Sketch the image of the circle $|z-1| \leq 1$ under the map $w = z^2$. Compute the area of the image.
Ok I know the image of the circle under the complex mapping $z^2$ is a cardioid, but I'm not sure how to set up the double integral and find the bounds. Are we even suppose to use a double integral for this or is there a better way to find the area?
Best Answer
That circle is the set $\{\cos(\theta)+1+\sin(\theta)i\mid\theta\in[0,2\pi]\}$. And you have, for each $\theta\in[0,2\pi]$,\begin{align}(\cos(\theta)+1+\sin(\theta)i)^2&=\cos^2(\theta)+2\cos(\theta)+1-\sin^2(\theta)+i\bigl(2\sin(\theta)+2\sin(\theta)\cos(\theta)\bigr)\\&=\bigl(2+2\cos(\theta)\bigr)\bigl(\cos(\theta)+\sin(\theta)i\bigr).\end{align}Computing the area that you are interested in in polar coordinates, you get\begin{align}\int_0^{2\pi}\int_0^{2+2\cos(\theta)}r\,\mathrm dr\,\mathrm d\theta&=\int_0^{2\pi}2\bigl(1+\cos(\theta)\bigr)^2\,\mathrm d\theta\\&=\int_0^{2\pi}3+4\cos(\theta)+\cos(2\theta)\,\mathrm d\theta\\&=6\pi.\end{align}