TITLE: Calculate Supremum of the taylor series of an holomorphic function
QUESTION [1 upvotes]: let $t\in [0,1[$ and the function: $f_t(z) = \frac{z-t}{1-zt}$ which is holomorphic in the Open Disc $D(0,1)$.
I am trying to find $\sup_{|z|<1}|\sum_{k=0}^{+\infty} a_n z^n|$ where $\sum_{k=0}^{+\infty} a_n z^n$ is the Taylor series at $0$ of $f_t(z)$
What I did:
I calculated Taylor series at $0$ of $f_t(z)$:
$f_t(z)=-t + \sum_{k \geq 1}^{+\infty}z^k(t^{k-1} - t^{k})$
From here I don't know how to follow-up, Any help, a hint is much appreciated.
Many thanks!

REPLY [1 votes]: It's holomorphic, so it can't have an interior maximum. Since $|t|<1$, the definition of $f_{t}$ extends perfectly well to $|z|=1$, so let $z = e^{i\theta}$ and investigate:
$$|f_{t}|^{2} = \frac{(z-t)(\bar{z}-t)}{(1-zt)(1-\bar{z}t)} = \frac{1-2\cos \theta t + t^{2}}{1-2\cos \theta t + t^{2}} = 1$$
so the supremum is $1$.    
Alternatively, note that $f_{t}$ is a Mobius map, and hence sends circles and lines to circles and lines, and is determined by the images of any three distinct points in $\mathbb{C}$. The map sends $1\mapsto 1$, $-1\mapsto -1$ and $i \mapsto \frac{i-t}{1-it}$ (which has modulus $1$). So it sends the unit circle to itself, and by continuity it sends the whole unit disk to itself. In particular, $|z| \le 1 \implies |f(z)| \le 1$ (showing the supremum is attained is straightforward).
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