# A: Reciprocal series, B: Laurent Series and Cauchy's Formula

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1. Jan 13, 2016

### sinkersub

Problem A now solved!

Problem B:
I am working with two equations:

The first gives me the coefficients for the Laurent Series expansion of a complex function, which is:

$f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n$

This first equation for the coefficients is:

$a_n = \frac{1}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz$

The second equation is Cauchy's Integral formula for the nth derivative of a complex function:

$f^{(n)}(z_0) = \frac{n!}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz$

My problem is as follows:

We can clearly re-arrange these two equations to get the following expression:

$a_n = \frac{f^{(n)}(z_0)}{n!}$

Now, in this instance, $f(z) = \frac{z^3 + 2z^2 + 4}{(z-1)^3}$.

If we try to calculate $a_n$ for this function, it keeps getting sent to zero, due to the function being evaluated at $z_0$!

What am I missing? Where have I gone wrong in this derivation/problem?

Ultimately in this problem I'm trying to find the Laurent Series expansion of the function.

Any help/tips would be much appreciated!

sinkersub

Last edited: Jan 13, 2016
2. Jan 13, 2016

### Staff: Mentor

You first take the derivative, then plug in a fixed z0. In general, the result should not be zero.