# Thread: Complex roots

1. ## Complex roots

This should be the easiest part of my module but i just cant figure it out...

Let $\displaystyle f(z) = \frac{e^{2 i \pi z}}{1 + z^3}$. Find the singularities and classify them.

So need to find $\displaystyle z$ when $\displaystyle z^3 = -1$.
One root is -1. The others we get from DeMoivres Thm...
I seem to think it should be something like $\displaystyle z = \cos(\frac{k \pi}{3}) + i\sin(\frac{k \pi}{3})$ for k = 1,2,3 but that gets me $\displaystyle i \frac{\sqrt(3)}{2} \pm \frac{1}{2}$ instead of $\displaystyle \frac{1}{2} \pm i \frac{\sqrt(3)}{2}$...

Also is there an easy way to classify the singularities? I know they are simple poles but im not sure how to prove it...

2. Originally Posted by Deadstar
This should be the easiest part of my module but i just cant figure it out...

Let $\displaystyle f(z) = \frac{e^{2 i \pi z}}{1 + z^3}$. Find the singularities and classify them.

So need to find $\displaystyle z$ when $\displaystyle z^3 = -1$.
One root is -1. The others we get from DeMoivres Thm...
I seem to think it should be something like $\displaystyle z = \cos(\frac{k \pi}{3}) + i\sin(\frac{k \pi}{3})$ for k = 1,2,3 but that gets me $\displaystyle i \frac{\sqrt(3)}{2} \pm \frac{1}{2}$ instead of $\displaystyle \frac{1}{2} \pm i \frac{\sqrt(3)}{2}$...
The roots are of the form $\displaystyle cos(\frac{(\pi+ k 2\pi)i}{3})+ i sin(\frac{(\pi+ k 2\pi)i}{3}$. You start with the $\displaystyle \pi i$ of -1 and add multiples of $\displaystyle 2\pi$, not $\displaystyle \pi$, and divide by 3.

Also is there an easy way to classify the singularities? I know they are simple poles but im not sure how to prove it...
You can factor the function into $\displaystyle \frac{e^{2i\pi z}}{(z-z_0)(z-z_1)(z-z_2)}$ where $\displaystyle z_0$, $\displaystyle z_1$, $\displaystyle z_2$ are the zeros of the denominator. Now you can see that, for example, in a neighborhood of $\displaystyle z_0$, you have a function analytic in that neighborhood ($\displaystyle \frac{e^{2i\pi z}}{(z-z_1)(z-z_2)}$ times $\displaystyle \frac{1}{z-z_0}$ which clearly has a pole of order 1 at $\displaystyle z_0$- write the first as a Taylor's series about $\displaystyle z-z_0$ and divide by $\displaystyle z-z_0$ getting as Laurent series having -1 as the only negative power.