# Laurent series

1. Jul 22, 2014

### MissP.25_5

Hello.

The question is:
Let {$z\inℂ|0<|z|$}, expand $\frac{e^{z^2}}{z^3}$ where the centre z=0 into Laurent series.

And the solution is:
$$\frac{e^{z^2}}{z^3}=\sum_{n=0}^{\infty}\frac{\frac{(z^2)^n}{n!}}{z^3}=\sum_{n=0}^{\infty}\frac{z^{2n-3}}{n!}$$

I don't understand the solution because isn't the formula for Laurent series
$$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$

where
$$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$

$$b_n=\frac{1}{2\pi{i}}\oint_{c}^{}(z-z_0)^{n-1}f(z^*)dz^*$$

2. Jul 22, 2014

### MissP.25_5

Why is this a laurent series? This looks like a maclaurin expansion of the exponential function.

3. Jul 22, 2014

### SammyS

Staff Emeritus
Does a MacLaurin expansion have negative exponents?

4. Jul 22, 2014

### MissP.25_5

No it doesn't but that definitely is a power series of the exponential function. Is Laurent series the same as power series? I am confused.

5. Jul 22, 2014

### Ray Vickson

Just write out the first 3 or 4 terms of your series. Does that not look like a Laurent series to you?

6. Jul 22, 2014

### MissP.25_5

Aha, now I see it. So, can I conclude that a Taylor/maclaurin series is a Laurent series with only positive exponents?

7. Jul 23, 2014

### HallsofIvy

Yes, of course. But what does that have to do with this problem? This series has negative powers.