Thanks again for Martin's new extension
What a great extension! But I still need to check $\int_{\alpha}^{\beta}w(s)ds=\int_{\gamma}\frac{1}{z-a}dz$ . The integrands are different only at "several" points. Done! Wow, I'm extremely grateful for his help. Ahlfors' Complex Analysis is a masterpiece without doubt. But sometimes it is not friendly to self-learners. Maybe we can write an elaborated notes which will benefit all the future self-learners.
Thanks for Martin's answer. But it leads to some more essential questions.
According to Martin's answer, I get:
$h(t)=\int_{t_{k-1}}^{t}\frac{z^{\prime}(t)}{z(t)-a}dt, t\in[t_{k-1},t_{k}), k=1,2,3,\cdots,n$ and $h(t_{n})=\int_{t_{n-1}}^{t_{n}}\frac{z^{\prime}(t)}{z(t)-a}dt$.
$h(t)$ is discontinuous at $t_{1},t_{2},\cdots,t_{n-1}$.
We have $h^{\prime}(t)=z^{\prime}(t)/(z(t)-a), t\in(t_{k-1},t_{k}), k=1,2,\cdots,n$.
$h^{\prime}(t)$ is not defined at $t_{1},t_{2},\cdots,t_{n-1}$.
But Ahlfors states “since this function(this function means $e^{-h(t)}(z(t)-a)) $ is continous (on $[t_0,t_n]$) it must reduce to a constant”. I think this statement is not true since $h(t)$ is discontinuous at $t_{1},t_{2},\cdots,t_{n-1}$ and $h^{\prime}(t)$ is not defined at $t_{1},t_{2},\cdots,t_{n-1}$! I'm confused!
My orignal question: mistakes in proving $\int_{\gamma}\frac{dz}{z-a}=2k\pi i$ in Ahlfors' Complex Analysis
On the third edition of Ahlfors' Complex Analysis, page 115 Lemma 1 it states: If the piecewise differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral $\int_{\gamma}\frac{dz}{z-a}$ is a multiple of $2\pi i$.
Then, he tries to prove this lemma. He says: If the equation of $\gamma$ is $z=z(t)$, $\alpha\leq t\leq \beta$, let us consider the function $h(t)=\int_\alpha^t \frac{z^\prime (t)}{z(t)-a}dt$. It is defined and continuous on the closed interval $[\alpha,\beta]$, ...
Here are my questions:
First, I think, in the lemma, the condition "differentiable" is not strong enough. $\gamma$ should be "continuously differentiable".
Second, I think $z^{\prime}(t)$ doesn't even exist at some $t\in[\alpha,\beta ]$ (as a resutl, $h(t)$ is not well defined), since $\gamma$ is "piecewise" continuously differentiable. For example, let $\gamma$ be line segments from the point $(0,0)$ to $(1,0)$, from $(1,0)$ to $(1,1)$, from $(1,1)$ to $(0,1)$, and then goes back to the point $(0,0)$. We set $z(t)=x(t)+iy(t), 0\leq t\leq 4$ for the curve $\gamma$. Then, we set $x(t)=t, 0\leq t\leq1$ and $x(t)=1,1\leq t \leq 2$. $x^\prime(t)$ doesn't exist when $t=1$. Finally, $z^\prime(t)$ doesn't exist when $t=1$.
I show you the whole context below. I don't know how to modify Ahlfors' proof to fix this "mistake".