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## Sum with complex numbers

- Tolaso J Kos
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### Sum with complex numbers

Let $z \in \mathbb{C}$ such that $z^{2017}=1$ and $z\neq 1$. Evaluate the sum $$\sum_{n=1}^{2017} \frac{1}{1+z^n}$$

**Imagination is much more important than knowledge.**

### Re: Sum with complex numbers

The sum is written as:

$$\sum_{n=1}^{2017}\frac{1}{1+z^n}= \frac{1}{1+z}+ \frac{1}{1+z^2}+\cdots + \frac{1}{1+z^{2017}}$$

Now we are "changing" the first $1008$ terms. We note that

$$\frac{1}{1+z}= \frac{z^{2017}}{z+z^{2017}}= \frac{z^{2016}}{1+z^{2016}}$$

and similarly $\displaystyle \frac{1}{1+z^2} = \frac{z^{2015}}{1+z^{2015}}$.

This way $1008$ pairs are created which sum to $1$. That is we get $1008$ ones and of course the last term is obviously $1/2$. Hence:

$$\sum_{n=1}^{2017}\frac{1}{1+z^n}= 1008 + \frac{1}{2} = \frac{2017}{2}$$

$$\sum_{n=1}^{2017}\frac{1}{1+z^n}= \frac{1}{1+z}+ \frac{1}{1+z^2}+\cdots + \frac{1}{1+z^{2017}}$$

Now we are "changing" the first $1008$ terms. We note that

$$\frac{1}{1+z}= \frac{z^{2017}}{z+z^{2017}}= \frac{z^{2016}}{1+z^{2016}}$$

and similarly $\displaystyle \frac{1}{1+z^2} = \frac{z^{2015}}{1+z^{2015}}$.

This way $1008$ pairs are created which sum to $1$. That is we get $1008$ ones and of course the last term is obviously $1/2$. Hence:

$$\sum_{n=1}^{2017}\frac{1}{1+z^n}= 1008 + \frac{1}{2} = \frac{2017}{2}$$

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$