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Question:

How to find the Laurent series expansion in powers of z of

a) $f(z)= \dfrac{e^{z^2}}{z^3}$ $\text{where} \left| z \right| > 0$

Attempt:

I know that the main idea is to rearrange the equation in such a way that you can use standard Tylor series such as:

$e^x$ which is given as $\displaystyle \sum \frac{x^n}{n!}$

But how do you do this, and what is the method to tackle L series questions?

Edit:

After looking at some of the comments, would this method be correct:

$f(z)= \dfrac{e^{z^2}}{z^3}$ $f(z)= \dfrac{1}{z^3} e^{z^2}$ $f(z)= \dfrac{1}{z^3} \displaystyle \sum_{n=0}^\infty \frac{(z^2)^n}{n!}$ L.series? $f(z)= \dfrac{1}{z^3} \biggl(1+z^2+\dfrac{z^4}{2!}+\dfrac{z^6}{3!}+...\biggr)$
GeoffChurch
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UniStuffz
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1 Answers1

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Yes your answer is correct. Since the taylor series of $e^{x}$ converges absolutely it is the same as the laurent series. Now if you want to you can simplify.

$\frac{1}{z^{3}}\sum\frac{z^{2n}}{n!}$

$\sum\frac{z^{2n-3}}{n!}$

Which expands as...

$$\frac{1}{z^3}+\frac{1}{z}+\frac{z}{2!}+\frac{z^3}{3!}...$$

Arbuja
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