# Complex number

1. Mar 14, 2009

### Perillux

I'm looking at a problem involving complex numbers and a proof. It shows the solution too, but I don't get how they did a certain step.
At one step they end up with this: (NOTE: the sigma should have 'n=0' on bottom and infinity on top, but I don't know how to do that in latex. If someone knows, please share):
$$Im(e^{i\theta}\sum(re^{2i\theta})^{n})$$
Then in the next step they get from the above to the below:
$$Im(e^{i\theta}\frac{1}{1-re^{2i\theta}})$$
If someone could please explain how they are equal I would greatly appreciate it.

I didn't include the default template for threads on this section of the forum because I don't think it really applies to this question, if I'm wrong then I appologize. I didn't really attempt much of a solution, other than just writing out the sums to see if I could do anything with that... no luck.

2. Mar 14, 2009

### jdougherty

The way to do limits on the sum is \sum^{\infty}_{0}. You can use ^{} or _{} to put things at top or bottom of anything you want, generally.

The reason that they can get rid of the sum is that it is a geometric series, which has the explicit formula
$\sum^{\infty}_{k=0}ar^{k} = \frac{a}{1-r}$
When $r<1$ (do you see why r must be less than 1?).

Cf. http://en.wikipedia.org/wiki/Geometric_series#Formula

3. Mar 14, 2009

### Perillux

Of course! It even says 0 < r < 1 in the problem, lol, can't believe I overlooked that.

Thank you