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
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*}