Compute the area of the image of the unit disk \[ \{z\in\mathbb C:|z|<1\} \] under \[ f(z)=z+\frac{z^2}{2}. \]
Solution
1. The map \(f\) is injective on \(\DD\).
For \(z,w\in\DD\), \[ f(z)-f(w) = (z-w)\left(1+\frac{z+w}{2}\right). \] If \(f(z)=f(w)\) and \(z\neq w\), then \(z+w=-2\). But \[ \abs{z+w} \le \abs{z}+\abs{w} < 2, \] which is impossible when \(z+w=-2\). Hence \(z=w\).
2. The real Jacobian determinant of \(f\) at \(z\in\DD\) is\[ J_f(z)=\abs{1+z}^2. \]
For a holomorphic map, the real Jacobian determinant is \(\abs{f'(z)}^2\). Here \[ f'(z)=1+z, \] so \[ J_f(z)=\abs{f'(z)}^2=\abs{1+z}^2. \] Since \(-1\notin\DD\), this Jacobian is positive throughout \(\DD\).
3. The area of \(f(\DD)\) is\[ \int_{\DD}\abs{1+z}^2\,dA(z). \]
By step 1, \(f\) is injective on \(\DD\), and by step 2 its Jacobian is positive there. The change-of-variables formula therefore gives \[ \operatorname{Area}(f(\DD)) = \int_{\DD}J_f(z)\,dA(z) = \int_{\DD}\abs{1+z}^2\,dA(z). \]
4. The required area is\[ \boxed{\frac{3\pi}{2}}. \]
Writing \(z=re^{i\theta}\), \[ \abs{1+re^{i\theta}}^2 = 1+2r\cos\theta+r^2. \] Hence step 3 gives \[ \begin{aligned} \operatorname{Area}(f(\DD)) &= \int_0^{2\pi}\int_0^1 \left(1+2r\cos\theta+r^2\right)r\,dr\,d\theta\\ &= 2\pi\int_0^1(r+r^3)\,dr\\ &= 2\pi\left(\frac12+\frac14\right)\\ &= \frac{3\pi}{2}. \end{aligned} \]
5. Q.E.D.
Step 4 gives the requested area.