I wanted to verify if the following proof is correct: If $\bar{\gamma}$ is $\gamma$ with reverse orientation then: $$\int_{\gamma}f(z)dz=-\int_{\bar{\gamma}}f(z)dz$$ where $z:[a,b]\rightarrow \mathbb{C}$

Proof: $$\int_{\gamma}f(z)dz=\int_a^b f(z(t))z'(t)dt$$ As a particular parametrization of $\bar{\gamma}$ we can take $\bar{z}:[a,b]\rightarrow \mathbb{R}^2$ defined by $\bar{z}(t)=z(b+a-t)$. Then:

$$\int_a^b f(z(t))z'(t)dt=\int_{a}^b f(z(t(s)))z'(t(s))t'(s)dt=-\int _a^b f'(\bar{z}(t))\bar{z}'(t)dt=-\int_{\bar{\gamma}}f(z)dz$$ where $t:[a,b]\rightarrow [b,a], t(s)=b+a-s$.


Yes, your proof is correct.

As an aside, I'll remark that there is another type of integral over curves, written $$\int_{\gamma}f(z)\,|dz|=\int_a^b f(z(t)) |z'(t)|\,dt$$ For this kind of integral, changing the orientation of the curve makes no difference.


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