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I've just learnt about the contour integral of a complex function, but I'm having trouble figuring out what it is calculating visually. I understand it is somewhat analogous to the line integral for real functions so should I be thinking about it like the $\mathbb{R}^2 \rightarrow \mathbb{R} $ version of that (as the complex plane is isomorphic to $ \mathbb{R}^2)? $

Watson
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bigfocalchord
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  • the definition is $\displaystyle \oint_C f(z) dz = \int_a^b f(\gamma(t)) d\gamma(t) = \int_a^b f(\gamma(t)) \gamma'(t) dt$ where $\gamma(t),t \in [a,b]$ is a (piecewise) $C^1$ parametrization of the contour $C$. for example if $C$ is the unit circle counter-clockwise, then you can choose $\gamma(t) = e^{it}, t \in [0,2\pi]$ so that $\displaystyle\oint_C f(z) dz = \int_0^{2 \pi} f(e^{it}) i e^{it} dt$ – reuns Jun 08 '16 at 09:42
  • and what you need to prove is that $\oint_C f(z) dz$ doesn't depend on the chosen parametrization – reuns Jun 08 '16 at 09:45
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    Asked many times: http://math.stackexchange.com/questions/111368/looking-for-explanation-of-complex-integration-in-plain-english, http://math.stackexchange.com/questions/110334/line-integration-in-complex-analysis, http://math.stackexchange.com/questions/446724/what-is-contour-integration, http://math.stackexchange.com/questions/904493/geometric-interpretation-of-complex-path-integral – Hans Lundmark Jun 08 '16 at 10:40

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