Question: Prove Laurent’s Expansion$$f(z)=\sum\limits_{i\geq0}a_n(z-z_0)^n+\sum\limits_{j\geq0}\frac {b_j}{(z-z_0)^j}$$where$$a_i=\frac 1{2\pi i}\oint\limits_{C_1}dw\,\frac {f(w)}{(w-z)^{i+1}}$$$$b_j=\frac 1{2\pi i}\oint\limits_{C_2}dw\,\frac {f(w)}{(w-z)^{-j+1}}$$
I’m having trouble proving this. If we let $C_1$ and $C_2$ be two concentric circles with the radius of $C_2$ less than $C_1$, then we have$$f(z)=\frac 1{2\pi i}\oint\limits_{C_2}dw\,\frac {f(w)}{w-z}-\frac 1{2\pi i}\oint\limits_{C_1}dw\,\frac {f(w)}{w-z}$$However after this, I’m not sure what to do. I don’t see how to get both contour integrals into an infinite series expansion.