# How to show that $\phi_a(z)=\frac{z-a}{1-\bar{a}z}$ is mapping $\phi_a(z):\mathbb{D} \rightarrow \mathbb{D}$?

How to show that $\phi_a(z)=\frac{z-a}{1-\bar{a}z}$ is mapping $\phi_a(z):\mathbb{D} \rightarrow \mathbb{D}$? So how to show that if $|z| \leq 1,\forall z\in \mathbb{C}$, then $|\phi_a(z)| \leq 1$?

I tried to make standard stuff, but got stuck. $|\frac{z-a}{1-\bar{a}z}| =\frac{|z-a|}{|1-\bar{a}z|}...$

Just some hint please, whether it is able to solve the way I showed or not.

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1. It is only possible if $|a|<1$ otherwise $|\phi_a(0)|\geq 1$.

You can also use properties of the Möbius transformation. Not it suffices to show that three points out of $\mathbb{S}_1(0):=\{z ||z|=1\}$ are mapped onto itself. Choose for example$z=1,i,-1$. Then you know that the whole sphere is mapped onto itself. Explotiong that and $\phi_a(0)\subset \mathbb{D}$ you obtain by a property of Möbius transformations that $\mathbb{D}$ is mapped onto itself

However it is quiet similar to the solution of K.Stm. but with a slightly different argumentation

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If $|\tfrac{0-a}{1-\overline{a}·0}| < 1$, then $ϕ_a(0) ∈ \mathbb{D}$. For $|z| = 1$, you have: $$|z-a|^2 = (z-a)\overline{(z-a)} = 1 - a\overline{z} - z\overline{a} + a\overline{a} = (1 - \overline{a}z)\overline{(1-\overline{a}z)} = |1-\overline{a}z|^2$$

The inverse of $ϕ_a$ is given by $ϕ_{-a}$, so $|z| = 1 ⇔ |ϕ_a(z)| = 1$ or $ϕ_a(S_1) = S_1$, this already somehow shows the assertion, just make it precise:

Since $ϕ_a$ is continuous, the connected component of $0$ in $ℂ \setminus S_1$, viz $\mathbb{D}$, must lie in the connected component of $ϕ_a(0)$ in $ϕ_a (ℂ \setminus S_1) = ℂ \setminus S_1$. But if $ϕ_a(0) ∈ \mathbb{D}$, then $ϕ_a(\mathbb{D}) ⊂ \mathbb{D}$.

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$(1-\bar{a}z)\bar{(1-\bar{a}z)}=(1-\bar{a}z)(1-\bar{z}a) =1(1-\bar{z}a)-\bar{a}z(1-\bar{z}a)=1-\bar{z}a-\bar{a}z+\bar{a}az\bar{z}=...$ok. –  alvoutila Mar 9 '13 at 12:57