# Summation of complex numbers

This is a series problem where the terms are complex numbers. I am looking for a better approach to solving this problem.

If $\displaystyle z = \frac{1+i}{\sqrt2}$, Evaluate $1 + z + z^2 + ... + z^{20}$

The way I solved this was to evaluate the terms upto the 8th term. like below,

\begin{align} z^2 &= i \\ z^3 &= -\frac{1}{\sqrt2} +\frac{1}{\sqrt2}i \\ z^4 &= -1 \\ z^5 &= -\frac{1}{\sqrt2} -\frac{1}{\sqrt2}i \\ z^6 &= -i \\ z^7 &= -\frac{1}{\sqrt2} +\frac{1}{\sqrt2}i \\ z^8 &= 1 \\ \end{align}

Then evaluating the first 8 terms, $$S_7 = 1 + z + ... + z^7 = 0$$

Which implies that, $$S_20 = 0 + 0 + z^{17} + z^{18} + z^{19} + z^{20} = 1 + z + z^2 + z^3$$

Thus giving the solution, $$S_20 = 1 + (1 + \sqrt2)i$$

1. Just looking at the series itself($1 + z + z^2 + ...$) makes me think that the series is similar to a binomial series. The way the terms cancelled out made me think of a telescopic series. Can you guys shed some light on this?

2. While this solution works it took me a while to get there. Is there a more elegant/less tedious way of doing this?

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@mathguy: the value of $z_{7}$ is wrong? there should be a $"+"$ sgin – user9413 Jun 2 '11 at 10:49
@mathguy, note that as soon as you got $z^2=i$, you could have gone right to $z^8=(z^2)^4=i^4=1$, so $z^8-1=0$, so $(z-1)(z^7+z^6+\dots+z+1)=0$, so $z^7+z^6+\dots+z+1=0$. – Gerry Myerson Jun 2 '11 at 11:24
@Gerry Myerson, very elegant and concise! Thank you. – mathguy80 Jun 3 '11 at 9:19
@Chandru, yup fixed that. – mathguy80 Jun 3 '11 at 9:20
@mathguy80: Thanks. PLease accept an answer :) – user9413 Jun 3 '11 at 9:21

1. If $z \ne 1$, we have

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

So it greatly reduces the number of computation

1. Note that in your case $z = e^{\frac{i \pi}{4}}$, which makes it faster to compute the powers of $z$.

Applying those two ideas gets you the result.

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Thanks. So it's a binomial series. Does $z$ being complex have any bearing on the series? – mathguy80 Jun 3 '11 at 9:23

Use the formula for the sum of a geometric series and the fact that $$\bigl(\cos\theta + i\sin\theta)^{n} = \cos{n\theta} + i \sin{n \theta}$$

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I considered that but converting to polar form would have been more tedious. :) – mathguy80 Jun 3 '11 at 9:24