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 $|z|=1 \text{ and } z\neq-1\implies z=e^{it}, t\in[0,2\pi]/\{\pi\}.$
 A: HINT: 
$$\frac{1 + i \tan \frac{\theta}{2}}{ 1 - i \tan \frac{\theta}{2}} = \cdots$$
(a picture can make it obvious)
A: Without appealing to polar form, using only complex algebra...
Solving for $z-1$ gives $z-1= it(1+z)$, and as $z \neq -1$ we can divide to get
$$
\frac{z-1}{z+1} = it.
$$
Write $z=a+bi$ and start doing the algebra on the left hand side.  It's not hard (conjugate by the denominator).  When you realize that $a^2+b^2=1$ you end up with
$$
\frac{z-1}{z+1} = \frac{2bi}{2a+2} = it.
$$
Hence $t = \frac{2b}{2a+2} = \frac{b}{a+1}$ should work, and you can check it. 
A: Like Randall, I would begin by solving that
$$
it=\frac{z-1}{z+1}.
$$
But from this point on I would use the fact that $1=|z|^2=z\overline{z}$. This allows a rewrite
$$
it=\frac{z-1}{z+1}=\frac{z-z\overline{z}}{z+z\overline{z}}=\frac{1-\overline{z}}{1+\overline{z}}=-\overline{\left(\frac{z-1}{z+1}\right)}=-\overline{it}.
$$
This implies that $t$ must be real.
A: $$
\frac{1+it}{1-it} = \frac{(1-t^2) + 2it}{1+t^2} = \frac{1-t^2}{1+t^2} + i \frac{2t}{1+t^2} = x+iy, \qquad x,y\in\mathbb R.
$$
Can you show that $t= \dfrac y {1+x}$?
That means if $x^2+y^2=1$ then $t=\dfrac y{1+x}$ will do it, and you see why they excluded $z=-1,$ since at that point the denominator $1+x$ is $0.$
A: 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.
A: Taking the modulus of both sides and applying the fact that $|z|=1$ gives
$$\frac{\left|1+\mathrm i t\right|}{\left|1-\mathrm i t\right|} = 1$$
This implies that $\left|1+\mathrm i t\right|=\left|1-\mathrm i t\right|$, i.e.
$\left|(\mathrm i)(t-\mathrm i)\right| = \left|(-\mathrm i)(t+\mathrm i)\right|$, i.e.
$$\left|t- \mathrm i\right|=\left| t+ \mathrm i \right|$$
This is the perpendicular bisector of $t=\mathrm i$ and $t=-\mathrm i$, i.e. the real axis.
This tells us that the Mobius transformation $\mathbb f : \widehat{\mathbb C}_t \to \widehat{\mathbb C}_z$, given by
$$z=\mathbb f(t) = \frac{1+\mathrm it}{1-\mathrm it}$$
takes the real axis $t \in \mathbb R$ onto the circle $|z|=1$, and vice-versa.
(Non-degenerate Mobius transformations are automorphisms of the Riemann Sphere $\widehat{\mathbb C}$) 
Your question mentions $z \neq -1$. That is not a problem on the Riemann Sphere: $\mathrm f(\infty)=-1$.
A: We have for $z\ne -1$, 
$$z=\frac{1+it}{1-it}\implies t=i\left(\frac{1-z}{1+z}\right)$$
Moreover, we can write
$$\begin{align}
i\left(\frac{1-z}{1+z}\right)&=i\left(\frac{1-|z|^2-2i\text{Im}(z)}{1+|z|^2+2\text{Re}(z)}\right)\\\\
&=2\left(\frac{\text{Im}(z)}{1+|z|^2+2\text{Re}(z)}\right)+i\left(\frac{1-|z|^2}{1+|z|^2+2\text{Re}(z)}\right)\tag 1
\end{align}$$
From $(1)$, it is trivial to see that $t=i\left(\frac{1-z}{1+z}\right)$ is purely real if and only if $|z|=1$, with $z\ne -1$.
And we are done!
