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To calculate the hyperbolic distance we use the formula $$\left|\frac{w-z}{1-\bar wz}\right|$$

I want to apply this to the following pair of points:

\begin{align*} w&=\frac{-1}{\sqrt{3}}\space +\space \frac{1}{\sqrt{3}}i \\ z&=\frac{1}{\sqrt{3}}\space +\space \frac{1}{\sqrt{3}}i \\ \left|\frac{w-z}{1-\bar wz}\right| &= \frac{\left(\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i\right)-\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i\right)}{1-\left(-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}i\right)\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i\right)} \\ &=\frac{2\sqrt{3}}{1-\frac{1}{3}\left(1+i\right)^2} \\ &=\frac{2\sqrt{3}}{\sqrt{13}} \end{align*}

I am not sure how you get from the penultimate step to the final step. Could someone explain in depth how this was done? Thanks

MvG
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Nessa
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    What model of hyperbolic geometry? Where did you get these formulas, both the generic one for the length and the specific ones where you don't understand the last step? To me it looks as if this is simply wrong. – MvG Feb 24 '15 at 13:56
  • `Poincare Disc Model' – Nessa Feb 24 '15 at 15:06
  • The example I was given in my notes uses the above formula (and the above example) but I'm not sure how they obtained the final answer... – Nessa Feb 24 '15 at 15:07

1 Answers1

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Wikipedia gives the formula for the distance in the Poincaré disk model:

$$d=\operatorname{arcosh}\left(1+2\frac{\lvert z-w\rvert^2}{\left(1-\lvert z\rvert^2\right)\left(1-\lvert w\rvert^2\right)}\right)= \operatorname{arcosh}(1+2\cdot12)=\operatorname{arcosh}25\approx3.9116$$

Your formulas are different from this, and the penultimate one has several errors as well: it lacks the absolute value bars, and it has a sign wrong in the denominator, and a wrong numerator as well.

\begin{align*} \left\lvert\frac{w-z}{1-\bar wz}\right\rvert &= \left\lvert \frac{\left(\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i\right)-\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i\right)}{1-\left(-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}i\right)\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}i\right)} \right\rvert \\&= \left\lvert\frac{-\frac2{\sqrt{3}}}{1+\frac{1}{3}\left(1+i\right)^2}\right\rvert = \left\lvert\frac{-\frac2{\sqrt3}}{1+\frac23i}\right\rvert = \left\lvert\frac{-\frac2{\sqrt3}\left(1-\frac23i\right)}{\left(1+\frac23i\right)\left(1-\frac23i\right)}\right\rvert \\&= \left\lvert\frac{-\frac2{\sqrt3}\left(1-\frac23i\right)}{1+\frac49}\right\rvert = \left\lvert\frac{\left(-\frac2{\sqrt3}\left(1-\frac23i\right)\right)\cdot9}{\left(1+\frac49\right)\cdot9}\right\rvert = \left\lvert\frac{-2\sqrt3\left(3-2i\right)}{9+4}\right\rvert \\&= \left\lvert\frac{2\sqrt3}{13}(-3+2i)\right\rvert = \frac{2\sqrt3}{13}\cdot\left\lvert-3+2i\right\rvert = \frac{2\sqrt3}{13}\sqrt{3^2+2^2} = \frac{2\sqrt3}{13}\sqrt{13} = \frac{2\sqrt3}{\sqrt{13}} \end{align*}

MvG
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