We had the following statement.
Let $D \subset \mathbb C$ be a domain, $f: D \to \mathbb C$ a continuous function and $\gamma : [\alpha, \beta] \to D$ a contour. Assume that $\int_\gamma f$ depends only on the initial and final points of $\gamma$. Fix a point $a \in D$ and define a function $F_a: D \to \mathbb C$ by \begin{align} F_a(z) := \int_{\gamma_{a,z}} f, \end{align} where $\gamma_{a,z}: [\alpha,\beta] \to D$ is a smooth contour with initial point $a$ and final point $z$. Then it is $F_a'(z) = f(z)$ in $D$.
Now I want to prove:
Let $a,b \in D$ and define $F_a, F_b : D \to \mathbb C$ as in the previous statement. Show that $F_a(z) - F_b(z) = F_a(b)$.
I started by plugging in the definition: \begin{align*} F_a(b) + F_b(z) &= \int_{\gamma_{a,b}} f + \int_{\gamma_{b,z}} f = \int_{\gamma_{a,z}} f = F_a(z). \end{align*}
Problems: I know that there is something wrong. I am not sure if I can assume that $\gamma_{a,b} + \gamma_{b,z} = \gamma_{a,b}$. Furthermore, in my exercise was the hint that $F_b(b) = 0$. Can one give me a suggestion how I can use that, please?
Indeed you cannot assume that. However,
says that doesn't matter. $\gamma_{a,b} + \gamma_{b,z}$ is a path from $a$ to $z$, and since the integral depends only on the initial and end-point of the path by assumption, you have
$$\int_{\gamma_{a,b} + \gamma_{b,z}} f = \int_{\gamma_{a,z}} f = F_a(z).$$
There really isn't anything more to prove.
The hint that $F_b(b) = 0$ was meant for the proof via differentiation. You have
$$\bigl(F_a(z) - F_b(z)\bigr)' = f(z) - f(z) = 0,$$
hence $F_a(z) - F_b(z)$ is constant (since $D$ is a domain), and
$$F_a(b) - F_b(b) = F_a(b) - 0 = F_a(b).$$