So, I've read a proof for path-independence in the case of the existence of an anti-derivative of a continous (in the domain $D$) complex function $f$, being the anti-derivative $F$ an analytic funtion in $D$. The proof starts:

In the case of the existence of $F$ $$ F'(z)dz = u_xdx -v_xdy + i(u_xdy + v_xdx) = u_xdx + u_ydy + i(v_xdx + v_ydy) $$

Since $du = u_xdx + v_xdx$ and the same goes for $dv$ we have $F'(z)dz = du + idv$.

Now we can integrate this

$$ \int_{\Gamma} F'z(dz) = \int_{\Gamma} du + i\int_{\Gamma} dv = u(x_2, y_2) - u(x_1, y_1) + i(v(x_2, y_2) - v(x_1, y_1)) = F(z_2) - F(z_1) $$

First of all, I feel like this only proves the necessity of the path independency, and not the sufficiency, for which I can't find the proof.

Now, the object of my question was, if this proof is correct, isn't it enough to prove that for an analytical function $f$ in $D$ then

$$ \oint_C f(z)dz = 0 $$


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