# Finding the laurent series around z = 0

I'm trying to find the Laurent series around $$z = 0$$ and specify the largest annulus where the expansion is valid. It's for the functions:

$$f(z) = 1/((z-a)(z-b))$$, for $$a,b, \in \mathbb{C}$$

and

$$f(z) = z^3 e^{1/z}$$

I've tried to look around the internet for help, but the tasks always differ a bit.

For the first function i tried a partial fraction expansion:

$$f(z) = \frac{1}{(b-a)(z-b)} + \frac{1}{(a-b)(z-a)}$$

and for function two, i think i need to find simpler expansions so i won't have to do a contour integral?

I would love some tips and inputs.

• $f(z) = \frac{1}{(b-a)(z-b)} + \frac{1}{(a-b)(z-a)}$ is not a Laurent series about $z=0$. All the terms should be powers of $z.$ It's just like Taylor series but negative powers are allowed. – saulspatz Mar 13 '19 at 13:13

For the first one, you did the right thing. The next step is to write:$$f(z)=\frac1{ab-b^2}\times\frac1{1-\frac zb}+\frac1{ab-a^2}\times\frac1{1-\frac za}.$$On the other hand$$(\forall z\in\mathbb C\setminus\{0\}):z^3e^{\frac1z}=z^3+z^2+\frac z{2!}+\frac1{3!}+\frac1{4!z}+\cdots$$
• Dear José, thank you very much for the reply. For the first, would you then find the coefiiccents for the laurent series, using the contour integral? Did you find the second using the infinete sum representation of $e^{1/x}$ ? – nammerkage Mar 13 '19 at 13:53
• For the second one, the answer is affirmative. For the first one: I would justa use the fact that$$\lvert z\rvert<\lvert c\rvert\implies\frac1{1-\frac zc}=1+\frac zc+\frac{z^2}{c^2}+\cdots$$ – José Carlos Santos Mar 13 '19 at 13:54
• $$\frac1{(b-a)(z-b)}=\frac1{a-b}\cdot\frac1{b-z}=\frac1{(a-b)b}\cdot\frac1{(b-z)\cdot\frac1b}.$$ – José Carlos Santos Mar 14 '19 at 18:29