Question : Let $\gamma:[a, b] \rightarrow \mathbb C $ be a differentiable curve and $z_0 \in \mathbb C$ such that $\gamma(t)\neq z_0$ for all $t \in [a,b].$ Let $g(t) := \int_a^t \frac{\gamma'(s)}{\gamma(s)-z_0}ds.$ Show that the function $e^{-g(t)}(\gamma(t)-z_0)$ is constant and show that $e^{g(t)}= \frac{\gamma(t)-z_0}{\gamma(a)-z_0}.$
I first calculated $g(t)=\ln(\frac{\gamma(t) - z_0}{\gamma(a)-z_0})$ by substituting $\gamma(s) - z_0 = u$ in given definition of $g(t)$ and integrating , then $e^{-g(t)}(\gamma(t)-z_0) = \gamma(a)-z_0.$
I'm asking this because I'm unsure if integration is allowed in any such question, or if I should differentiate $e^{-g(t)}(\gamma(t)-z_0)$ and show it to be $0$.
I would like to know in general when one is allowed to integrate complex functions.
You cannot say that $g(t)=\ln\left(\frac{\gamma(t)-z_0}{\gamma(a)-z_0}\right)$. What is $\ln$? Every non zero complex number has infinitely many logarithms.
However, it follows from the definition of $g$ that $g'(t)=\frac{\gamma'(t)}{\gamma(t)-z_0}$. Therefore\begin{align}\left(e^{-g(t)}\bigl(\gamma(t)-z_0\bigr)\right)'&=-g'(t)e^{-g(t)}\bigl(\gamma(t)-z_0\bigr)+e^{-g(t)}\gamma'(t)\\&=0.\end{align}Therefore, $e^{-g(t)}(\gamma(t)-z_0)$ is constant and so\begin{align}e^{-g(t)}\bigl(\gamma(t)-z_0\bigr)&=e^{-g(a)}\bigl(\gamma(a)-z_0\bigr)\\&=\gamma(a)-z_0.\end{align}