Find the Laurent series expansion of $f(z)=\frac{z^2-1}{(z+2)(z+3)^2}$ at $0<|z+3|<1$

I have a couple of doubts in how to handle this problem: First of all, should I do it with partial fractions? Because now I have a polynomial in the numerator, so I don't know how to proceed if that's the case.

If not, can I use the general term for the Laurent series?

$$a_n=\frac{1}{2\pi i}\int \frac{f(\zeta)d\zeta}{(\zeta +3)^{n+1}}$$

But what happens with the $(z+3)^2$ in the numerator? Im pretty mixed up.


Doing partial fractions as in real integration:




Observe that the development of $\;\frac{-3}{1-(z+3)}\;$ in the above power series is valid since we re given $\;|z+3|<1\;$ . You get here $\;z=-3\;$ is a double pole with residue equal to $\;-2\;$ .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.