frac{1}{z} = \frac{\bar{z}}{|z|^2}$ (1). So let $z = 1 - e^{\frac{i \pi}{7}}$. Then from (1), $$\frac{1}{1 - e^{\frac{i \pi}{7}}} = \frac{\bar{z}}{|z|^2}.$$
Now, by Euler's formula, $e^{\frac{i \pi}{7}} = \cos(\frac{\pi}{7})+i\sin(\frac{\pi}{7}),$ so $z = (1- (\cos(\frac{\pi}{7})+i\sin(\frac{\pi}{7}))) = (1 - \cos(\frac{\pi}{7})) - i\sin(\frac{\pi}{7})$. 
Now this means that $\bar{z} = (1 - \cos(\frac{\pi}{7})) + i\sin(\frac{\pi}{7})$. Moreover, $|z|^2 = (1 - \cos(\frac{\pi}{7}))^2 + \sin(\frac{\pi}{7})^2$, therefore, $|z|^2 = 1 + (\cos^2(\frac{\pi}{7}) + \sin^2(\frac{\pi}{7})) - 2\cos(\frac{\pi}{7}) = 2 - 2\cos(\frac{\pi}{7})$, so; $$\frac{1}{z} = \frac{1 - \cos(\frac{\pi}{7})}{2 - 2\cos(\frac{\pi}{7})} + i\frac{\sin(\frac{\pi}{7})}{2-2\cos(\frac{\pi}{7})}.$$ 
So, $Re(\frac{1}{z}) = \frac{1 - \cos(\frac{\pi}{7})}{2 - 2\cos(\frac{\pi}{7})} = \frac{1 - \cos(\frac{\pi}{7})}{2(1 - \cos(\frac{\pi}{7}))} = \frac{1}{2}$. I hope that helps :P
