Post by Admin on Oct 19, 2016 17:54:29 GMT
Gary
Refering to figure 35 on page 316 we can see that the centre of the circle $K$ is at $-i$, and from the right angled triangle with vertices at $-i,0,-1$ we can see that the radius of $K$ is $\sqrt{2}$.
The net transformation from the Poincaré upper half-plane to the Poincaré disc is the composition of $z\mapsto\mathcal{I}_K(z)$ and $z\mapsto\bar{z}$.
So if we use (4) on page 125, the formula for inversion in $K$, we can write $R^2=2$ and $q=-i$ and so
$\mathcal{I}_K(z)=\frac{q\bar{z}+(R^2-|q|^2)}{\bar{z}-\bar{q}}=\frac{-i\bar{z}+(2-1)}{\bar{z}-i}=\frac{-i\bar{z}+1}{\bar{z}-i}$
If we now apply $z\mapsto\bar{z}$, we obtain
$D(z)=\overline{\mathcal{I}_K(z)}=\frac{\overline{-i\bar{z}+1}}{\overline{\bar{z}-i}}=\frac{iz+1}{z+i}$. Done
Vasco
PS The above are all overstrikes.
Refering to figure 35 on page 316 we can see that the centre of the circle $K$ is at $-i$, and from the right angled triangle with vertices at $-i,0,-1$ we can see that the radius of $K$ is $\sqrt{2}$.
The net transformation from the Poincaré upper half-plane to the Poincaré disc is the composition of $z\mapsto\mathcal{I}_K(z)$ and $z\mapsto\bar{z}$.
So if we use (4) on page 125, the formula for inversion in $K$, we can write $R^2=2$ and $q=-i$ and so
$\mathcal{I}_K(z)=\frac{q\bar{z}+(R^2-|q|^2)}{\bar{z}-\bar{q}}=\frac{-i\bar{z}+(2-1)}{\bar{z}-i}=\frac{-i\bar{z}+1}{\bar{z}-i}$
If we now apply $z\mapsto\bar{z}$, we obtain
$D(z)=\overline{\mathcal{I}_K(z)}=\frac{\overline{-i\bar{z}+1}}{\overline{\bar{z}-i}}=\frac{iz+1}{z+i}$. Done
Vasco
PS The above are all overstrikes.