# 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.

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Now posted to MO, mathoverflow.net/questions/138491/… --- Michael, you know enough not to do this without posting a link. – Gerry Myerson Aug 4 '13 at 0:34
I posted a link within the m.o. posting. Now someone's migrated that to m.s.e. as a duplicate of this despite the explicit statement at the top that it was earlier posted to m.s.e. – Michael Hardy Aug 4 '13 at 2:03

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).
+1. But approximately this reasoning is where I got $(3)$ in the first place (except that I used the line $x=0$), so that part is largely the same as what I had already done. If you're right about $i$ going to $e^{-i\beta}$ then I was right to think I might not have been nitpicking enough with "$+$" and "$-$", since I intended $i$ to go to $e^{i\beta}$; I'll check through that again. The part with $f(f(x))$ surprised me and it was so simple to derive it that I thought it must be well within the beaten paths, but I don't see anyone saying so here. – Michael Hardy Aug 8 '13 at 0:15
OK, now I'm getting that $i$ goes to $-e^{i\beta}$ (so neither the $e^{i\beta}$ that I had intended, nor the $e^{-i\beta}$ that you wrote, is correct). I should have written $z-\tan\frac\beta2$ rather than $(\tan\frac\beta2) - z$ in the denominator. – Michael Hardy Aug 8 '13 at 0:20
Well, weirdly I'm now getting that $i$ goes to $-i e^{-i\beta}$. – user43208 Aug 8 '13 at 0:46