Why must a meromorphic function, bounded near infinity, have the same number of poles and zeros?

Question

Writing down some easy rational functions to check this, I don't see why this must be the case.

Although if the function had 3 simple zeros and 2 simple poles its rational form would be in the form of a $\frac{cubic}{quadratic}$, and this function doesn't stay bounded near infinity.

However, flipping the above, i.e., let's say $f$ has 2 simple zeros and 3 simple poles, and is required to stay bounded near infinity.

Doesn't this function exist? It would be a $\frac{quadratic}{cubic}$, which stays bounded near infinity.

A solution that I am reading claims that such a function cannot exist, unless the number of zeros and poles are the same.

Any ideas are welcome.

Thanks,

Answer 1

The idea is that the zero or pole at infinity cancels out the finite zeroes and poles. For example, in your $quadratic/cubic$ example, there are two finite zeroes, three finite poles, and a simple zero at infinity. The proof is as follows: If $f(z)=\dfrac{z^2+az+b}{z^3+cz^2+dz+e}$, then $f(1/z)=\dfrac{1+az+bz^2}{1+cz+dz^2+ez^3}z$ which clearly has a simple zero at $z=0$. This proof generalizes easily to any rational function, and thus any meromorphic function on the Riemann sphere.

Answer 2

The number of zeroes is equal to the number of poles (counted with multiplicities) for any meromorphic function on a (compact, connected) Riemann surface.

Taken from these notes on Rieman-Roch theorem.