# Thread: Help with Laurent Series

1. ## Help with Laurent Series

I'm trying to construct Laurent series around $z = 0$ and $z= \infty$ for $f(z) = 1/{z-2}$ with $z_{0} = 2$ the only pole of this function.

I have $f(z) = \sum_{n=-\infty}^{\infty}a_{n}(z-c)^n$

with $a_{n} = \oint_{\gamma}f(z)/(z-c)^{n+1}dz$

and $\oint_{\gamma}g(z)dz = 2i\pi res_{z_{0}}g(z)$

and $res_{z_{0}}g(z) = \lim_{z \to \infty}(z - z_{0})g(z)$

so am I correct in assuming that around z=c:

$f(z) = \sum_{n=-\infty}^{\infty}a_{n}(z-c)^n = \sum_{n=-\infty}^{\infty} (z - c)^{n}\lim_{z \to z_{0}}f(z)\frac{z - z_{0}}{(z - c)^{n+1}}$

or did I make a mistake?

And how the hell do I evaluate the function around z=infinity?

2. I am confused about what you are doing here. $f(z)= \frac{1}{z- 2}$ has a pole at z= 2, as you say, but is analytic at z= 0. A function's "Laurent series" about any point at which it is analytic is just its Taylor series.
In this case, $\frac{1}{2-z}= \frac{1/2}{1- z/2}$ which is the sum of a geometric series with first term a= 1/2 and common ratio r= z/2.

As for "at z=infinity", the Laurent series of any function, f(z), about $z= \infty$ is just the Laurent series of f(1/z) about z= 0.

If $f(z)= \frac{1}{2-z}$ then $f(\frac{1}{z})= \frac{1}{2- \frac{1}{z}}= \frac{z}{2z-1}$.

3. Thank you.

Yeah, I was kinda wondering why only this problem had different values for c and z0 so it makes sense this is a special case. But if I were told to look for the Laurent Series near c=2=z0 would my last equation be correct?
I looked up the answer using an online calculator (for z0 = 2 and c=0) and it corresponded to what I found using my method above, except for the sign: the calculator found: -1/2 -z/4 -(z^2)/8 etc.. while I found 1/2 z/4 (z^2)/8 etc... is it a coincidence that I got so close?