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Let $\mathbb{D}= \{ z\in \mathbb{C}: |z|<1\}$. For $t\in \mathbb{R}$, let $f_t$ denote the holomorphic function on $\mathbb{D}$ defined by $f_t(z)= (\frac{1+z}{1-z})^{it}$, $z\in \mathbb{D}$ with respect to the principal branch of the logarithm.

Show that there is $C>0$ such that for all $t\in \mathbb{R}$, we have $\sup_{z\in \mathbb{D}}|f_t(z)|<C^t$

I know that $\frac{1+z}{1-z}$ is a mobius transformation.

Also, we have $f_t(z)= (\frac{1+z}{1-z})^{it} = \exp(it\frac{\log(1+z)}{\log(1-z)})) = \exp(it\log (1+z))\exp(-it\log (1-z))) = (1+z)^{it} (1-z)^{-it}$.

From here, I'm not sure how to proceed to bound such function.

Thanks in advance!

Toasted_Brain
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1 Answers1

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You have $$ f_t(z) = \exp\left(it \log \left(\frac{1+z}{1-z}\right)\right) = \exp\left( it \log \left | \frac{1+z}{1-z}\right| - t \arg \frac{1+z}{1-z} \right) $$ where $\arg$ is the principal value of the argument. $z \mapsto \frac{1+z}{1-z}$ maps the unit disk to the right half-plane, so that its argument is in the range $(-\pi/2, \pi/2)$. It follows that $$ |f_t(z)| = \exp\left(- t \arg \frac{1+z}{1-z} \right) \le \exp\left( |t| \frac \pi2\right) = C^{|t|} $$ with $C= e^{\pi/2}$. That bound is sharp, because $\arg \frac{1+z}{1-z}$ can be arbitrarily close to $+\pi/2$ and to $-\pi/2$.

Martin R
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  • Why is the argument in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$, not $(-\pi, \pi)$? Are we still using the principal value of the argument $(-\pi, \pi)$ to begin with, or do we just have $(-\frac{\pi}{2}, \frac{\pi}{2})$ starting from the first line? Thanks for your help! – Toasted_Brain Aug 13 '22 at 23:04
  • @Toasted_Brain: $\frac{1+z}{1-z}$ is in the right half-plane, therefore its argument is in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$. – Martin R Aug 15 '22 at 16:45