Ahlfors' lemma of winding number On page 115 of Ahlfors Complex Analysis, he defines 
$$h(t) = \int_\alpha^t \frac{z'(t)}{z(t)-a} dt$$
Later on the page he states that $e^{h(t)} = \frac{z(t) - a}{z(\alpha)-a}$, but I don't see how to arrive at that. Is there a result outside of complex analysis this relies on or am I just not seeing the algebra of it?
 A: Does he really use $t$ for both a limit of integration and the variable of integration?
I'll rewrite as
$$h(t)=\int_\alpha^t\frac{z'(u)}{z(u)-a}\,du.$$
Then
$$h'(t)=\frac{z'(t)}{z(t)-a}.$$
Define
$$H(t)=\exp(-h(t))\frac{z(t)-a}{z(\alpha)-a}.$$
Then $H(\alpha)=1$ and
\begin{align}
H'(t)&=-h'(t)\exp(-h(t))\frac{z(t)-a}{z(\alpha)-a}
+\exp(-h(t))\frac{z'(t)}{z(\alpha)-a}\\
&=-\exp(-h(t))\frac{z'(t)}{z(\alpha)-a}
+\exp(-h(t))\frac{z'(t)}{z(\alpha)-a}=0.
\end{align}
Therefore $H(t)$ is constant: $H(t)=H(\alpha)=1$.
A: I see it like that
$$
h'(t)=\displaystyle\frac{z'(t)}{z(t)-a}\,\,\Rightarrow\,\,
(z(t)-a)\cdot h'(z)=z'(t)
$$
therefore
$$
\begin{array}{rcl}
(z(t)-a)\cdot h'(z)\cdot e^{h(t)}&=&z'(t)\cdot e^{h(t)}\\
(z(t)-a)\cdot h'(z)\cdot e^{h(t)}-z'(t)\cdot e^{h(t)}&=&0\\
(z(t)-a)\cdot\left( e^{h(t)}\right)'-(z(t)-a)'\cdot e^{h(t)}&=&0\\
\displaystyle\left(\frac{e^{h(t)}}{z(t)-a}\right)'&=&0\\
\displaystyle\frac{e^{h(t)}}{z(t)-a}&=&k\\
e^{h(t)}&=&k\cdot (z(t)-a)\\
\end{array}
$$
Given that $h(\alpha)=0$, then $k=\displaystyle\frac{1}{z(\alpha)-a}$. Hence
$$e^{h(t)}=\frac{z(t)-a}{z(\alpha)-a}.$$
