Sum with complex numbers
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
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:
$$\sum_{n=1}^{2017}\frac{1}{1+z^n}= 1008 + \frac{1}{2} = \frac{2017}{2}$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 8 guests