Let $\Omega = \{ z + z^2/2 : |z| < 1\}$ be the interior of a shape in the complex plane, and let $f(z)$ be an analytic function. Show the following formula:

$$ \int_\Omega f \, dA = \frac{3\pi}{2} \, f(0) + \frac{\pi}{2} \, f'(0)$$

One strategy is to notice that both sides are linear in $f$ so we can assume a power series expansion, $f(x)=\sum a_n \, z^n$. This is not very geometric. Stokes theorem is another possibility.

It's slightly odd to integrate complex function with area? Perhaps I'm a bit rusty, I calculated the area $dA$ should be:

$$(dx + idy)\,(dx - idy) = 2i\, dx \,dy $$

I plotted an image of the interior region. It's simply connected but it's not quite a circle.

As stated in the comments, the map $z \mapsto z + z^2/2$ seems to be injective on the unit disk $\mathbb{D} = \{ |z| < 1 \}$:

enter image description here

Originally I had posted $z \mapsto z/2 + z^2$ which has another interesting shape and analogous question could be posed.

enter image description here

  • $\begingroup$ $dA$ should be $dx\wedge dy = \frac{1}{2i} d\overline{z} \wedge dz$. Since $g \colon z \mapsto z + \frac{z^2}{2}$ is injective on the unit disk, a change-of-variables suggests itself. $\endgroup$ – Daniel Fischer Dec 6 '17 at 14:20
  • $\begingroup$ @DanielFischer No it's not, the image curve is clearly overlapping. $\endgroup$ – cactus314 Dec 6 '17 at 14:24
  • $\begingroup$ We have $\operatorname{Re} g'(z) = \operatorname{Re} (1+z) > 0$ on the unit disk, so $g$ is injective there. I'm not sure what you plotted, I agree that that doesn't look injective. $\endgroup$ – Daniel Fischer Dec 6 '17 at 14:31
  • $\begingroup$ @DanielFischer I have the corret picture now. Could you explain where the $f'(0)$ term happens? $\endgroup$ – cactus314 Dec 6 '17 at 14:41
  • $\begingroup$ From the change of variables, you get $$\frac{1}{2i}\int_{\mathbb{D}} f(g(z)) (1+z)(1+\overline{z})\,d\overline{z}\wedge dz.$$ Change to polar coordinates, and all terms containing an $e^{ik\varphi}$ with $k\neq 0$ integrate to $0$. From $\overline{g'(z)}$ you have a term $e^{-i\varphi}$, and that gets you the $f'(0)$ term by the grouping $f'(0)\cdot re^{i\varphi}\cdot (1 + \dotsc)(\dotsc + e^{-i\varphi})\, rd\varphi\,dr$. $\endgroup$ – Daniel Fischer Dec 6 '17 at 14:48

We must of course assume that $f$ is integrable over $\Omega$. Then, since $g$ is injective, we have

\begin{align} \int_{\Omega} f\,dA &= \frac{1}{2i} \int_{\Omega} f(w) \, d\overline{w} \wedge dw \\ &= \frac{1}{2i} \int_{\mathbb{D}} f(g(z))\, d\overline{g(z)} \wedge dg(z) \\ &= \frac{1}{2i} \int_{\mathbb{D}} d\bigl(\overline{g(z)}\cdot f(g(z))\,dg(z)\bigr). \end{align}

If $f\circ g$ is continuous on $\overline{\mathbb{D}}$, we can directly apply Stokes' theorem to that last integral, otherwise we apply it to the integral over $r\cdot \mathbb{D}$ for $0 < r < 1$ and let $r\to 1$. Let's do that anyway. Then Stokes' theorem yields

\begin{align} \frac{1}{2i} \int_{r\mathbb{D}}&\, d\bigl(\overline{g(z)}\cdot f(g(z))\,dg(z)\bigr) \\ &= \frac{1}{2i} \int_{\lvert z\rvert = r} \overline{g(z)}\bigl(f(g(z))g'(z)\bigr)\,dz \\ &= \frac{1}{2i} \int_{\lvert z\rvert = r} \biggl(\overline{z} + \frac{\overline{z}^2}{2}\biggr) f(g(z))(1+z)\,dz \\ &= \frac{1}{2i} \int_{\lvert z\rvert = r} \biggl(\frac{r^2}{z} + \frac{r^4}{2z^2}\biggr)f(g(z))(1+z)\,dz \tag{$\overline{z} = r^2/z$} \\ &= \pi r^2 \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{f(g(z))(1+z)}{z}\,dz + \frac{\pi r^4}{2}\cdot \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{f(g(z))(1+z)}{z^2}\,dz\\ &= \pi r^2 f(g(0))(1+0) + \frac{\pi r^4}{2} \frac{d}{dz}\biggr\rvert_{z = 0}\bigl(f(g(z))(1+z)\bigr) \\ &= \pi r^2 f(0) + \frac{\pi r^4}{2}\bigl(f(0) + f'(0)\cdot g'(0)^2\bigr) \\ &= \pi f(0)\biggl(r^2 + \frac{r^4}{2}\biggr) + \frac{\pi r^4}{2} f'(0)\,. \end{align}

Taking the limit $r \to 1$ then gets us to

$$\int_{\Omega} f\,dA = \frac{3\pi}{2} f(0) + \frac{\pi}{2} f'(0).$$

  • $\begingroup$ Right. It doesn't just "fall out" of the change of variables. I think the $f'(0)$ term has to do with the cusp of the shape at $g(-1)=-1/2$. Because of that one point, the interior of the cardioid is not diffeomorphic to the disk. $\endgroup$ – cactus314 Dec 6 '17 at 16:11

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