# Easiest/shortest proof of the following theorem

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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Isn't this clear just from factoring the numerator and denominator then cancelling any common factors? The number of zeros and poles is then visible (being careful about those at infinity) – m_t_ Oct 9 '11 at 11:08
Just so I can clarify what your saying: $f(z)=\frac{(z-1)}{(z-2)(z-3)}$ has a zero at 1 and a pole at 2,3 each of multiplicity 1. How can I work out the multiplicity of the zero at infinity without invoking the theorem mentioned above? – Jam Baxter Oct 9 '11 at 11:34
A factor $z-\alpha$ has a zero of order $1$ at $\alpha$ and a pole of order $1$ at $\infty$. – Christian Blatter Oct 9 '11 at 11:54
Ok thanks guys. Really helped. – Jam Baxter Oct 9 '11 at 12:29

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.