Exponential map in the Poincaré upper half plane I have a question regarding the Poincaré upper half plane. Is there a simple way to express the exponential map? I have been looking unsuccessfully on internet for an expression...
Thanks for any help on how to get one.
 A: EDIT: essay questions: what is the exponential map in the ordinary plane? What is the exponential map on the ordinary unit sphere, say at point $(a,b,c)$ with $a^2 + b^2 + c^2 = 1,$ in tangent direction $(d,e,f),$ so that $ad+be+cf=0?$
ORIGINAL: There are two types of geodesics, vertical lines and semicircles with center on the real axis.  All you need for the exponential map is the way to parametrize these at unit speed.
$$ A + i e^t  $$
$$ A + B \tanh t + i B \operatorname{sech } t  $$
for real constants $A,B$ with $B> 0.$
$$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= $$
$$  \cosh^2 x - \sinh^2 x = 1 $$
$$ 1 - \tanh^2 x = \mbox{sech}^2 \; x    $$
$$ \cosh x + \sinh x = e^x  $$
$$ \cosh x - \sinh x = e^{-x}  $$
$$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y   $$
$$ \cosh  (x+y) = \cosh x \cosh y + \sinh x \sinh y $$
$$  \tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}  $$
$$ \sinh \frac{x}{2} = \pm \sqrt { \frac{\cosh x - 1}{2}   } $$
$$ \cosh \frac{x}{2} =  \sqrt { \frac{\cosh x + 1}{2}   } $$
$$ \tanh \frac{x}{2} = \pm \sqrt { \frac{\cosh x - 1}{\cosh x + 1}   } = \frac{\cosh x - 1}{\sinh x} = \frac{\sinh x}{\cosh x + 1} $$
$$  \mbox{arcsinh} \; x = \log \left( x + \sqrt{x^2 + 1} \right) $$
$$  \mbox{arccosh} \; x = \log \left( x + \sqrt{x^2 - 1} \right), \; \; \; x \geq 1 $$
$$  \mbox{arctanh} \; x =  \frac{1}{2} \; \log \left(  \frac{1 + x}{1 - x} \right), \; \; \; |x | < 1 $$
