- #1
MissP.25_5
- 331
- 0
Hello.
I need explanation about this Laurent series.
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
$$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$
$$b_n=\frac{1}{2\pi{i}}\oint_{c}^{}(z-z_0)^{n-1}f(z^*)dz^*$$
I need explanation about this Laurent series.
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
$$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$
$$b_n=\frac{1}{2\pi{i}}\oint_{c}^{}(z-z_0)^{n-1}f(z^*)dz^*$$