the area of the image under a specific holomorphic function of the unit disk Let $f(z)=z^3+\frac{z^2}{2}$. Let $D$ be the unit disk in $\mathbb{C}$. How to compute $$ Area(f(D))? $$ In the case that $f:D\to \mathbb{C}$ is injective, \begin{align*} Area(f(D))&= \int_D |f'(z)|^2 ~ dx\,dy \\\ &= \int_0^1 \int_0^{2\pi} r\left|f'(re^{i\theta})\right|^2 ~ d\theta \, dr \\\ &= \int_0^1 \int_0^{2\pi} r \left(\sum_{n = 1}^\infty nc_nr^{n-1} e^{i(n-1)\theta}\right)\left(\sum_{n = 1}^\infty n\overline{c_n}r^{n-1} e^{-i(n-1)\theta}\right) ~ d\theta \, dr \\\ &= \int_0^1\left( \sum_{n=1}^\infty 2\pi n^2|c_n|^2 r^{2n-1} \right)~dr \\\ &= \sum_{n=1}^\infty \frac{2\pi n^2}{2n} |c_n|^2 \\\ &= \pi \sum_{n=1}^\infty n|c_n|^2. \end{align*} But now $f$ is not injective. How to compute? Thanks.

I would use Green's formula for the area: $$A= \frac12 \int (y\,dx-x\,dy) \tag1$$ where the integral is taken over the boundary. Here the boundary is parametrized by $x=\cos 3t+\frac12 \cos 2t$ and $y=\sin 3t+\frac12 \sin 2t$. The integrand in (1) should simplify nicely. The only unpleasant part is finding the interval of integration: since the curve self-intersects, it is not $[-\pi,\pi]$.

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Integrate over the outmost part of this curve. The domain of integration is $[-\theta,\theta]$ where $\theta$ is the smallest positive root of $\sin 3t+\frac12 \sin 2t=0$. It can be found analytically with Wolfram Alpha, and presumably by hand too.