I don't understand the path independence in the Cauchy integral theorem which $0$ for a closed curve.
For example, consider Cauchy integral theorem on the closed curve $C$, which is created by gluing $2$ non-identical or asymmetrical looking curves $C_1, C_2$. Since, it is on complex plain, then some values (images) of an arbitrary holomorphic function $f$ on both $C_1, C_2$ alog real and imaginary axis might be same (for example $a=4+1i$ is on $C_1$, and $b=4+4i$ on $C_2$, so the real part of $f(a)$ = real part of $f(b)$), but it os not always true, thus clockwise sum, i.e. integration will not be equal to anticlockwise sum.
so while integrating on the closed curve** $C$, all the real x-components will be cancelled out (since $C_1, C_2$ are clockwise anti-clock-wise curve, respectively), but the y-axis or the complex component will not be a $0$!
What am I missing?
I am not saying Cauchy integral theorem is wrong, I am just trying to understand as newbie. I have seen other posts ( like, 1, 2,, 3) on this forum, but didn't help.
EDIT: Consider a situation, where all conditions for Cauchy integral theorem are met.
If you know about Green's theorem, here's a quick way to relate Cauchy's Integral Theorem and the Cauchy-Riemann equations. Define a holomorphic function $f(z)=u(x,y)+iv(x,y)$, then the integral over a loop $C$ is
$$\oint_C f dz = \oint (u + i v) d(x+iy) = \oint (u dx - v dy) + i \oint ( v dx + u dy )$$
Using Green's theorem, we can rewrite this as
$$\oint_C f dz = \iint_S \left(-\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right) dxdy + i \iint_S \left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) dxdy $$
where $S$ is the domain enclosed by the loop $C$. And since the function is holomorphic, both of these integrands are zero by the Cauchy-Riemann equations.
So, the reason Cauchy's integral theorem is true, is because of the strong restriction that the requirement of holomorphicity imposes on a function.