Sum of complex series : $1+z+z^2 + \cdots + z^{n-1} = 1 + i \cot \left(\frac{\pi}{2n}\right)$

Question

Show that $$1+z+z^2 + \cdots + z^{n-1} = 1 + i \cot \left(\frac{\pi}{2n}\right)$$ where $z = \cos\left(\frac{\pi}{n}\right) + i \sin\left(\frac{\pi}{n}\right)$ where $n \in \mathbb{Z^+}$

This question has confused me slightly and has left me rather stuck.

My attempt so far has been:

Note that $z = e^{i\frac{\pi}{n}}$ and that $$1+z+z^2+\cdots + z^{n-1} = \frac{1 - z^n}{1-z} \\ = \frac{1- e^{i\pi}}{1-e^{i\frac{\pi}{n}}} = \frac{2}{1-e^{i\frac{\pi}{n}}}$$

but I don't see a way to go from here...

Any help is appreciated.

Answer 1

$$1+z+z^2+\cdots + z^{n-1}$$

$$ = \frac{1 - z^n}{1-z} $$

$$ = \frac{1- e^{i\pi}}{1-z} $$

$$= \frac{2}{1-z }$$

$$=\frac{2(1-\bar z)}{(1-z)(1-\bar z) }$$

$$=\frac{2(1-cos(\pi/n)+i\sin(\pi/n))}{2-2\cos (\pi/n) }$$

$$=1+i\frac {\sin(\pi/n)}{1-\cos(\pi/n)}=1+i\cot(\pi/{2n})$$

In the last step half angle identities were used.

Answer 2

First notice $$1-e^{i\theta} = 2\sin(\theta/2)e^{-i\theta/2}.$$

Indeed, we know from trigonometry that $$1-\cos(\theta)=2\sin^2(\theta/2) \quad\text{and}\quad \sin(\theta) = 2\sin(\theta/2)\cos(\theta/2)$$ and therefore $$1-e^{i\theta} = (1-\cos(\theta)) - i\sin(\theta) = 2\sin(\theta/2)e^{-i\theta/2}.$$

With that in mind we have $$\frac{2}{1-e^{i\theta}} = \frac{e^{i\theta/2}}{\sin(\theta/2)} = 1+ i\cot(\theta/2).$$

Taking $\theta = \pi/n$ we obtain the result you want.

Answer 3

You have already found that $$1 + z + x^2 + \dots + z^{n-1} = \frac{2}{1-e^{i \pi / n}}$$ Now note that $$\begin{align} \frac{2}{1-e^{i \pi / n}} - 1 &= \frac{1 + e ^{i \pi /n}}{1 - e ^{i \pi /n}} \\ &= \frac{1 + e ^{i \pi /n}}{1 - e ^{i \pi /n}} \cdot \frac{e^{-i \pi /2n}} {e^{-i \pi /2n}} \\ &= \frac{e^{- i \pi / 2n} + e^{i \pi / 2n}} {e^{- i \pi / 2n} - e^{i \pi / 2n}} \\ &= \frac{2 \cos(\pi / 2n)}{-2i \sin(\pi / 2n)} \\ &= i \cot \left( \frac{\pi}{2n} \right) \end{align}$$