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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.

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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.

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  • $\begingroup$ "Since, it is on complex plain, then all values of an arbitrary function f1 on both γ1,γ2 of x-axis is same but the complex component is not" This sentence doesn't make sense at all. Can you explain more clearly what you mean. I think your mistake is here. $\endgroup$ Jul 22, 2021 at 14:51
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    $\begingroup$ Welcome to MSE. It's hard to understand your description of $\gamma_1$ and $\gamma_2$. Can you draw a picture? Also, “arbitrary” function is probably too broad. Do you mean holomorphic? $\endgroup$ Jul 22, 2021 at 14:52
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    $\begingroup$ OK, I'm still not understanding the situation you're describing. Please give an explicit example of $\gamma_1$, $\gamma_2$ and holomorphic $f$ such that $\int_{\gamma_1 - \gamma_2} f\,dz = bi$ for some $b \neq 0$. $\endgroup$ Jul 22, 2021 at 15:30
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    $\begingroup$ "(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 $f(b)$)" This statement is false. consider $f(z)=z^2$. Then, the real part of $f(a)$ is different from the real part of $f(b)$. $\endgroup$ Jul 22, 2021 at 18:48
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    $\begingroup$ Rather than speculate about some arbitrary $f$, why don't you pick out a concrete example like $f(z) = z^2$ and work it out fully. You'll see that there is no contradiction anywhere. Then you can start to think about how it works in general. $\endgroup$ Jul 22, 2021 at 18:51

1 Answer 1

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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.

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