# Homework Help: [Complex analysis] Coefficients of Laurent series

1. Apr 10, 2012

### mick25

1. The problem statement, all variables and given/known data
I have some past exam questions that I am confused with

2. Relevant equations

$a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\, dz$

3. The attempt at a solution

I'm not sure how to approach this, I'm completely lost and just attempted to solve a few:

a) it says $f(z)$ has a pole of order 5, so $f(z) = \frac{g(z)}{z^5}, g(z)\neq0$

so then I guess the condition is $a_{4} = \frac{g^{(4)}(0)}{4!}$? But that's just applying the formula for the coefficients...

c) $f(\frac{1}{z}) = \frac{g(\frac{1}{z})}{z^5} => f(z) = z^5g(z)$

so the coefficients are $a_{n} = \frac{1}{2\pi i} \oint_\gamma z^5g(z) dz$?

d) $\frac{1}{f(z)} = \frac{g(z)}{z^5} => f(z) = \frac{z^5}{g(z)}$

so, $a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{z^5}{g(z)} dz$

g) $a_{-1} = \frac{1}{2\pi i} \oint_ \gamma f(z) dz = \frac{1}{2\pi i} = Res(f; c)*I(\gamma; c) = -Res(f; c)$

h) $\frac{a_{n}}{16} = 4^{n}a_{n} => 0 = a_{n}(4^{n} - 4^{-2}) => a_{n} = 0$ or $n = -2$

for e) and f), I'm not sure what the relevance of the essential singularity is

Well, I think you can see I'm clearly lost, would appreciate if you could help me out.

2. Apr 10, 2012

### jackmell

You need to look at these by sayin' for example, (a), if
$$f(z)=\sum_{n=0}^{\infty} a_n z^n+\sum_{n=1}^{\infty}\frac{b_n}{z^n}$$
and it has a pole of order 5, then that must mean it at least has a non-zero term for the $\frac{1}{z^5}$ term and all the other terms $\frac{1}{z^n}$ for n>5 must be zero else the pole would have a higher order. Ok, I'll do some of (b) same dif:

If:
$$f(z)=\sum_{n=0}^{\infty} a_n z^n+\sum_{n=1}^{\infty}\frac{b_n}{z^n}$$
and $f(z)-7e^{1/z}$ has a pole of order 5 and I know that $e^{1/z}=\sum_{n=0}^{\infty} \frac{1}{z^n n!}$, then that must mean the terms of f(z) would have to be what to cancel all that out except for at least the $\frac{1}{z^5}$ term?

Ok then, do that same thing with all the rest.