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 $\mathbb{D} = \{ |z| < 1 \}$:
Originally I had posted $z \mapsto z/2 + z^2$ which has another interesting shape and analogous question could be posed.
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
$$\int_{\Omega} f\,dA = \frac{3\pi}{2} f(0) + \frac{\pi}{2} f'(0).$$