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More specifically:

Suppose that $f(z)$ is meromorphic on a disk $\{\lvert z\lvert , show that the Laurent decomposition of $f(z)$ on the annulus $\{s-\epsilon<\lvert z\lvert has the form $f_0(z) + f_1(z)$, where $f_1(z)$ is the sum of the principal parts of $f(z)$ at its poles.

Since $f(z)$ is meromorphic, the number of poles it has must be finite. However, I don't know how to define the disks surrounding each pole and add them up... Any help much appreciated.

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As $f$ is meromorphic, it has a finite number of poles with finite orders. Let us say $c_1,\ldots, c_k$ with orders $m_1,\ldots, m_k$. Therefore $f(z)\prod_{j=1}^k(z-c_j)^{m_j}$ is a holomorphic $g(z)\in\mathcal{O}(\{|z|. Hence $f(z)=g(z)\left(\frac{1}{(z-c_1)^{m_1}}\cdots\frac{1}{(z-c_k)^{m_k}}\right)=g(z)\sum_{j=1}^k\sum_{h=1}^{m_j}\frac{a_{j,h}}{(z-c_j)^{h}}$ where $a_{j,h}$ are suitable complex numbers.

So, the Laurent series of $f(z)$ is the sum of the Laurent series of $g(z)\frac{a_{j,h}}{(z-c_j)^{h}}$.

The coefficients of the Laurent series of $g(z)\frac{a_{j,h}}{(z-c_j)^{h}}$ can be calculated by integration on every curve surrounding $c_j$, because the function is holomorphic for $z\neq c_j$. On a very small disk around $c_j$, the difference $f(z)-\sum_{h=1}^{m_j}g(z)\frac{a_{j,h}}{(z-c_j)^{h}}$ is holomorphic around $c_j$, therefore the principal parts of these two functions around $c_j$ coincide.

And the thesis should follow.

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    It isn't exactly easy to write these two functions in my setting. The idea is that we can find a small disk D_j=\{|z-c_j|<\epsilon\} such that the only pole of $f$ in $D_j$ is $c_j$, therefore in D_j^*=\{0<|z-c_j|<\epsilon\} the principal parts (pp) of $f$ and $\phi_j=\sum_{h=1}^mg(z)a_{j,h}(z-c_j)^{-h}$ coincide. On the other hand, the pp of $\phi_j$ is the same if calculated on $D_j^*$ or on \{0<|z|, because the function $\phi_j$ is holomorphic outside $c_j$. So,the pp of $f$ is the same of the sum of the pp of $\phi_j$ which are the pp of $f$ around each $c_j$.2012-12-07