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Let $D = \{ z \in \Bbb{C} : |z| < 1\}$ be the unit disc. Let $d: D \times D \to \Bbb{R}$ be the function defined by

$$d(z_1, z_2) = \frac{1}{2} \ln \frac{|1−z_3 z_2|+ |z_1−z_2|}{|1−z_3 z_2|-|z_1−z_2|},$$

where $z_3$ is the conjugate of $z_1$.

We will take for granted that $(D, d)$ is a metric space.

is $(D, d)$ complete?

  • 2
    Hi there. Welcome to MSE. Without some editing, this post will probably get closed, so you'll want to take a look at this tutorial to see how to type mathematics on this site. http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-04
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    Let $\rho(z,w) = \Bigl\lvert\dfrac{z-w}{1-\overline{z}w}\Bigr\rvert$. Note that $$d(z,w) = \frac{1}{2}\ln \frac{1 + \rho(z,w)}{1-\rho(z,w)}.$$ Show that for every $d$-Cauchy sequence $(z_n)$ in $D$, there is an $r < 1$ with $\lvert z_n\rvert \leqslant r$ for all $n$. Then show that on the disk $\overline{D_r(0)} = \{ z : \lvert z\rvert \leqslant r\}$ $d$ is (Lipschitz-)equivalent to the Euclidean distance. – 2017-02-04
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    could you please elaborate more? – 2017-02-04

1 Answers 1

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Outlines only.

Let $(z_n)$ be a Cauchy sequence in $(D,d)$, then $$ \forall \varepsilon >0,\, \exists N;\, m,n> N\implies d(z_n,z_m)<\varepsilon .$$ I.$\,$ $ d(z_n,z_m)<\varepsilon$ implies $$\left|\frac{z_n-z_m}{1-\overline{z_n}z_m}\right|<\frac{e^{2\varepsilon}-1 }{e^{2\varepsilon }+1}=\varepsilon ^\prime.$$ II. From this we can deduce that $$ |z_n-z_m|<\varepsilon ^\prime (1+|z_n||z_m|)<2\varepsilon ^\prime.$$ In other words, $(z_n)$ is a Cauchy sequence in the usual Euclidean distance.

III. Let $n_0=N+1.$ Then for all $m> N$, $$ \left|\frac{z_{n_0}-z_m}{1-\overline{z_{n_0}}z_m}\right|<\varepsilon ^\prime\implies |z_m|<\frac{|z_{n_0}|+\varepsilon ^\prime}{1+|z_{n_0}|\varepsilon ^\prime}.$$ Let $r=\max \{|z_1|,|z_2|,...,|z_{n_0}|, \frac{|z_{n_0}|+\varepsilon ^\prime}{1+|z_{n_0}|\varepsilon ^\prime}\}<1$. Then $|z_n|