Consider the following question
If $w \in \mathbb{C}$, $r > 0$, and $\gamma$ is a closed curve which does not meet $D(w, r)$, show that $I(\gamma; w) = I(\gamma; z)$ for every $z \in D(w, r)$.
First I thought that this is just an exercise of decomposing a path into bits on which the winding number stays the same by considering the "interior" of closed segments (i.e. of the loops) and mapping it to the unit disc by the Riemann Mapping theorem and dealing with that.
However, this gets ugly fast and it assumes that $\gamma$ is $C_1$. At this point in the course we have not even talked about functions that are nowhere differentiable and I am not aware how one can use homotopy to "smooth them out" as doing so will probably use the result of this question.
Does anybody have an idea how to do this question for such general curves as described?
Without loss of generality we can assume that $w=0$, this simplifies the calculations a bit. Now fix a point $z \in D(0, r)$. We want to show that $I(\gamma, z) = I(\gamma, 0)$.
For all $\zeta $ in the image of $\gamma$ is $|\zeta| \ge r > |z|$ and therefore, using the geometric series: $$ \frac{1}{\zeta - z} = \frac{1}{\zeta(1 - z/\zeta)} = \sum_{n=0}^\infty \frac{z^n}{\zeta^{n+1}} \, . $$ The series converges uniformly with respect to $\zeta \in \operatorname{Im}(\gamma)$ since $$ \sum_{n=0}^\infty \left|\frac{z^n}{\zeta^{n+1}} \right| \le \sum_{n=0}^\infty \frac{|z|^n}{r^{n+1}} = \frac{1}{r-|z|} $$ so that the order of integration and summation can be interchanged: $$ I(\gamma, z) = \frac{1}{2 \pi i}\int_\gamma \frac{d\zeta}{\zeta - z} = \frac{1}{2 \pi i}\sum_{n=0}^\infty \left(z^n \int_\gamma\frac{d\zeta}{\zeta^{n+1}} \right) \, . $$ But for $n \ge 1$ is $$\int_\gamma\frac{d\zeta}{\zeta^{n+1}} = 0$$ because the integrand has an antiderivative in $\Bbb C \setminus \{ 0 \}$. It follows that $$ I(\gamma, z) = \frac{1}{2 \pi i}\int_\gamma\frac{d\zeta}{\zeta} = I(\gamma, 0) $$ and this finishes the proof.
On a more general level one can argue that the function $$ z \mapsto I(\gamma, z) $$ is continuous and integer-valued in $\Bbb C \setminus \operatorname{Im}(\gamma)$, which implies that $I(\gamma, z)$ is constant on every connected component of $\Bbb C \setminus \operatorname{Im}(\gamma)$. In particular, since $D(w, r)$ is a connected subset of the complement of the image, $I(\gamma, z)$ is constant on $D(w, r)$.