= Solution
Write $B=\operatorname{diag}(\lambda,\lambda^{-1})$ and $q=\lambda^2>0$, so its <Möbius transformation> is $z\mapsto qz$, with $q\ne1$. For $z=u+iv$, the <Hyperbolic distance in the Poincare half-plane> satisfies
$$
\cosh\rho(z,qz)
=1+\frac{|z-qz|^2}{2\operatorname{Im}z\operatorname{Im}(qz)}
=1+\frac{(q-1)^2}{2q}
\left(1+\frac{u^2}{v^2}\right).
$$
At $z=i$, the final parenthesis is $1$. Since $\cosh$ is strictly increasing on nonnegative arguments, every $z$ off the imaginary axis has
$$
\boxed{\rho(z,Bz)>\rho(i,Bi).}
$$
Back to article page
