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I would like to solve part b) of the problem below. I have proven a) by rewriting $f(x,y)=f(z)$, where $z=re^{i\theta}$, and this gave me that $f(re^{i\theta})=r^2e^{2i\theta}$. Proofs for this can be found here $\phantom{}$ $f(x,y)=(x^2-y^2,2xy)$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$. What is the set $f(A)$

Since $r^2e^{2i\theta}=r^2\cos2\theta+r^2\sin\theta$, and because in my proof i had to restrict $r>0$ and $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, in order to satisfy the conditions of set $A$, as described in part a) of the question, How can I describe $f(A)$? In the solution i linked I saw a solution that restricted $f(A)$ to the complex plane not including $y=0$ and $x\leq 0$ but I'm unsure of how to arrive at this restriction on the $\mathbb{C}$?

Thanks! enter image description here

john fowles
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enter image description here

Note that this mapping of $ \mathbb{R}^2 \rightarrow \mathbb{R}^2$ represents the map $z \rightarrow z^2$ of the complex plane. It maps the positive half plane to the complex plane minus the negative real axis.

Donald Splutterwit
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  • So you got that sketch of the domain, by rewriting $(x,y)=(x+iy)$, and considering the restriction $x>0$. And the image is $(x+iy)^2=x^2-y^2+i2xy$, and since $x^2-y^2$ is the real part, and because $x>0$, we get all points on the $x$-axis besides $x\leq 0$? and the imaginary part is unrestricted. Is my interpretation of the map correct? – john fowles Jul 03 '18 at 00:45