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.
, 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