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,
Best Answer
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.