# prove an analytic function takes two values the same number of times

Can anyone please help me with this question: Let $G$ in $\mathbb{C}$ be a bounded region and $f$ a function analytic on $G$. Let $E= f(\partial G)$. If a and b are in the same component of $\mathbb{C}$ \ $E$ ; show that $a$ and $b$ are taken the same number of times by $f$:

I really dont know how to even start solving this question. Please help me with it.

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The number of times $f$ takes on the value $a$ is the number of zeros of $f-a$. Do you have a formula --- a contour integral, perhaps? --- that gives you the number of zeros of $f$? – Gerry Myerson Oct 23 '12 at 5:01
but how can I relate that with the number of zeros of f-b? – Danny Oct 23 '12 at 5:13
Well, maybe you have a formula for that, too, and maybe you can use the hypotheses to show that the two formulas must give the same number. Or, maybe not. Worth a try, don't you think? If you don't have any other ideas, you have nothing to lose by checking out this one. – Gerry Myerson Oct 23 '12 at 5:19
Im not really sure how to do that. I really dont know. Can you elaborate more? – Danny Oct 23 '12 at 12:31
Why don't you start by writing down that formula you have for the number of zeros of a function. – Gerry Myerson Oct 23 '12 at 21:31

Consider the argument principle $$\frac{1}{2\pi i}\oint_{\partial G}\frac{f'(z)}{f(z)}\ dz = N-P$$ where $N$ is the number of zeroes and $P$ the number of poles. We know that $f$ is holomorphic inside $G$ so $P=0$. Equivalently, $$\frac{1}{2\pi i}\oint_{\partial G}\frac{f'(z)}{f(z)-a}\ dz = N_a$$ is the number of times $f(z) - a$ takes on $0$ in $G$, i.e. the number of times $f(z)$ takes on the value $a$ in $G$. But this is the same as rewriting the integral as $$\frac{1}{2\pi i}\oint_{f(\partial G)}\frac{1}{t-a}\ dt$$ which is the winding number of $f(\partial G)$ about the point $a$. What can you say about the winding numbers of points in a connected component?

Proof of the fact that the winding number is constant on a connected component.

Theorem: The winding number is locally constant. Let $\gamma: [t_1,\ t_2]\rightarrow U$ be a (piecewise) continuously differentiable curve in an open set $U$. The map $g:\ U\setminus{\mathrm{Im}(\gamma)}\rightarrow \mathbb{Z}$ given by $$g(w) = \mathrm{Wnd}(\gamma,\ w) = \frac{1}{2\pi i}\oint_{\gamma}\frac{dz}{z-w}$$ is constant on each connected component of $U\setminus{\mathrm{Im}(\gamma)}$. The notation $\mathrm{Wnd}(\gamma,\ w)$ denotes the winding number of the curve $\gamma$ about the point $w$.

Proof: The interval $[t_1,\ t_2]$ is closed and bounded and hence compact. Since $\gamma$ is continuous, that means $\mathrm{Im}(\gamma)$ is also compact. Therefore $U\setminus{\mathrm{Im}(\gamma)}$ is open. Choose $r>0$ so that the $D(a,\ 2r)$, the open disk centered at $a$ of radius $2r$ is fully contained in $U\setminus{\mathrm{Im}(\gamma)}$. Choose $w$ in $D(a,\ r)$. Then \begin{aligned}\left|g(w) - g(a)\right| &= \left|\frac{1}{2\pi i}\oint_{\gamma}\frac{dz}{z-w} - \frac{1}{2\pi i}\oint_{\gamma}\frac{dz}{z-a}\right| \\ & =\frac{1}{2\pi}\left|\oint_\gamma \frac{w-a}{(z-a)(z-w)}\ dz\right| \\ & \le \frac{1}{2\pi}\mathrm{length}(\gamma)\sup_{z\in \mathrm{Im}(\gamma)}\frac{|w-a|}{|z-a||z-w|} \\ & \le \frac{1}{2\pi}\mathrm{length}(\gamma)\frac{|w-a|}{2r^2}\end{aligned} It is clear then that $|g(w)-g(a)| \rightarrow 0$ as $|w \rightarrow a|$ and hence $g$ is a continuous function on each connected component of $U\setminus{\mathrm{Im}(\gamma)}$. Since $g$ is continuous, if $V$ is a connected component, then $g(V)$ is also connected. But $g$ is a map to the integers, so that necessarily means that $g$ is constant for each connected component. $\square$

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can I say that the winding number are the same in the connected component? Im not very sure about the justification though.. – Danny Oct 23 '12 at 23:16
There is a rather simple proof I can include if you really want. But it's rather standard. Are you sure you haven't seen a proof somewhere in your course or text? – EuYu Oct 23 '12 at 23:22
no, not at all. I was actually searching that today and for sure we didnt prove that in class. I think I saw something similar online, but Im not able to find it now. – Danny Oct 24 '12 at 0:00
@Danny I've included a proof. – EuYu Oct 24 '12 at 1:48