For real integration, to show the line integral is independent it's enough to show that the integrand is conservative field and show that the curl is zero.
But what about complex line integral? What is their difference? Are they the same? It's just that the complex line integral is line integral in the complex plane?
Actually i'm referring to this question:
Path independence of complex functions
What i've learned so far from that link, if the function has primitive function, then the complex line integral is independent of the path, namely joining $a$ and $b$. (Maybe i got a big misunderstanding about it, please verify this, thanks)
But how does it related with independent of path? Independent path means the integral on simple closed curve is $0$ right? (Based on Cauchy's Theorem) What i got from that link is the function just has the anti-derivative, that's all. And we can apply Fundamental Theorem of Calculus.
Where is the relation between the function has
Please explain to me with easy language.
Thanks.
Anyway, i'm new in complex variables course.
First the line integral of a complex function $f = u+iv\colon U\subset \mathbb{C} \rightarrow \mathbb{C}$ along a path $\gamma$ is just (heuristically) $\int_\gamma (u+iv)(dx+idy)$ and you can separate real and imaginary parts going back to the real line integrals. The precise definition is that for $f$ integrable and $\gamma : [a,b] \rightarrow U$ picecewise smooth $$ \int_\gamma f(z)dz = \int_a^b f(\gamma(t)) \gamma'(t) dt. $$
Suppose that $f$ has a (holomorphic) primitive $F\colon U \rightarrow \mathbb{C}$ i.e. $F'(z) = f(z)$ for every $z\in U$. One can use the chain rule and the Fundamental Theorem of Calculus (for real functions, of course) and some simple manipulation to prove that for $\gamma \colon [0,1]\rightarrow U$ $$ \int_\gamma f(z)dz = F(\gamma(1))-F(\gamma(0)) $$ It is really simple but a good exercise. You just need to prove that $(F\circ \gamma)'(t) = f(\gamma(t))\gamma'(t)$ .
On the other direction suppose $U$ is a star-shaped open set i.e. there exists $z_0$ such that every $z\in U$ can be joined to $z_0$ by a straight line segment $L(z)$ inside $U$. Then one may define $$ F(z) = \int_{L(z)} f(\zeta) d\zeta $$ It follows from Goursat's theorem that if $f$ is holomorphic then $F$ is a holomorphic primitive for $f$. This is a particular case of the Poincaré Lemma.