Why is the complex conjugate function (on the upper half plane) not a Möbius transformation?

Question

I want to show that the map $z\mapsto -\overline z$ is not a Möbius transformation. I've seen the answer to do something with analytic maps, which we haven't learned, so is there a more basic proof to this?

Answer 1

Although it seems a little misguided to prove that $z\to -\overline{z}$ is not a Mobius transformation without using the idea of analytic/holomorphic functions (the proof would be quick: $z\to -\overline{z}$ is not holomorphic, while all Mobius maps are holomorphic... done) we can do so.

Suppose $-\overline{z}={az+b\over cz+d}$ for some $a,b,c,d$, for all $z$, and plug in various $z$'s to see the impossibility... For example, from $z=0$, we have $b=0$. From $z=1$, we have $c+d=-(a+b)$. From $z=i$, ...

Answer 2

Whereas you can easily see that a nontrivial Möbius transformation has at most two fixed points, your transformation is fixed on all points of the imaginary axis.