Image of a set under a mapping

Question

I need to show that the image of the closed unit ball in $\mathbb{C}$, under the polynomial mapping $p(x) = (1-x)^2$ is the cardioid: $$\{re^{i\theta} : 0 ≤ \theta < 2π, 0 ≤ r ≤ 2 + 2 \cos\theta\}$$ Can anyone see how to get this?
(It's to calculate the spectrum of an operator with the spectral mapping theorem)

Answer 1

I assume that by $\Bbb R^2$ you mean the complex plane, as otherwise, polynomial maps don't make sense. Let me give you an approach you might take. I leave several details (and verification of several of my claims) to you.

Note that $p=f\circ g,$ where $g(z)=1-z$ and $f(z)=z^2$. It is trivial to show that $g$ maps the closed unit disk into and onto the closed disk $|z-1|\le 1$. You need only show, then that $f$ maps the latter disk into and onto the cardioid-bounded figure you described. Now, $f(0)=0$, and $f(z)\neq 0$ for $z\neq 0$. If we delete $0$ from your cardioid-bounded figure, then we are left with $$C:=\{re^{i\theta}:-\pi<\theta<\pi,0<r\le2+2\cos\theta\},$$ and it will suffice to show that $f$ maps $$\{z:|z-1|\le1,z\ne 0\}$$ into and onto $C$.

Writing $z=\rho e^{i\phi}$ with $\rho>0$ and $\phi$ real, we have $$\begin{align}|z-1|^2 &= \left(\sqrt{(\rho\cos\phi-1)^2+(\rho\sin\phi)^2}\right)^2\\ &= \rho^2\cos^2\phi-2\rho\cos\phi+1+\rho^2\sin^2\phi\\ &= \rho^2-2\rho\cos\phi+1,\end{align}$$ so $|z-1|\le 1$ if and only if $|z-1|^2\le1$ if and only if $$\rho^2-2\rho\cos\phi+1\le1$$ if and only if $$\rho^2\le2\rho\cos\phi$$ if and only if $\rho\le2\cos\phi$ (since $\rho>0$). From this, we can show that $$\{z:|z-1|\le1,z\ne 0\}=\left\{\rho e^{i\phi}:-\frac\pi2<\phi<\frac\pi2,0<\rho\le2\cos\phi\right\}.$$ (I leave the proof to you.) At this point, it follows readily that $f$ maps this set into and onto $C$. Indeed, given $-\frac\pi2<\phi<\frac\pi 2$ and $0<\rho\le 2\cos\phi$, we have $f(\rho e^{i\phi})=\rho^2e^{i2\phi},$ and putting $\theta=2\phi$ and $r=\rho^2$, we have $-\pi<\phi<\pi$ and $$0<r\le4\cos^2\phi=4\cdot\frac{1+\cos2\phi}2=2+2\cos2\phi=2+2\cos\theta,$$ proving the "into" part. For the "onto" part, take $-\pi<\theta<\pi$ and $0<r\le2+2\cos\theta,$ then put $\rho=\sqrt{r},$ $\phi=\frac12\theta,$ and show that $f(\rho e^{i\phi})=re^{i\theta}$ and $|\rho e^{i\phi}-1|\le 1$.