Find the area of the image of $D (0, 1)$ under tha map $$f (z) = z +\frac {z^2}{2}$$ Here's a solution for this: $f(w)=f(z)$ if and only if $$(w-z)(1+\frac{w+z}{2})=0$$
so f is injective. Then area of its image is given by $$\int_{\Bbb D}|f'(z)|^2dxdy=\int_{\Bbb D}(1+2Re(z)+ |z|^2)dz$$ $$=\int_{\Bbb D}(1+2x+ x^2+y^2)dxdy$$ $$=\int_{0}^{2\pi}\int_{0}^{1}(1+2r\cos(\theta)+ r^2)drd\theta$$ $$=\frac{3\pi}{2}$$
Here, I did not understand:
$1.$ What is $f(w)$ and from where did $(w-z)(1+\frac{w+z}{2})=0$ come?
$2.$ How come it is $\frac{3\pi}{2}$? Because after calculation it is $\frac {8\pi}{3}$