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I have trouble understanding the definition of a line integral over $\mathbb C$. If we take a line $\gamma$ that is $C^{1}$ or $C^1$ by pieces included in an open and connex set $\Omega$ for instance, the textbook definition of the integral of a continuous complex function function is that by taking a $C^1$ parametrisation of $\gamma$, let's call it $z : [a,b]\rightarrow \mathbb \gamma$ then we have

$$\int_{\gamma}f(z)dz=\int_{a}^{b} f(z(t))z'(t)dt$$.

Now i understand that if we take another equivalent parametrisation of $\gamma$ (that is to say $y:[c,d]\rightarrow \gamma$ such that a bijective, orientation-preserving function $t$ exists such that $z(t(s))=y(s)$ and $t'>0$) then $$\int_{\gamma}f(z)dz=\int_{c}^{d} f(y(t))y('t)dt$$.

However i have don't see why by taking another parametrisation $w$ of $\gamma$ that is non equivalent to $z$ we can't end up with another value of $$\int_{\gamma}f(z)dz$$. I have tried to prove that two $C^1$ parametrisations of same orientation of the same line $\gamma$ are equivalent to no avail. Thanks in advance for your help.

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  • Check this: https://math.stackexchange.com/q/1820326/42969 – Martin R Sep 20 '23 at 09:18
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    @MartinR I don’t see how that answers the question. Seems like OP’s question is why if $y,z$ are two orientation-preserving parametrizations of a point-set curve $\gamma\subset \Bbb{C}$, then $y,z$ differ by the composition of a change of parameters $t:[a,b]\to [c,d]$ with $t’>0$. Essentially, one would like to take $t=y^{-1}\circ z$ or the other way around, but the smoothness issue requires say the inverse/implicit function theorem to prove properly. I remember having this conversation once but I can’t seem to find my answer/comments on it lol. – peek-a-boo Sep 20 '23 at 09:40
  • @peek-a-boo: You are right, that answers a different question. Here the question is whether two parametrizations are necessarily “equivalent” in the sense that $z(t(s))=y(s)$ for some $t:[a,b]\to [c,d]$. – Martin R Sep 20 '23 at 09:47
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    this is a bit ambiguous because first one has to define what one means a curve to be - one definition is just an equivalence set of parametrizations where equivalence is defined precisely as required here; another is the zero set of a suitable two variable function and then the implicit function theorem shows that at non singular points the definitions are same etc; so first I would say for this question is for the OP to define what they mean by $C^1$ curve – Conrad Sep 20 '23 at 13:10

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