Winding number of composition of curve

Question

Let $f$ be analytic in a "simply connected domain" D $\subseteq \mathbb{C}$ and let $\gamma$ be a closed piecewise differentiable path in $D$. Set $\beta = f \circ \gamma.$ Show that $\frac{1}{2\pi i}\int_{\beta}\frac{dz}{z-c} = I(\beta,c) = 0$ ($I(\beta, w)$ is usually called the winding number of $\beta$ around a point $w$) for all $c \in \mathbb{C}\setminus f(D).$

Here this what I try :

Let $\gamma$ be a closed pdp (piecewise differentiable path) in $D$ parameterized by $[0,1]$. Since $\gamma$ is closed and $f$ is a function, $f(\gamma(0)) = f(\gamma(1))$, that is, $\beta$ is a closed curve. Moreover, $\beta$ is a closed pdp since $f$ is analytic on $D$ and range $\gamma \subseteq D$. Let $c \in \mathbb{C}\setminus f(D).$ Since range $\beta \subseteq f(D)$ and $c \notin f(D),$ $\frac{1}{z-c}$ is defined on $f(D).$

I want to show that $I(\beta, c) = 0$ which will immediately true if $c$ is in unbounded component $C^*$ of $\mathbb{C}\setminus \mbox{range} \ \beta$ (that is, $c$ is outside the region enclosed by the curve $\beta$). And the proof will finish if I can prove that $C^* = \mathbb{C}\setminus \mbox{range} \ \beta$. It is equivalent to say that $f(D)$ has no hole in it (if $f(D)$ has holes, then at least $I(\beta,c) \neq 0$ for all $c$ in the holes). So, finally it is reduced to show that $f(D)$ is simply connected, which I doubt if it is true. I do not think analytic function will always map simply connected domain to simply connected domain. So I am not sure that my idea in solving this problem is suitable.

Answer 1

Usually, $f(D)$ is not simply connected. e.g. $D=\mathbb{C}, f=e^{z}$.

Answer 2

This is an old thread in need of a simple answer: use the Argument Principle.

Select arbitrary $c \in \mathbb C - f\big(D\big)$ and define $g$ as translation by $-c$ composed with $f$ i.e.
$g(z) := f(z) -c$. Now compute winding number $n\big(f\circ \gamma,c\big)$ by application of the Argument Principle

$n\big(f\circ \gamma,c\big)=n\big(g\circ \gamma,0\big)=\sum_{z, g(z)=0}v_g(z)\cdot n\big(\gamma, z\big)=0$
because (i.) $z\in D\implies g(z)\neq 0$ and (ii.) $z\not \in D\implies n\big(\gamma, z\big)=0$ since $D$ is simply connected.