I want to prove that if $\phi_a: B(0,1) \to \Bbb C$ is given by $\phi_a(z) = (z-a)/(1-\overline{a}z)$ with $|a| < 1$, then $|\phi_a(z)| < 1$.
Resist the itch on your finger urging you to close the question: I already took a look at this question and this one.
I'm supposed to prove things in the most elementary way possible (not by choice, sadly). Meaning: no Möbius transformations (I haven't studied them yet, anyway), no exponential maps, etc.
My attempt so far: $$ \begin{align} |\phi_a(z)| &< 1 \iff \left| \frac{z-a}{1-\overline{a}z}\right| <1 \iff \\ \iff |z-a| &< |1-\overline{a}z| \iff |z-a|^2 < |1-\overline{a}z|^2 \iff \\ \iff |z|^2-&2\,{\rm Re}(\overline{a}z)+|a|^2 < 1 -2\,{\rm Re}(\overline{a}z)+|\overline{a}z|^2 \iff \\ \iff &|z|^2 + |a|^2 < 1 + |a|^2|z|^2\end{align}$$
I'm stuck here. If $|a|^2 < |a|^2|z|^2$, then I could go back on these implications, but this is false (would give $|z|^2 > 1$).
Is there a way to save this, or there is another (elementary) approach? Thanks!
Best Answer
Hint: rearrange $|z|^2 + |a|^2 < 1 + |a|^2 |z|^2$ by putting the terms with $|z|^2$ on one side and the others on the other side, and divide...