# Möbius Transforms that preserve $\mathbb{H}$

I know that every möbius transform that preserves the upper half plane is of the form

$m(z) = \frac{az+b}{cz+d}$, where $a,b,c,d \in \mathbb{R}$, or $m(z) = \frac{a\bar{z} + b}{c\bar{z} + d}$, where $a,b,c,d$ are all purely imaginary. In both cases $ad-bc = 1$.

However, I could be wrong but I see that most books in the literature tend to exclude the latter case involving the complex conjugate. Is there any reason why??

It is pretty important that I understand this, as I need to use these facts when calculating the distance between two points in the upper half plane model of 2 dimensional hyperbolic space.

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Your initial condition is necessary, but not sufficient. For example, $m(z)=-z$ is of this form, but it send $i$ to $-i$, so does not preserve $\mathbb{H}$. – Thomas Andrews Apr 21 '11 at 3:15
@ThomasAndrews The möbius transform $m(z) = \frac{-z+0}{0z + 1}$ has $ad - bc = -1$, which is not of the form above. – user38268 Apr 21 '11 at 3:25
Ah, I missed that condition somehow. Brain freeze. – Thomas Andrews Apr 21 '11 at 14:16

The formula involving conjugate is not a Möbius transformation. Another reason is that it is not holomorphic.

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Ok Thanks very much. That makes sense now. – user38268 Apr 21 '11 at 3:03