I have the following problem:
Let $f:\Omega \rightarrow \Bbb{C}$ be a meromorphic function in a region $\Omega$ with finately many zeros $z_j$ of order $m_j$ and finately many poles $p_k$ of order $m_k$. I define $$F(z)=\frac{\prod_k (z-p_k)^{m_k}}{\prod_j (z-z_j)^{m_j}}f(z)$$
I don't see why $F$ has removable singularities at $p_k$ and $F(z)\neq 0$ for all $z\in \Omega$.
Could maybe someone explain me that? It would be nice if you could use the definition below
We had the following definition of a removable singularity:
Let $f$ be an analytic function on $\{0<|z-z_0|<\delta\}$ for some $\delta>0$. Then $f$ has a laurent series expansion $$f(z)=\sum_{z\in \Bbb{Z}} a_n (z-z_0)^n$$ If $a_n=0$ for $n<0$ then $f(z)=a_0+a_1(z-z_0)+...$ we define $f(z_0)=a_0$ then $f$ is analytic on $|z-z_0|<\delta$. In this case $z_0$ is a removable singularity.
Let $n$ be the number of poles and take $l\in\{1,2,\ldots,n\}$. Then the function$$z\mapsto\frac{\prod_{k\ne l}(z-p_k)^{m_k}}{\prod_j(z-z_j)^{m_j}}$$is a rational function which has no zero and no pole at $p_l$. So, near $p_l$, its Taylor series has the form $a_0+a_1(z-p_l)+a_2(z-p_l)^2+\cdots$, with $a_0\ne0$. On the other, near $p_l$ you have$$f(z)=\frac{b_0+b_1(z-p_l)+b_2(z-p_l)^2+\cdots}{(z-p_l)^{m_l}},$$with $b_0\ne0$, and therefore$$\frac{\prod_{k\ne l}(z-p_k)^{m_k}}{\prod_j(z-z_j)^{m_j}}f(z)=\frac{c_0+c_1(z-p_l)+c_2(z-p_l)^2+\cdots}{(z-p_l)^{m_l}},$$with $c_0=a_0b_0(\ne0)$, $c_1=a_0b_1+a_1b_0$, $c_2=a_0b_2+a_1b_1+a_2b_0$, and so on… But then, near $p_l$,\begin{align}\frac{\prod_{k}(z-p_k)^{m_k}}{\prod_j(z-z_j)^{m_j}}f(z)&=(z-p_l)^{m_l}\frac{c_0+c_1(z-p_l)+c_2(z-p_l)^2+\cdots}{(z-p_l)^{m_l}}\\&=c_0+c_1(z-p_l)+c_2(z-p_l)^2+\cdots,\end{align}So,$$\dfrac{\prod_{k}(z-p_k)^{m_k}}{\prod_j(z-z_j)^{m_j}}f(z)\tag1$$has no zero and no pole at $p_l$.
A similar argument shows that, if $m$ is the number of zeros, and if $l\in\{1,2,\ldots,m\}$, then $(1)$ has no zero at $z_l$.