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My teacher gave us this statement to study for a question on an exam. We've asked for specific examples, but instead he gives examples of analytic functions and tell us to change the power. So .. I just want a concrete example of how to do something like this .. is that possible?

Contour integral of a non-analytic function (something like $\overline{z}^3$ or to another power), requiring parametrization.

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2 Answers 2

A concrete example like this? Let $\Gamma$ be the positively oriented circle $|z-a|=r$, which we parametrize as $\gamma(t) = a + r e^{it}$, $0 \le t \le 2 \pi$. Then $dz = r i e^{it}\ dt$ so

$$ \oint_\Gamma \overline{z}^3\ dz = \int_0^{2 \pi} (\overline{a} + r e^{-it})^3\ r i e^{it}\ dt = \int_0^{2 \pi} \left(\overline{a}^3 r i e^{it} + 3 \overline{a}^2 r^2 i + 3 \overline{a} r^3 i e^{-it} + r^4 i e^{-2it}\right)\ dt = 6 \pi i \overline{a}^2 r^2$$

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What you need is the Stoke's formula, which asserts that $$\int_{\partial S}w=\int_{S}dw$$ for a bounded open manifold $S$. Therefore we have $$\int_{\partial S}wdz=\int_{S} dw\wedge dz=\int_{S}\frac{\partial w}{\partial \overline{z}}dz\wedge d\overline{z}$$ using the fact that $$dw=\frac{\partial w}{\partial \overline{z}}d\overline{z}+\frac{\partial{w}}{\partial z}dz$$

Thus for your example we have $$\int_{\partial S}\overline{z}^{3}=3\int_{S}\overline{z}^{2}dz\wedge d\overline{z}$$

This integral seems can be perfectly non-zero in appropriate regions. I am not sure how the integral works in practice (say, giving a retangle or a circle) without breaking into real and imaginary components. We at least have $$dz\wedge d\overline{z}=(dx+idy)\wedge (dx-idy)=-idx\wedge dy+idy\wedge dx=-2idx\wedge dy$$ While $$\overline{z}^{2}=x^{2}-y^{2}-2ixy$$

Thus separating the real and imaginary components we should have $$-12\int_{S}xydxdy-6i\int_{S}(x^{2}-y^{2})dxdy$$

The real and complex part can easily be nonzero I guess.

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