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

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.
