Winding Numbers of unit circle under $f(z)=\frac{z}{e^z-1}$

Question

I've been working on the following question, and am stumped by part c.

I know how to find the winding number if I have the graph, and I tried using the special case of the Cauchy Integral formula $$v(C,0)=\frac{1}{2\pi i}\oint_C \frac{1}{z}dz,$$ but I don't know how to first take the map into account. Would I try to compute $\frac{1}{2\pi i}\int_0^{2\pi} \frac{1}{f(\Gamma(t))}dt$? Is this even possible?

Thanks!

Answer 1

The function $f$ has no zeros inside the circle $|z|=10$ and two poles (at $\pm 2\pi i$) inside the circle $|z|=10$. Thus, the index is $\nu(f(\Gamma),0)=0-2=-2$ according to the principle of argument.

Update The OP requested a plot (part (d) of the exercise). Here you can see what is going on. In the left picture I plotted the image $f(\gamma(t))$ for $0<t<2\pi$. However, it is difficult to see what happens close to the origin. Thus, in the right plot I plotted in the interval $-1<t<1$ (note that this is the same as plotting from $0 to 1$ and then also $\pi-1$ to $\pi$). This might not be the best way of showing the situation, but I won't spend more energy on it for the moment.

Answer 2

To express the integral in explicit terms, let $$\gamma(t) = 10 e^{it}.$$ Then compute $$ \frac{1}{2\pi i} \int_0^{2\pi} \frac{1}{f(\gamma(t))} \gamma'(t) dt. $$

This may or may not be workable in a nice way algebraically. One would more usually use @mickp's approach to actually evaluate such an integral.