Map $z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it },t\in\mathbb{R}$ maps the real axis $(-\infty,\infty)$ onto the unit circle $\mid z\mid=1$.

Question

Show that the mapping $$ z^2 = \frac{\frac{1}{2}+it}{\frac{1}{2}-it}, \quad t\in\mathbb{R} $$ maps the real axis $(-\infty,\infty)$ to the unit circle $|z|=1$.

My try- $$z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it }$$ $$|z^2|=\frac{|\frac{1}{2}+it|}{ |\frac{1}{2}-it| }$$ $$|z^2|=\frac{\sqrt{1/4+t^2}}{ \sqrt{1/4+t^2} }$$ $$|z^2|=1$$ $$|z|^2=1$$ $$|z|=1$$...

Answer 1

You argument is partially correct. You did show that $f$ maps $\mathbb{R}$ into a subset of the unit circle, but the problem requires to show the entire unit circle is covered by the image. In other words, if you look at $f:\mathbb{R} \to D_1(0)$, it would be onto (surjective).

To do this, let $z \in D_1(0)$ and show that there is $t \in \mathbb{R}$ such that $f(t)=z$.

Answer 2

The map is multivalued as $z^2=1$ at $t=0$ gives $z=\pm 1$ lying on unit circle $\lvert z\vert=1$.

Added 1-The partial proof you have done shows that the map is onto. But it cannot be a bijection as it is not a single valued function.

Added 2- For another proof, note that

$t=\frac{z^2-1}{2i(z^2+1)}$, where $z=x+iy, x,y\in\mathbb R$ After splitting the RHS into real and imaginary parts, you can equate the imaginary part of $RHS=0$ (Why?) to get the result $x^2+y^2=1$.

Answer 3

Let the unit circle be $z=e^{i\theta}$, $\theta\in[0,2\pi]$. Then,

$$t= \frac1{2i} \frac{z^2-1}{z^2+1} = \frac1{2i} \frac{e^{i 2\theta}-1}{e^{i 2\theta} +1} =\frac12\tan\theta $$ which reveals that $\theta\in[0,\pi)$ and $\theta\in[\pi,2\pi)$ separately map to $t\in(-\infty,\infty)$, thus, not bijective.