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I could use some help in calculating $$\int_{\gamma} \bar{z} \; dz,$$ where $\gamma$ may or may not be a closed curve. Of course, if $\gamma$ is known then this process can be done quite directly (eg. Evaluate $\int \bar z dz$), though that is not the case here.

For instance, if $\gamma$ is indeed a closed curve then I can show the above integral is purely imaginary, but still don't know how to explicitly calculate it.

Thanks!

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4 Answers 4

Let $\gamma$ has parametric representation $$\gamma=\{(x, \ y)\in\mathbb{R^2}\colon\quad x=\varphi(t), \, y=\psi(t),\; t\in[a, \, b] \}.$$ Then for $z=x+iy \in \gamma$ \begin{gather} \int\limits_{\gamma}{\bar{z} \ dz}=\int\limits_{\gamma}{(x-iy) \ (dx+i\:dy)}=\int\limits_{\gamma}{(x\ dx +y\ dy)}+i\int\limits_{\gamma}{(x \ dy - y \ dx )}= \\ =\int\limits_{a}^{b}{\left(\varphi(t) \varphi'(t)+\psi(t) \psi'(t)\right)dt}+i \int\limits_{a}^{b}{\left(\varphi(t) \psi'(t)-\psi(t) \varphi'(t)\right)dt}= \\ =\dfrac{1}{2}\int\limits_{a}^{b}{d\left( \varphi^2(t)+\psi^2(t) \right)dt}+i \int\limits_{a}^{b}{\left(\varphi(t) \psi'(t)-\psi(t) \varphi'(t)\right)dt}= \\ =\varphi^2(b)-\varphi^2(a)+\psi^2(b)-\psi^2(a)+i \int\limits_{a}^{b}{\left(\varphi(t) \psi'(t)-\psi(t) \varphi'(t)\right)dt}. \end{gather}

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You have $$ \int_{\gamma} \bar{z} dz = \int_{\gamma} (x - iy) (dx + idy) = \int_{\gamma} xdx + ydy + i \int_{\gamma} x dy - ydx.$$ The integrand of the real part is an exact differential, and so it depends only on the end points of $\gamma$ and vanishes if $\gamma$ is closed. You can write an explicit formula for this part. The imaginary part is not exact, and so in general, the integral depends on the path. Like WimC mentions above, if $\gamma$ is a simple closed curve, you can use Green's theorem to relate this to the enclosed area. If $\gamma$ is not closed, I don't think you can expect any more information.

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For a closed curve it is $2 i$ times the oriented area of the region enclosed by $\gamma$. Here "oriented area" means that the winding number around a point determines how that point contributes to the area. For example, for any circle of radius $r$ traversed in counter clockwise direction the integral equals $2i\, \pi r^2$. To see the general formula note that if $\gamma: [0,1] \to \mathbb{C}$, $\gamma(t) = x(t) + iy(t)$ then $$\operatorname{Im} \int_{\gamma} \overline{z} dz = \int_0^1 \det \begin{pmatrix} x(t) & x'(t) \\ y(t) & y'(t) \end{pmatrix} dt $$ and this is indeed twice the area of the region enclosed by $\gamma$. You already know that the real part is zero.

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As the integrand is not holomorphic, the integral will depend on the whole path $\gamma$ and not only on the endpoints. In that sense your expression is already the most compact expression one can write without knowledge of the path $\gamma$. What kind of final (closed form) expression do you have in mind?

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