Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be defined by equation $f(x,y)=(x^2-y^2, 2xy)$

Question

Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be defined by equation $$f(x,y)=(x^2-y^2, 2xy)$$ Take $S$ to be the set $$S=\{(x,y): x^2+y^2\leq a^2 \text{ and } x\geq 0 \text{ and } y\geq 0\}$$

(a) Calculate $Df$ and $\det Df$.

(b) Sketch the image under $f$ of the set $S$.

[Hint: Parametrize part of the boundary of $S$ by setting $x = a \cos t$ and $y = a \sin t$; find the image of this curve. Proceed similarly for the rest of the boundary of $S$.]

We remark that if one identifies the complex numbers $\mathbb{C}$ with $\mathbb{R}^2$ in the usual way, then $f$ is just the function $f(z) = z^2$.

For (a), $Df(x,y)=\begin{bmatrix}2x & -2y\\2y & 2x\end{bmatrix}$ and $\det \begin{bmatrix}2x & -2y\\2y & 2x\end{bmatrix}=4x^2+4y^2=4(x^2+y^2)$

I would like to know what $\det Df$ is for, could someone please tell me?

I do not understand what I have to do in (b), could someone help me please? Thank you

Answer 1

Note that if $z=r(\cos\theta+i\sin\theta)$, then $z^2=r^2\bigl(\cos(2\theta)+i\sin(2\theta)\bigr)$. So, since $S$ is the intersection of

• the closed disk centered at $(0,0)$ with radius $a$;
• the first quadrant

its image under $f$ is the top half of the closed disk centered at $(0,0)$ with radius $a^2$.