Let the curve $\gamma$ on $z_1$ to $z_2$ in complex plain
Let the function, $f(z) : D(\subset \mathbb{C}) \to \mathbb{C}$
It is definitely true that
'$\exists F(z)$ $s.t. F(z)$ is antiderivative of the $f$' $\iff$ '$ \int _ \gamma f(z) dz = F(z_1) - F(z_2)$'
There is a someone (1) who claim $f$ is continuous on $D$ $ \iff$ $\exists F(z)$ ' And plus he said if The D is simple connected, $\exists F(z)$ $ \iff$ $f$ is analyitic on $D$
But the other one (2) clam $\int_{\vert z \vert =1}(1-{1 \over {z^2}}) e^{z+ {1 \over z}} = 0 $
since the closed curve $\vert z \vert =1$ so $z_1 = z_2$, Hence $F(z_1) - F(z_2) $=0
I'm really confused who is the correct. Because the (2) case $f(z)= (1-{1 \over {z^2}}) e^{z+ {1 \over z}} $ is not continuous at $z=0$ so it is the contradict with the (1)'s claim.
So here is the question
First, Which one is correct? And What is the reason for the (1) and (2) is right or not?
Second, what the condition need for saying the $\exists F(z)$ for $f(z) : D \to \mathbb{C}$ ?
Any help would be appreciated, thanks.
Most of the statements you have made are imprecise and lacking crucial assumptions. For example, you cannot fix the path $\gamma $ in advance for the equivalence in 1).
What is true is: Every analytic function in an open connected set $D$ has an anti-derivative iff $D$ is simply connected. When this condition is satisfied an anti-derivative $F$ can be defined by fixing a point $c$ and defining $F(z)$ as $\int_{\gamma} f(\zeta)d\zeta$ where $\gamma$ is any path in $D$ from $c$ to $z$. The integral does not depend on the path.