Show that if \(f(z) \sum_{k\in \ZZ} c_k (z-z_0)^k\), then \[ c_k = {1\over 2\pi i}\int_\gamma {f(z) \over (z-z_0)^{n+1}}\dz ,\] and that this converges in an annulus \(D_R(z_0)\sm \bar{D_r(z_0)}\) where \[ r=\limsup _{n \rightarrow \infty} \sqrt[n]{\left|a_{-n}\right|} \text { and } R=\frac{1}{\limsup _{n \rightarrow \infty} \sqrt[n]{\left|a_{n}\right|}} \text {. } .\]
Hint: start with \[ f(z)=\frac{1}{2 \pi i} \oint_{\left|w-z_{0}\right|=s_{2}} \frac{f(w)}{w-z} d w-\frac{1}{2 \pi i} \oint_{\left|w-z_{0}\right|=s_{1}} \frac{f(w)}{w-z} d w ,\] and try to obtain a geometric series to obtain \[ f(z)=\sum_{j=-\infty}^{\infty}\left(\frac{1}{2 \pi i} \oint_{\left|w-z_{0}\right|=r} \frac{f(w)}{\left(w-z_{0}\right)^{j+1}} d w\right)\left(z-z_{0}\right)^{j} .\]
Solution
The statement has three notation errors: the opening display is missing an equality sign, the coefficient formula uses \(n\) where it must use \(k\), and the radii are written with \(a_{\pm n}\) instead of the coefficients \(c_{\pm n}\). With these corrections, if \[ f(z)=\sum_{k\in\ZZ}c_k(z-z_0)^k, \] then for any positively oriented circle \(\gamma\) centered at \(z_0\) and lying in the annulus of analyticity, \[ \boxed{c_k={1\over2\pi i}\int_\gamma {f(z)\over(z-z_0)^{k+1}}\,dz.} \] Indeed, termwise integration gives \[ {1\over2\pi i}\int_\gamma {f(z)\over(z-z_0)^{k+1}}\,dz =\sum_jc_j{1\over2\pi i}\int_\gamma(z-z_0)^{j-k-1}\,dz=c_k. \]
For the positive-power part, Cauchy–Hadamard gives outer radius \[ R={1\over\limsup_{n\to\infty}|c_n|^{1/n}}. \] For the negative-power part, put \(w=(z-z_0)^{-1}\); then \[ \sum_{n\ge1}c_{-n}(z-z_0)^{-n}=\sum_{n\ge1}c_{-n}w^n \] has radius \(1/\limsup|c_{-n}|^{1/n}\) in \(w\), hence converges exactly when \[ |z-z_0|>r, \qquad r=\limsup_{n\to\infty}|c_{-n}|^{1/n}. \] Therefore the Laurent series converges on \[ \boxed{r<|z-z_0|<R.} \] The two-circle formula in the hint is the usual Cauchy formula on an annulus; expanding each kernel geometrically yields the same Laurent expansion and coefficient formula.