Let $f:C- {0}$ $\longrightarrow C $ be analytic function such that for any closed contour $\gamma $ the line integral over $\gamma $ is $0$. Then which of the following are true?
(1)$ f $ has removal singularity at $0$
(2)$f $ can't have essential singularity at $0$
(3) There is a holomorphic function $F:C-{0}$$\longrightarrow C $ such that $F'(z)=f(z) $ for all $z\in C-{0} $
(4) There is a holomorphic function $F:C \longrightarrow C $ such that $F' (z)=f(z) $ for all $z$$\in C $
My try: I am unable to do rigorously. If $f $ has a pole at $0$ then the vector field $V= u-iv ; $where$ f=u+iv $ seems like all the vectors near to $0$ scales up. Physically say , I have nice fluid flow on some surface except at one hole through which external fluid enters and it's flow increases as the fluid is near to that point. So to divergence is not $0$ and hence the line integral along the closed contour is not $0$. So the pole can't be there at $0$.
If $f $ has essential singularity at $0$ then $lim f $ as $z \longrightarrow 0$ are different (in fact it takes any value of C) So it seems that there is a chaosnees in the vector filed $u-iv $ near to $0$ , So the Curl doesn't vanishes. So line integral $f $ on closed contour doesn't vanishes. So $f $ can have at most removal singularity at $0$. So option 1 and 2 are true.
(3) is true as $f $ is diffentiable implies $f'$ is so. (4)is true As $f $ has removal singularity so we can extend it to an holomorphic function to $C $ and then follow the argument of (3).
Am I correct? I need more rigorous answer or hint to convince myself.
(3) is indeed true, but your argument is wrong. You have to prove that $f$ has an antiderivative $F$ if $\int_\gamma f(z) \, dz = 0$ for any closed contour $\gamma$ in $\Bbb C \setminus \{ 0 \}$.
A correct argument would be to define $F$ in $\Bbb C \setminus \{ 0 \}$ as $$ F(z) = \int_{z_0}^z f(w) \, dw $$ where $z_0$ is fixed, and the integral is taken along any curve in $\Bbb C \setminus \{ 0 \}$ from $z_0$ to $z$. Then show that this definition is independent from the choice of the curve (so that $F$ is well-defined), and that $F' = f$.
The argument is in the wrong direction. If (4) is satisfied then $\int_\gamma f(z) \, dz = 0$ for any closed contour $\gamma$ in $\Bbb C \setminus \{ 0 \}$, but not the other way around.
In fact (1), (2) and (4) are wrong. As a counterexample take $f(z) = F'(z)$ where $F$ is any holomorphic function in $\Bbb C \setminus \{ 0 \}$ with an essential singularity at $z=0$, for example $$ f(z) = \left( e^{1/z}\right)' = -\frac{1}{z^2} e^{1/z} \, . $$
Then $\int_\gamma f(z) \, dz = 0$ for any closed contour $\gamma$ in $\Bbb C \setminus \{ 0 \}$ because $f$ is the derivative of $F$, but $f$ has an essential singularity at $z=0$.