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[SOLVED] Laurent series

dwsmith

Well-known member
Feb 1, 2012
1,673
I have never done a Laurent series nor have we went over it in class but I guess I am supposed to know it perfectly already. The explanation in the book isn't that great.

Find the Laurent series of the form $\sum\limits_{n = -\infty}^{\infty}c_nz^n$ for $f(z) = \dfrac{33}{(2z - 1)(z + 5)}$ that converges in an annulus containing the point $z = -3i$

So I have used partial fractions
$$
f(z) = \frac{33}{(2z - 1)(z + 5)} = \frac{6}{2z - 1} - \frac{3}{z + 5}
$$

What am I supposed to do next?

Is this it

$$
\frac{-6}{1-2z}=-6(1+2z+(2z)^2+\cdots )\quad\text{and}\quad
\frac{3/5}{1-\left(-z/5\right)} = \frac{3}{5}\left(1+\frac{-z}{5}+\left(\frac{-z}{5}\right)^2+\cdots \right)
$$
 
Last edited:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
340
There are some cases to be considered. The function is analytic when \( |z| < \frac{1}{2} \), \( \frac{1}{2} < |z| < 5 \) and \( 5 < |z| < \infty \). The only annulus that contains \( z = -3i \) is the middle. So, we need that \( |z| > \frac{1}{2} \) which means \( \left| \frac{1}{2z} \right| < 1 \) and \( |z| < 5 \) with \( \left| \frac{z}{5} \right| < 1 \).

Then $$\frac{6}{2z-1} = \frac{3}{z} \cdot \frac{1}{1 - \frac{1}{2z}} = \frac{3}{z} \sum_{n=0}^{\infty} 2^{-n} z^{-n} = 3 \sum_{n=0}^{\infty} 2^{-n} z^{-(n+1)}$$ and $$\frac{-3}{z+5} = \frac{-3}{5} \cdot \frac{1}{1 - \left( - \frac{z}{5} \right) } = \frac{-3}{5} \sum_{n=0}^{\infty} \frac{(-1)^n z^n}{5^n} = -3 \sum_{n=0}^{\infty} \frac{(-1)^n z^n}{5^{n+1}}.$$

Finally, $$ f(z) = 3 \left( \sum_{n=0}^{\infty} 2^{-n} z^{-(n+1)} - \sum_{n=0}^{\infty} \frac{(-1)^n z^n}{5^{n+1}} \right).$$

Hope it's clear. By the way, which book are you using?
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673