The value of the itegral $\int_{\gamma} \dfrac{dz}{z-a}$ is a multiple of $2\pi i$

Question

I am reading Ahlfors' proof of the lemma:

Lemma If the piecewise differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral $$\int_{\gamma} \dfrac{dz}{z-a}$$ is a multiple of $2\pi i$

I have doubts on some steps of his demonstration which I would like to clear up.

First, Alfohrs' proof:

Proof If the equation $\gamma$ is $z=z(t)$, $\alpha\leq t \leq \beta$, let us consider the function $$h(t)=\int_{\alpha}^t \dfrac{z'(t)}{z(t)-a}dt.$$

It is defined and continuous on the closed interval $[\alpha,\beta]$ and it has the derivative $$h'(t)=\dfrac{z'(t)}{z(t)-a}$$

whenever $z'(t)$ is continuous. From this equation it follows that the derivative of $e^{-h(t)}(z(t)-a)$ vanishes except perhaps at a finite number of points, and since this function is continuous it must reduce to a constant.

We have thus $$e^{h(t)}=\dfrac{z(t)-a}{z(\alpha)-a}.$$

Since $z(\beta)=z(\alpha)$ we obtain $e^{h(\beta)}=1$, and therefore $h(\beta)$ must be a multiple of $2\pi i$. This proves the lemma.

My doubts

My two doubts are:

1) How is it that one deduces that $(e^{-h(t)}(z(t)-a))'$ vanishes except perhaps at a finite number of points? I've calculated its derivative:$$(e^{-h(t)}(z(t)-a))'=-h(t)e^{-h(t)}(-h'(t))(z(t)-a)+e^{-h(t)}z'(t)$$

The right member of the equality being equal to $$e^{-h(t)}z'(t)\int_{\alpha}^t\dfrac{z'(t)}{z(t)-a}dt+e^{-h(t)}z'(t)$$

I can't see how this expression "vanishes except perhaps at a finite number of points".

2) Why $e^{h(\beta)}=1 \implies h(\beta)$ must be a multiple of $2\pi i$?

I would be grateful if someone could explain these two parts of the proof, thanks in advance.

Answer 1

The derivative is $$ - h'(t){e^{ - h(t)}}(z(t) - a) + {e^{ - h(t)}}z'(t)$$

which you can write as $$e^{-h(t)}(z'(t)-h'(t)(z(t)-a))$$

But you know that $$h'(t)=\frac{z'(t)}{z(t)-a}$$ whenever $z'(t)$ is continuous. So...?

It is also a fact that $e^z=1$ iff $z=2\pi i k$. If you don't know this, prove it! Set $z=h(\beta)$.

Answer 2

If $z=x+iy$ and $x,y\in\mathbb R$, then $$ e^z = e^{x+iy} = e^x(\cos y + i\sin y). $$ This can be equal to $1$ only if its absolute value is $1$. The absolute value of $\cos y + i\sin y$ is $1$ regardless of which real number $y$ is, so the absolute value is that of $e^x$, and that is $e^x$ itself. Hence $x$ must be $0$. In order that $\cos y+i\sin y$ be $1$, it is necessary that $\cos y$ be $1$ and $\sin y$ be $0$. That happens only if $y$ is an integer multiple of $2\pi$.

One thing to remember in the other part is that $e^{h(t)}$ never vanishes.