I'm doing some exercises on complex integration, and stumbled over this:

$$\int_\gamma \bar{z} dz,$$

Where $\gamma$ is any closed $C^1$ curve which is the boundary of a bounded, connected, open $U \subset \mathbb{C}$, and $\bar{z}$ denotes the complex conjugate of $z$. One should express the result in terms of the quantity

$$\int_\gamma x dy ,$$

Which coincides with the area enclosed by $\gamma$.

I had no idea how to do this, so I checked the solution. It sais:

$$\int_\gamma \bar{z} dz = \int_\gamma z + \bar{z} dz = 2 \int_\gamma x dz.$$

This part is clear (the identity is holomorphic). Then they assume that $t \mapsto x(t) + iy(t), \quad t \in [0,1]$ is a parametrization of $\gamma$. They compute

$$\int x dx = \int_0^1 x(t)\dot{x}(t) dt = \frac{1}{2}[x(1)^2 - x(0)^2] = 0,$$

since $\gamma$ is closed. I do get that what's stated is true, however I don't see the motivation to compute this part. Then comes the conclusion

$$\int_\gamma \bar{z} dz = 2i\int_\gamma x dy.$$

I really don't see what's done here in the last part, no idea how that follows.

Any help?

  • $\begingroup$ Do you know what the integral $\int \bar{z}dz$ means? $\endgroup$ – Hui Yu Jul 17 '13 at 13:47
  • $\begingroup$ I don't think I understand what you mean. I see it as the integral of the complex conjugate of the identity. $\endgroup$ – Pascal Engeler Jul 17 '13 at 14:03
  • 1
    $\begingroup$ Regarding motivation, maybe it's helpful simply to write $\bar{z}=x-iy$ and $dz=dx+i\,dy$? (The real part of the product is exact, and Green's theorem (e.g.) gets you the rest of the way.) $\endgroup$ – Andrew D. Hwang Jul 17 '13 at 14:20
  • $\begingroup$ You don't even need top use Green's theorem. Since you know that $\int_\gamma \bar zdz=2\int_\gamma x dz$, just replace $dz$ with $dx+idy$ in the right-hand sid and use the fact that $\int_\gamma xdx=0$. $\endgroup$ – Etienne Jul 17 '13 at 17:02
  • $\begingroup$ That helped! I didn't know I could replace dz by dx + idy and then just treat it like a normal product. I had thought you couldn't do that. What's the justification for this? $\endgroup$ – Pascal Engeler Jul 18 '13 at 8:51

Let the curve $\gamma$ be parameterized with $z(t)=x(t)+iy(t)\in C^1$, $t\in [0,1]$. Then, we can write the integral as

$$\begin{align} \oint_\gamma \bar z\,dz &=\int_0^1 \bar z(t)\frac{dz(t)}{dt}\,dt\\\\ &=\int_0^1 (x(t)x'(t)+y(t)y'(t))\,dt+i\int_0^1 (x(t)y'(t)-y(t)x'(t))\,dt\\\\ &=\frac12 \int_0^1 (x^2(t)+y^2(t))'\,dt+\color{blue}{i\int_0^1(x(t)y(t))'\,dt}-\color{red}{i2\int_0^1 y(t)x'(t)\,dt} \\\\ &=0+\color{blue}{0}-\color{red}{i2(-\text{Area enclosed by}\,\,\gamma)}\\\\ &=i2\times (\text{Area enclosed by}\,\,\gamma) \end{align}$$


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