# If a function maps the contour of an open set to a closed curve in the range, then it maps to the region bounded by the curve?

Imagine we have an analytic function on $\mathbb{C}$. Then if the image of the unit circle under the function is a closed curve, say a circle of radius $r$, then is it immediately true that every point in the open unit disk is mapped to one point in the circle?

Put more mathematically: Let $C$ denote the unit circle and $\mathbb{D}$ denote the open unit disk. Then if $$f(C)=\{z\in\mathbb{C}: |z|= r \}$$ then $$f(\mathbb{D})=\{z\in\mathbb{C}: |z|< r\}$$

I saw a very convoluted example of this (well at least I think so, hence the question), which then stated that since the image of the contour of the open set (in the case above the unit disk) is a closed curve, the winding number of every point inside the closed curve is $1$ and the winding number for every point outside the closed curve is $0$. This I understand.

However it then states immediately that as a result the function takes every value in the region bounded by the closed curve exactly once and no value outside of the region (in our case the circle of radius $r$). The example mentioned the argument principle which I have researched and cannot establish a link to this example. If anyone could shed light on this, I'd be grateful.

• FYI: The image of a differentiable curve is a differentiable curve if $f$ is analytic, so the image of a circle can't be a semi-circle. Mar 24, 2016 at 2:03
• Ah, I was just using it for an example. Is there a better one to use? Maybe just change the semi-circle to a disk of radius $r$. Mar 24, 2016 at 2:05

It's not true that the winding number is 1 or 0. It is an integer. For example, $z^2$ maps the unit circle to a curve that wraps around the origin twice, which gives winding number of $2$. This corresponds to a double $0$ at $z=0$. More generally, suppose $f$ does not take on the value $z$ for any $|w| = 1$, but that it does take on the value of $z$ at $w_1,w_2,\cdots,w_n$ inside this circle. Then there are positive integers $r_1,r_2,\cdots,r_n$ such that $$f(w)-z = (w-w_1)^{r_1}(w-w_2)^{r_2}\cdots(w-w_n)^{r_n}g(w)$$ where $g$ is non-vanishing in the closed unit disk $\overline{D}$. The curve $\gamma(\theta) = e^{i\theta}$ maps $[0,2\pi]$ to the closed unit disk, and is positively oriented. The image of this curve $\gamma$ under $f$ is a closed curve with winding number about $z$ that is given by the following (and is reduced to something manageable by applying the logarithmic derivative): \begin{align} w_{f\circ \gamma}(z) & =\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{d_{\theta}f(e^{i\theta})}{f(e^{i\theta})-z} \\ & = \frac{1}{2\pi i}\int_{\gamma}\frac{\frac{d}{dw}(f(w)-z)}{f(w)-z}dw \\ & = \frac{1}{2\pi i}\int_{\gamma}\left[\frac{r_1}{w-w_1}+\cdots+\frac{r_n}{w-w_n}+\frac{g'(w)}{g(w)}\right]dw \\ & = r_1+r_2+\cdots+r_n. \end{align} The integral of $g'/g$ vanishes because $g$ is non-vanishing on $\overline{D}$, which makes it holomorphic. So the winding number $\omega_{f\circ\gamma}(z)$ of $f\circ\gamma$ around $z$ is the sum of the orders of the zeros of $h(w)=f(w)-z$ for $w\in D$, counted according to multiplicity. If $f$ does not take on the value of $z$ in $\overline{D}$, then the winding number of $f\circ \gamma$ around $z$ is $0$ because all you get for the integral is the $g'/g$ term.
As an example, the function $z^2$ takes on every value in the open disk $D$ twice in the $D$, corresponding to the fact that the image of the unit circle under $z^2$ winds around everything in the open unit disk twice, and winds around everything outside the closed disk $0$ times.
• Wow, very good answer, very clear. For your example for $z^2$ does this extend to any function? i.e. if the image of a contour winds round everything in a given region $n$ times, then the image of the interior of the contour under the function takes every value in the given region $n$ times, and nothing outside of that? i.e. does this work for any function? Mar 24, 2016 at 3:32
• @George1811 : For $z^2$, $0$ is a double root. But $1/2$ has two roots in $|z| < 1$--namely $1/\sqrt{2}$, $-1/\sqrt{2}$, and each of these roots is of order $1$. That's representative of the type of thing that can happen. The image of the unit circle under $z^{n}$ circles $n$ times on the unit circle. Nothing outside the unit circle is in the imagine under $z^n$ of the unit disk, and everything inside is taken on as value with total multiplicity $n$. Mar 24, 2016 at 6:21
Let $|\zeta| < r$ and assume that $f(z)\neq\zeta$ for all $z\in\mathbb D$. Then $\log(f(z)-\zeta)$ is defined on $\mathbb D$ and is holomorphic. Hence, $$0 = \int_C \log(f(z)-\zeta)f'(z)\,dz = \int_0^{2\pi}\log(f(e^{it})-\zeta)ie^{it}f'(e^{it})\,dt = \int_\gamma \log(w-\zeta)\,dw,$$ where $\gamma$ is $\{z : |z|=r\}$ (possibly run through several times). But this is obviously a contradiction.
• That zero must be in the image of $f$. Mar 24, 2016 at 3:50