Suppose that $f : G \longrightarrow \mathbb C$ is analytic in $G$, where $G$ is a region. Let $\gamma$ be a closed contour in $G$. Let $G' = G \setminus \{\gamma \}$. For each $z \in G'$ define a map $g$ by $g(z) = n(\gamma ; z)$ for all $z \in G'$, where $n(\gamma ; z)$ represents the winding number or index of $z$ w.r.t. the closed contour $\gamma$.
Now my question is $:$
Is $g$ differentiable in $G'$?
It has been proved in J.B. Conway's ''Function of One Complex Variable'' that $g$ is continuous in $G'$. Though it doesn't say a single word about $g$'s differentiability there.
What I have done is that if I take a $z \in G'$. Then either it is inside $\gamma$ or ouside $\gamma$. WLOG let us assume that it is inside $\{\gamma \}$ then $\exists$ suitable $r>0$ such that $B(z;r)$ is still inside $\{\gamma \}$. Take $h \in \mathbb C$ in such a way that $|h| < \delta < r$. Then for $0<|h|<\delta$ we have $\frac {g(z+h) - g(z)} {h} =0.$ $\implies$ $g'(z) = 0$. Similarly if $z$ is outside $\{\gamma \}$ then also $g'(z) = 0$. So $g' \equiv 0$ in $G'$.
Is it correct or not? Please check it.
Thank you in advance.
EDIT $:$
However Conway used this concept to find Cauchy's theorem for higher order derivatives but he doesn't prove it anywhere in his book.
Please see Theorem 5.7 of page no. 85 and Theorem 5.8 of page no. 86 in the second edition of his book.
It does not matter if $z$ is "inside" or "outside" $\gamma$. All that matters is that $z \in G'$, so that it is in some connected component of $G'$.
You make the true assertion that $\frac{g(z+h) - g(z)}{h} = 0 \implies g'(z) = 0$. But you don't justify $g(z+h) - g(z) = 0$.
Are you sure your argument need be any more complicated than "The winding number of $\gamma$ is constant for all points in any connected component of $G'$."