Argument Principle - from J.Bak and D.Newman textbook

Question

The next example is proposed in the mentioned book, in order to light up the origin for the name of the theorem: $$ \frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)} \mathrm{d}z= \frac{\log f(z(1))-\log(f(z(0)))}{2\pi i} $$ where the parametrisation of $\gamma$ is given by $z(t)$.
Here is what confuses me: since the curve is closed, $z(1)=z(0)$. Doesn't this imply that the numerator is $0$?

Answer 1

It doesn't, because the branch of the complex logarithm when evaluated at $f(z(0))$ and $f(z(1))$ may differ. This can be seen through the following computation. One has that $$\log(z)=\log |z|+i\arg(z).$$ Hence, $$\begin{align*}\frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)}~dz &= \frac{\log(f(z(1)))-\log(f(z(0)))}{2\pi i}\\&=\frac{\log|f(z(1))|-\log|f(z(0))|+i\arg(f(z(1)))-i\arg(f(z(0)))}{2\pi i}\\&=\frac{i\arg(f(z(1)))-i\arg(f(z(0)))}{2\pi i}\\&=\frac{1}{2\pi}\Delta_{\gamma} \arg(f(z)), \end{align*}$$ which is not necessarily zero (the last quantity tells you the winding number of the curve $f\circ \gamma$ around the origin).