# Must a meromorphic function on a compact set have same number of zeros and poles?

Let $f:X\rightarrow\mathbb{C}\cup\{\infty\}$ be a meromorphic function while $X$ is compact.

Must $f$ have same number of zeros and poles?

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What about $f(z) = 0$ or $f(z) = z$? –  Carl Mummert Jan 9 '13 at 13:57
It looks like it could be true if you specify that $X$ is not only compact but is actually the entire Riemann sphere, and you count zeroes and poles with multiplicity, and you exclude the constant zero function. –  Henning Makholm Jan 9 '13 at 14:10
Function $f(z)=\dfrac{z-2}{(z^4-1)^3}$ has one zero and four poles (with multiplicity 3) in $\mathbb{C}$ –  M. Strochyk Jan 9 '13 at 14:22
@M.Strochyk $f(\infty)=0$ and the zero has multiplicity 3 ;) –  N. S. Jan 9 '13 at 15:02
@N. S. Precisely ;) –  M. Strochyk Jan 9 '13 at 15:35

I think that you mean $X$ is a compact Riemann surface rather that a compact subset of the complex plane. If $X$ is a compact Riemann surface, and $f:X\to \mathbb C\cup \{\infty\}$ is a meromorphic function, then indeed $f$ has the same number of zeros and poles, with multiplicity counted. In the more general case, let $X, Y$ be Riemann surfaces, $X$ compact and $f:X\to Y$ be a holomorphic map and $f(x)=y$. Then in properly chosen local coordinates around $x$ and $y$, we can write $f(z)=z^k$. Then we define the ramification index at the point $x$ to be $v_x=k$. It's easy to see that $f^{-1}(y)$ is a finite set for all $y\in Y$. The function $d(y)=\sum\limits_{f(x)=y}v_x$ is defined. Now I will show that $d$ is a locally constant function on $Y$ and hence it's a constant. Then your question will be answered.
For any $y\in Y$, let $f^{-1}(y)=\{x_1,\dots,x_m\}$. Choose a coordinate chart $U$ around $x$ and coordinate charts $V_i$ around $x_i$ such that $f(V_i)\subset U$. Via replacing $U$ by $\cap f(V_i)$ and replacing $V_i$ by $V_i\cap f^{-1}\left(\cap f(V_i)\right)$, we may assume that $f(V_i)=U$ for each $i$. Note that here I used the fact the holomorphic maps are open. By replacing $U$ by $U-f(X-\cup V_i)$, and intersecting $V_i$ with the preimage of the new $U$, we may assume that $f^{-1}(U)=\cup V_i$. (Here I used the fact that $f$ is a closed map.) Since locally a holomorphic map is $z\mapsto z^k$, by shrinking $U$ and intersecting $V_i$ and the preimage of the new $U$, finally we may assume that for each $y\neq y'\in U$, $y'$ has exactly $v_{x_j}$ preimages in $V_j$. Now clearly $d$ is constant in $U$.
Beautiful proof! But I still have a question, what if $X$ is just a compact subset of $\mathbb{C}$ rather a compact Riemann surface, what part of the proof fails? –  hxhxhx88 Jan 10 '13 at 2:04
@hxhxhx88: Dear hx, if $X$ is just a compact subset of $\mathbb C$, there will be a big problem on the boundary of $X$. A nonempty compact subset of $\mathbb C$ is a closed bounded subset. It cannot be open. How do you define "holomorphic" on such sets? –  Yang Zhou Jan 10 '13 at 15:12