Winding number and zeroes of a holomorphic map

Question

Let $f\in C^\infty(\bar{\mathbb{D}})\cap\mathcal{O}(\mathbb{D})$ (where $\mathbb{D}$ is the open unit-disk in the complex plane and $\mathcal{O}$ denotes holomorphic functions) and suppose $f(z)\neq 0$ for $\vert z\vert =1$, such that the winding number $w(f)=[f\vert_{S_1}]\in \pi_1(\mathbb{C}^\times)=\mathbb{Z}$ is well defined. I have seen suggested somewhere that $$ w(f)=0\quad \Leftrightarrow\quad f(z)\neq 0\text{ for all } z\in \mathbb{D}, $$ but I fail to see how to prove the "$\Rightarrow$" direction. (For "$\Leftarrow"$ note that the absence of zeroes means that the loop $f\vert_{S_1}$ can be filled in with a disk in $\mathbb{C}^\times$ and thus $[f\vert_{S_1}] = 0 \in H_1(\mathbb{C}^\times,\mathbb{Z}) \cong \pi_1(\mathbb{C}^\times)$.)

Question. Suppose the winding number $w(f)=0$, why does $f$ not have any zeroes?

• Note that $w(f)=0$ is equivalent to the existence of a smooth function $g:\bar{\mathbb{D}}\rightarrow \mathbb{C}^\times$ with $g=f$ on $S^1$. It would thus suffice to prove that $h=g^{-1}f$ is free of zeroes. Now $h(z)=1$ for $\vert z \vert =1$, but $h$ is not holomorphic - it only satisfies $(\partial_{\bar z}+g^{-1}\partial_{\bar z}g)h=0$.
• If we would know that $f(S^1)\subset\{\Re >0\}$ (the right half-plane), then $u=\Re f$ would be harmonic and positive on the boundary, such that the maximum principle implies that $u>0$ on $\mathbb{D}$, which means that $f$ cannot have zeroes.
• The zeroes of $f$ are obviously the poles of $f^{-1}$, which leave residues when integrating on $S^1$ - however the might cancel each other, so I don't know whether this is useful.

Answer 1

the idea is that zeroes of holomorphic functions have positive degree so they cannot cancel out so to speak when computing the degree (or if you want the winding number on the boundary); one way to easily prove this in this case (without going through the machinery of homotopy and the argument principle) is to note that the hypothesis implies that $f$ has only finitely many zeroes in the unit disc so assuming that number to be $n \ge 1$ there is a Blaschke product $B$ of degree $n$ st $f=gB$ for some non zero $g$; but then $B$ winds precisely $n$ times around the unit circle (it actually induces a cyclic group of order $n$ action on the unit circle since it has no critical points there and takes each value $n$ times in distinct points); since the winding number of $g$ is $0$ we get that $f$ has winding number $n \ge 1$ and that is a contradiction

Edit later - to show that the result is highly non-trivial and strongly depends on properties of holomorphic functions consider the harmonic polynomial $f(z)=z+ \bar z+iz^2$; it is fairly easy to see that $f$ has zeroes only at $0, \omega_1=\sqrt 2e^{i\pi/4},\omega_2=\sqrt 2e^{3i\pi/4}$, so in particular $f$ has a zero inside the unit disc and no zeroes on the unit circle, but its degree wr the unit disc is zero or if you want the winding number wr the unit circle is zero (to see that show that the image of $f(\mathbb D)$ avoids a ray!) so $0$ is a degree (multiplicity) zero root of $f$ by the generalized argument (or degree) principle for continuos functions with isolated zeroes!

Answer 2

As mentioned in Ted Shifrin's comment and Conrad's answer the result is really just a special case of the argument principle. As I found this viewpoint most illuminating, I will briefly spell out some details. We have $$ \mathrm{w}(f) = \left\langle f\vert_{S^1}, \frac1{2\pi i}\frac{d w}{ w}\right\rangle_{H_1(\mathbb{C}^\times,\mathbb{Z})\times H^1(\mathbb{C}^\times,\mathbb{Z})} = \frac1{2\pi i}\int_{S^1} f^*\left(\frac{d w}{ w}\right) = \int_{S^1}\frac{f'(z)}{f(z)} d z = \#\text{ of zeroes}, $$ where the first three equalities are definitions or simple computations and the last inequality is the argument principle as stated on wikipedia.

Ted's comment refers to the fact that $f$ in principle needs to be holomorphic in a neighbourhood of $\bar {\mathbb{D}}$ to apply the argument principle. To see that the stated reguality suffices, one could indeed just run the whole argument for $f\vert_{\mathbb{D}(1-\epsilon)}$, the restriction to a slightly smaller disk, such that none of the (isolated, finite) zeroes are contained in the missed annulus. The boundary restrictions will then be homotopic and thus share winding numbers.