The Fundamental theorem of calculus (FTC) of complex function.
Consider:
$$\int_{\gamma } f(z) \ dz \ ,$$
$\gamma$ is a smooth curve included in a domain $G$. Does $f(z)$ continuous on $G$, imply $f$ has a anti-derivative? (In this case, it holds true for a real-valued function).
If not, then for the Fundamental Theorem of Integral Calculus for Line Integrals of a complex function: Suppose $G$ is an open subset of the plane with $p$ and $q$ (not necessarily distinct) points of $G$.
Suppose $\gamma$ is a smooth curve in $G$ from $p$ to $q$. Then for any function $F$ analytic on $G$:
$$\int_{\gamma}F^{'}(z)dz=F(q)-F(p) \ .$$
But how do I explain the following process?
$$ \int_{\gamma}f(z) \ dz = \int^{b}_{a}f(z(t)) \ z^{'}(t) \ dt , \\ = \int^{b}_{a}dW(t) \\ =W(b)-W(a) $$
$f(z(t)) \ z^{'}(t)$ is continuous, according to FTC of a real-valued function. It has anti-derivative. It just uses "$f(z)$ is continuous on $G$". The analytic function $F(z)$ is not be needed.