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I want to solve part b) of the problem below. Another user has provided the picture below the problem which describes the map transformation. I am still uncertain about how to produce this image. I understand the picture depicting the domain, since $A$ is restricted by $x>0$, but how do we then get the image $f(A)$ to wrap around the way it does. Since the image consists of those points $(x^2-y^2,2xy)$, and in complex form, $(x^2-y^2+i2xy)$, why do we get that $y\neq 0$ for $x\leq 0$ in the image? it's only $x$ that's restricted in the image, not the real part $(x^2-y^2)$, right?

Thanks!

enter image description here

enter image description here

john fowles
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2 Answers2

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Since you mentioned the complex equivalent of this map, let’s write it more succinctly as $z\mapsto z^2$. Squaring a complex number squares its norm and doubles its argument, so this map “fans out” the half-plane $\Re z>0$ to cover most of the complex plane. The origin is obviously excluded from the image of this half-plane, but also since the $y$-axis, i.e., pure imaginary numbers, is excluded from the domain, the image of the half-plane doesn’t include the negative real axis: both square roots of a negative real number are pure imaginary.

amd
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Any complex number $z$ can be written as $$ re^{i\theta}=r(\cos \theta+i\sin \theta)=z $$ with $r\geq 0$ and $-\pi<\theta\leq \pi$, $\theta$.

Your restriction tells us that for $z$ in your domain, $\Re(z)>0$, i.e. $r>0$ and $-\pi/2<\theta< \pi/2$. Then, the image is the set $$ \{ r^2e^{2\theta i}:r>0,-\pi/2<\theta< \pi/2\}= \{ se^{\phi i}:s>0,-\pi<\phi<\pi\} $$ or the set of complex numbers with positive radius and argument any of the usual values but $\pi$. So this excludes the set $(-\infty,0]$, as I believe you found.

operatorerror
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