Prove that every complex number with modulus 1 and is not -1, has this property I'm currently self-studying "Theory of Functions of a Complex Variable" by A. I. Markushevich and now I've encountered several difficult problems in the book from the start.
This is just one of them.
Prove that every complex number $z$ where $|z|=1$ but $z\not=-1$ can be represented as
$$z=\frac {1+it}{1-it}$$
where $t \in \mathbb{R}$.
I've tried writing
$$z=e^{i(t+\frac{\pi}{2}-\frac{\pi}{2})}$$
$$=\cos(t+\frac{\pi}{2}-\frac{\pi}{2})+i\sin(t+\frac{\pi}{2}-\frac{\pi}{2})$$
and then reducing by the addition formula but it doesn't seem to go anywhere.
If someone would give me a hint to start, not the full answer, I would appreciate it very much.
 A: We have  $\frac{1+it}{1-it} = \frac{1-t^{2}}{1+t^{2}} + i\frac{2t}{1+t^{2}}$ and then recall that $$\cos(\alpha) = \frac{1-\tan^{2}(\frac{\alpha}{2})}{1+\tan^{2}(\frac{\alpha}{2})}$$ and $$\sin(\alpha) = \frac{2\tan(\frac{\alpha}{2})}{1+\tan^{2}(\frac{\alpha}{2})}$$
A: I suppose you meant $\;t\in\Bbb R\;$ :
$$\frac{1+it}{1-it}=\frac{1-t^2+2it}{1+t^2}=\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}i$$
Now
$$|z|=1\;,\;\;z\neq -1\iff z=e^{ix}=\cos x+i\sin x\;,\;(2k+1)\pi\neq x\in\Bbb R\;,\;\;k\in\Bbb Z$$
But since
$$\left|\frac{1+it}{1-it}\right|=\frac{(1-t^2)^2+(2t)^2}{(1+t^2)^2}=1$$
the question reduces to show that if $\;\alpha^2+\beta^2=1\;,\;\;\alpha\,,\,\beta\in\Bbb R\;$ , then there exists
$$x\in\Bbb R\;\;s.t.\;\;\cos x=\alpha\;,\;\sin x=\beta\;$$ 
Remembering your trigonometry (including the trigonometric functions, of course), take it now from here
A: The equation 
$$ \tag 1
z=\frac {1+it}{1-it}
$$
is equivalent to
$$ \tag 2
 t = \frac 1i \frac{z-1}{z+1} \, ,
$$ 
therefore it suffices to show that for $|z|=1$, $z \ne -1$, $t$
is a real number. That follows easily from $z \overline z = |z|^2 = 1$:
$$
 \overline t = \frac 1{-i} \frac{\overline z-1}{\overline z+1}
 = - \frac 1i \frac{1/z-1}{1/z+1} = - \frac 1i \frac{1-z}{1+z} = t
$$
So for each $|z|=1$, $z \ne -1$, $(2)$ defines a real number $t$
such that $(1)$ holds.

Alternative solution: For $z=x+iy$ with  $|z|=1$, $z \ne -1$,
define
$$ \tag 3
 t = \frac{y}{x+1} 
$$
Then
$$
\frac {1+it}{1-it} = \frac{(x+1) + iy}{(x+1)-iy}
= \frac{(x+1)^2 -y^2 + 2i(x+1)y}{(x+1)^2+y^2} = ... = x + iy = z \, .
$$
$(3)$ has a geometric interpretation:
For $z = x+iy = e^{i \varphi}$, draw the line from $-1$ to $z$.
Then $t$ is the y-coordinate of the intersection of that line with the imaginary axis:

$(3)$ follows from the similarity of the triangles $(-1, 0, it)$ and 
$(-1, x, x+iy)$.
Also the angle at $z=-1$ is $\frac \varphi 2$, and therefore $t = \tan \frac \varphi 2$.
A: You could replace $t$ with $\frac{i+b-i a}{1+a+i b}$ and reduce the expression until you reach $a+i b$ 
A: For $1=|z|=z \bar z \ne -1$ we have  $z=\frac {(1+i t)}{(1-i t)} \iff$ $$t=i\cdot \frac {(1-z)}{(1+z)}=$$    $$=i\cdot \frac {(1-z)(1+\bar z)}{(1+z)(1+\bar z)}=$$     $$=i\cdot \frac {(1+\bar z-z -z \bar z)}{(1+z+\bar z +z \bar z)}=$$    $$=i\cdot \frac {(1+\bar z-z -1)}{(1+\bar z +z+1)}=$$     $$=i\cdot \frac {(\bar z-z)}{(2+\bar z+z)}=$$ $$=i\cdot \frac {(-2i Im(z))}{(2+2 Re (z))}=\frac {-2 i^2 Im(z)}{(2+2 Re(z))}=\frac {Im(z)}{1+Re(z)} .$$ This last value is a real number. Since the implication is in both directions ,we have $$t=Im(z)/(1+Re(z))\implies z=\frac {(1+i t)}{(1-i t)}.$$
