Tangent half-angles and linear fractional transformations Suppose $z=x+iy$, $x$ and $y$ are real, and $|z|=x^2+y^2=1$ so that $z=e^{i\alpha}$ for some real $\alpha$.  Then for some real $\gamma$,
$$
\begin{align}
e^{i\gamma} = f(e^{i\alpha}) = f(z) & \overset{\text{def}}{=} \frac{\left(\tan\frac\beta2\right)z - 1}{\left(\tan\frac\beta2\right) - z} =\frac{(\sin\beta)-x}{1-(\sin\beta)x} + i \frac{(\cos\beta)y}{1-(\sin\beta)x} \tag{1}. \\[10pt]
& = f(f(x)) + i \frac{(\cos\beta)y}{1-(\sin\beta)x}. \tag{2} \\[15pt]
\text{and }\tan\frac\gamma2 & = \tan\frac\beta2\cdot\tan\frac\alpha2.\tag{3}
\end{align}
$$
PS: OK, now I hope I've fixed the problem of the correct trivial choices between "$+$" and "$-$".  Attention to detail and all that . . . . .
QUESTIONS:


*

*Is $(1)$ yet another known tangent half-angle identity?

*The real part of $f(z)$ is $f(f(x))$, where $x$ is the real part of $z$.  Can anything similar be done with the imaginary part?

*(Have I got all my pluses and minuses right?)

*The right side of $(3)$ is symmetric in $\alpha$ and $\beta$.  Is there some way to see that symmetry by staring at $(1)$?  (My suspicion: Either this is trivial and tomorrow I'll see that it's obvious, or else it's not.)

*Discuss applications to geometry and turn it in at 3:59 AM tomorrow. [No, this is not homework, and you are hereby instructed not to read this sarcastic final item unless you want to.]


(I'm not sure I've crossed all the "$t$"s and dotted all the "$i$"s$\ldots\ldots$  I'll come back later and check things again.)
PS: Now I'm guessing that the answers will emerge from looking at what happens in $(1)$ and $(2)$ if in $(3)$ one multiplies $n$ tangent half-angles instead of only two.
 A: For some of this, it is useful to view this in terms of the stereographic projection $\phi: S^1 \to L$, mapping the unit circle to the real projective line $L = \mathbb{P}^1 \mathbb{R}$ given by $x = 1$, by projecting from $z = -1$. By elementary geometry, $\phi$ takes $z = e^{i\alpha}$ to the point with $y$-coordinate $2\tan\frac{\alpha}{2}$. 
The transformation $\phi: S^1 \to \mathbb{P}^1 \mathbb{R}$ is actually given by restricting a complex fractional linear transformation to the unit circle (it takes $z = 1$ to $y = 0$, $z = -1$ to $y = \infty$, and $z = i$ to $y = 2$, so $\phi: z \mapsto \frac{2}{i}\frac{z-1}{z+1}$). The function $f$ is manifestly fractional linear, so that the analysis of $f$ is equivalent to the analysis of $g = \phi \circ f \circ \phi^{-1}$ as a fractional linear transformation on $\mathbb{P}^1 \mathbb{R}$. The function $f$ fixes $z = 1$ and $z = -1$, so $g$ fixes $y = 0$ and $y = \infty$, and thus is given by multiplication by some scalar. It is easy to check that $f$ takes $i$ to $e^{-i\beta}$; on the $\mathbb{P}^1 \mathbb{R}$ side, $g$ is given by multiplying by the scalar that takes $2$ to $-2\tan (\frac{\beta}{2})$. So $g(y) = -\tan(\frac{\beta}{2}) \cdot y$, and this explains formula (3) (except that I get a different sign). 
I'd have to think before answering the other questions. -- Todd Trimble
