<s>[INST] If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form $ z=\frac{1+it}{1-it}$ where $t\in \mathbb{R}$. If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form
$$ z=\frac{1+it}{1-it},$$
where $t\in \mathbb{R}$.
I don't know, how to begin. I started with the given conditions. Given that $|z|=1 \text{ and } z\neq-1\implies z=e^{it}, t\in[0,2\pi]/\{\pi\}.$
 [/INST] Note that
$\dfrac{1 + it}{1 - it} = \dfrac{(1 + it)^2}{(1 + it)(1 - it)} = \dfrac{(1 + it)^2}{1 + t^2} = \dfrac{(1 - t^2) + 2it}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2} + \dfrac{2it}{1 + t^2}; \tag 1$
since
$(1 - t^2)^2 + (2t)^2 = 1 - 2t^2 + t^4 + 4t^2 = 1 + 2t^2 + t^4 = (1 + t^2)^2, \tag 2$
there is a unique $\theta \in (-\pi, \pi)$ such that
$\cos \theta = \dfrac{1 - t^2}{1 + t^2}, \tag 3$
and 
$\sin \theta = \dfrac{2t}{1 + t^2}; \tag 4$
then we may take
$z = \cos \theta + i \sin \theta = e^{i\theta} \tag 5$
and thus
$\vert z \vert = 1. \tag 6$
We show every such $z \ne -1$ may be so represented.  Observe that $z = -1$ is excluded since then $\sin \theta = 0$, implying $t = 0$, whence $\cos \theta = 1$ forcing $z = 1$; so $\theta \ne \pi, -\pi$.  For all other $\theta \in (-\pi, \pi)$ the equation
$\cos \theta = \dfrac{1 - t^2}{1 + t^2} \tag 7$
may be written
$(1 + \cos \theta)t^2 = 1 - \cos \theta, \tag 8$
and we thus have
$t = \pm \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}, \tag 9$
yielding the two possible values for $\sin \theta$ via (4), showing every $z \ne -1$ with $\vert z \vert = 1$ may be so expressed.
Note:  The correspondence $e^{i \theta} \longleftrightarrow (1 + it)/(1 -it)$ is important in the theory of operators on Hilbert space.  See this wikipedia page on the Cayley Transform.  End of Note.
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