2
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As it is mentioned in a previous question of mine, one can construct a holomorphic bijection between the (open) unit disc $D=\{z\in{\bf C}: |z|<1\}$, and a domain $D\setminus\overline{B_{1/2}(-1/2)}$ where $B_r(z_0)$ denotes the ball of radius $r$ centered at $z_0$. One of such function is given by $$ F(z)=g\left(\frac{Log(g^{-1}(z))}{\pi}\right) $$ where $g(z)=\dfrac{i-z}{i+z}$.

Experiments suggest that $F$ maps the interval $(-1,1)$ to the boundary of the open ball $B_{2/3}(-1/3)$ and one can check numerically by taking sample points from the interval:

(*Definition of the function F*)
g[z_] := (I - z)/(I + z);
g1[z_] := InverseFunction[g][z];
F[z_] := g[Log[g1[z]]/Pi];

Rationalize[Abs[F[Range[-0.9, 0.9, 0.1]] + 1/3]]

Here is my question:

Can I do this symbolically?


The formula Mathematica returns by Abs[F[x]+1/3] is $$ \left| \frac{1}{3}+\frac{i-\frac{\log \left(-\frac{i (x-1)}{x+1}\right)}{\pi }}{\frac{\log \left(-\frac{i (x-1)}{x+1}\right)}{\pi }+i}\right| $$

Is there way to put a "constraint" (which shows that $x$ is a real number and $x\in(-1,1)$) to the expression Abs[F[x]+1/3] so that it would return the desired number $2/3$?

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4
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You need to define proper Assumptions in any of these forms:

Assuming[
 -1 < x < 1,
 FullSimplify[
  Abs[F[x] + 1/3]
  ]
 ]
(* 2/3 *)

Or

FullSimplify[
 Abs[F[x] + 1/3]
 , Assumptions -> (-1 < x < 1)
 ]
(* 2/3 *)

Or

 FullSimplify[
  Abs[F[x] + 1/3]
  , -1 < x < 1
 ]
(* 2/3 *)

You can read the documentation about Assumptions and Domains

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  • 1
    $\begingroup$ You can drop the Assumptions -> part for the second solution (and use the FullSimplify[expr, assum] form) $\endgroup$ – Lukas Lang Sep 18 '17 at 14:04
  • 2
    $\begingroup$ Altho it's more verbose, I prefer the first one to the others; it reads like a sentence to me. $\endgroup$ – J. M. will be back soon Sep 18 '17 at 14:12

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