It is well known that every Mobius map over the complex numbers can be written as the composition of $z+a$ , $1/z$ and $\lambda z$ where $a,\lambda$ are complex (perhaps using multiple compositions of the same type). However is it possible that every Mobius map can be written as the composition of maps of the form $z+a$ , $1/z$?
I am getting two different answers depending on how I think about it, I don't know which one is right.
Firstly as the group of Mobius maps is isomorphic to $\text{PSL}_{2}(\mathbb{C})$ we see if we can generate $\text{PSL}_{2}(\mathbb{C})$ using matrices of the form $\pmatrix{1 & a \\ 0 & 1}$ and $\pmatrix{0 & 1 \\ 1 & 0}$. However $\text{PSL}_{2}(\mathbb{C})$ has matrices of the form $\pmatrix{p & 0 \\ 0 & p^{-1}}, p > 0$ however $\pmatrix{1 & a \\ 0 & 1}$ and $\pmatrix{0 & 1 \\ 1 & 0}$ are similar to $\pmatrix{1 & 1 \\ 0 & 1}$ for $ a \neq 0$ and $\pmatrix{1 & 0 \\ 0 & 1}$ for $a = 0$ and no combination of those two will multiply to give $\pmatrix{p & 0 \\ 0 & p^{-1}}$. So this can't be done.
However I can also do the following: $$z \to \dfrac{1}{z} \to \dfrac{1}{z} + \sqrt{\lambda} = \dfrac{z\sqrt{\lambda}+1}{z} \to \dfrac{z}{z\sqrt{\lambda}+1} \to \dfrac{z}{z\sqrt{\lambda}+1} -\dfrac{1}{\sqrt{\lambda}} = \dfrac{1}{\sqrt{\lambda}(z\sqrt{\lambda}+1)} \to \sqrt{\lambda}(z\sqrt{\lambda}+1) = \lambda z+\sqrt{\lambda} \to \lambda z$$
This means I can write $\lambda z$ as a combination of $1/z$ , $z+a$ so they do indeed generate the Mobius group.
Can anyone tell me which one is the right way?
Thanks in advance.