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If $f$ is a rational function defined on the complex plane. Then the number of the zeros is equal to the number of the poles (counting multiplicity) and considering points at infinity.

I can imagine a proof using the argument principle. Is this the simplest route to the theorem above? Is there a name for the theorem above?

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    Ok thanks guys. Really helped.2011-10-09

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Summary of comments by mt_ and Christian Blatter: every rational function $f$ that is not identically zero can be written as $ f(z)= c \frac{\prod_{j=1}^m (z-z_j)}{\prod_{k=1}^n (z-p_k)} \tag1 $ where $c\ne 0$ and $z_j\ne p_k$ for all $j,k$. (Neither $z_j$ nor $p_k$ are required to be distinct.) In the complex plane $\mathbb C$, this function has $m$ zeroes and $n$ poles, counting multiplicities. As $z\to\infty$, we have $f(z)\sim cz^{m-n}$ which means a zero of order $n-m$ (if $n>m$) or a pole of order $m-n$ (if $m>n$). Either way, the number of zeroes is equal to the number of poles.