# 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.

• Note: what we actually have is $$\frac{1}{z}=\frac{\bar z}{|z|^\color{red}2}$$ and $$e^{i\pi/7}=\cos\left(\frac{\pi}{7}\right)+i\sin\left(\frac{\pi}{7}\right).$$
– Dave
Commented Jun 8, 2018 at 15:16
• @Dave Oops, sorry, thank you for the corrections. Commented Jun 8, 2018 at 15:24

Let $$z=\frac1{1-e^{it}}$$ where $t$ is real and $e^{it}\ne1$. Then $$z+\overline z=\frac1{1-e^{it}}+\frac1{1-e^{-it}} =\frac1{1-e^{it}}+\frac{e^{it}}{e^{it}-1}= \frac{1-e^{it}}{1-e^{it}}=1.$$ Therefore the real part of $z$ equals $1/2$.

• This is very nice, interesting that if the original question was more general (with $x$ instead of $\pi/7$) it probably would have been easier for me to find the solution. Commented Jun 8, 2018 at 15:21
• Nice. I'm sure there's a geometric way to phrase this proof since you're just manipulating the unit circle but I can't quite see it. Commented Jun 8, 2018 at 15:48

$$\frac{1}{1-e^{i\pi/7}}=\frac{e^{-i\pi/14}}{e^{-i\pi/14}-e^{i\pi/14}}=\frac12\frac{ie^{-i\pi/14}}{\sin\frac\pi{14}}=\frac12+\frac i2\frac{\cos\frac{\pi}{14}}{\sin\frac{\pi}{14}}=\frac{1}{2}+\frac i2\cot\frac{\pi}{14}$$

Try this,

$$z=\frac{1}{1+e^{\frac{i\pi}{7}}}=\frac{1}{1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}}$$

$$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{(1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7})\times(1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7})}$$ $$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{(1+\cos\frac{\pi}{7})^2+\sin^2\frac{\pi}{7}}$$ $$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{2(1+\cos\frac{\pi}{7})}$$ $$=\frac{1}{2}-\frac{i\sin\frac{\pi}{7}}{2(1+\cos\frac{\pi}{7})}$$

And so $Re(z)=\frac{1}{2}$

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