## Franklin's Notes

### Counting zeros of holomorphic functions

Suppose $f\ne 0$ is holomorphic in some disk $\Delta$. Given some simple closed curve $\gamma$ oriented counterclockwise, we may want to count the number of zeros of $f$ lying inside of the region determined by $\gamma$. We may immediately say that there are at most finitely many zeros, since $\gamma$ is contained in some closed disk $\Delta'\subset \Delta$ and $\Delta'$ must be compact, meaning that any infinite set of zeros would have to have a limit point - but this contradicts that the zeros of an analytic function must be isolated, as proven in Proposition 1 here . Knowing that there are finitely many zeros, we might consider the problem of determining exactly how many there are.

If we consider its logarithmic derivative $f'/f$, this function is meromorphic with poles precisely at the zeros of $f$. Further, since we know $f$ has at most finitely many zeros, we may write $f$ in the form where $\zeta_1,\cdots,\zeta_n$ are the zeros of $f$ with multiplicity, and $g(z)$ is a function with no zeros in the disk $\Delta$. This means that the logarithmic derivative of $f$ can be written in the form where $g'/g$ is holomorphic, since $g$ has no zeros. Hence, we have that and since $g'/g$ is holomorphic, by Cauchy's Theorem on disks , we have However, notice that the former integral can also be obtained by performing a change of variables $w=f(z)$ on the integral for some curve $\Gamma$, so that we obtain Certainly there is a neighborhood $N$ of $0$ which has no intersection with $\Gamma$, so that $\mathrm{ind}(\Gamma,a)=\mathrm{ind}(\Gamma,0)$ for all $a$ in this neighborhood. But this means that for all $a\in N$, we have and therefore, by applying the substitution $w=f(z)$, we have and hence the sum of the indices of $\gamma$ about the roots of $f$ equals the sum of the indices of $\gamma$ about the points where $f$ equals $a$. In particular, if $\gamma$ is a simple closed curve oriented counterclockwise, these integrals count the number of solutions (with multiplicity) to the equations $f(z)=0$ and $f(z)=a$ in the region determined by $\gamma$, so for all $a\in N$, these counts must be equal.

This gives the following theorem:

Proposition 1. If $f(z)$ is analytic in a disk $\Omega$ and has a zero of order $n$ at $z=z_0\in\Omega$, and no other zeros in $B_\epsilon(z_0)\subset \Omega$ for some $\epsilon > 0$, then there is a ball $B_\delta(0)$ such that for all $a\in B_\delta(0)$, the equation $f(z)=a$ has exactly $n$ solutions (with multiplicity).

Less formally stated, this means that if a holomorphic function takes a value at a certain point, then in a sufficiently small neighborhood of that value, $f$ takes on all other values in that neighborhood an equal number of times.