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

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

• You'll see what $det Df$ is for when you learn the change of variables theorem for multiple integrals. – saulspatz Dec 23 '17 at 20:06

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 $a$;
its image under $f$ is the top half of the closed disk centered at $(0,0)$ with radius $a^2$.
• $S$ is not a disk, but a quarter of a disk. – Jose27 Dec 23 '17 at 19:21
• @JoséCarlosSantos why "its image under $f$ is the top half of the closed disk centered at $(0,0)$ with radius $a^2$"? – user482152 Dec 23 '17 at 19:34
• @user482152 Because the elements of $S$ are those $z\in\mathbb C$ which can be written as $r\bigl(\cos\theta+i\sin\theta\bigr)$ with $0\leqslant r\leqslant a$ and $\theta\in\left[0,\frac\pi2\right]$. Since, for each such $z$, $f(z)=r^2\bigl(\cos(2\theta)+i\sin(2\theta)\bigr)$, the elements of $f(S)$ are those complex numbers $z$ which can be written as $r(\cos\mu+i\sin\mu)$ with $0\leqslant r\leqslant a^2$ and $\mu\in[0,\pi]$. That's the top half of the closed disk centered at $(0,0)$ with radius $a^2$. – José Carlos Santos Dec 23 '17 at 20:07