Fixed Points in Mobius Transformation

Question

Let $$w(z)=\frac{e^z-1}{e^z+1}; \ z=x+iy$$ and $$w=u+iv$$ Then show that the image of the $y$-axis in the domain is $v$-axis in the co-domain. Does this contradict the result that a Mobius transformation with at least $3$ fixed points is a identity mapping ? Why ?

I have already proved the fact that $y$-axis is mapped onto $v$-axis by showing that $$w(iy)=i\frac{\sin y}{1+\cos y}$$

Now I need to show that this mapping does not contradict the property of a Moebius transformation. For that I would either have to show that $$\frac{\sin y}{1+\cos y}=y$$ has less than $3$ solutions or that $w(z)$ is not a Moebius transformation. Which one is it and how do I prove that ?

Answer 1

Your transformation is not a Mobius transformation.

Answer 2

A mobius transformation $w(z)=\frac{az+b}{cz+d}, ad-bc\ne 0$ has a unique simple pole at $z=-d/c$ whereas in your problem there are infinite poles $z=(2k+1)i\pi, k=0,1,2,...$ so their equivalence in $\mathbb C$ is impossible.