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(i) Construct a conformal map $T$ such that $T(\mathbb{C}-[0,1])\subset B(0,1).$

(ii) If $f$ is entire such that $f(\mathbb{C})\subset \mathbb{C}-[0,1]$, show that $f$ is constant.

My solution for (ii). If $f$ is entire, $T\circ f$ is entire and bounded, by Liouville theorem, $T\circ f$ is constant. If $f$ is not constant, by open mapping theorem, $f$ is open map therefore $T\circ f$ is open map but $T\circ f$ is constant a contradiction. This is correct?

How find $T$ in (a)?

Actualization 1.

From If $f(\mathbb{C})\subset \mathbb{C}-[0,1]$ then $f$ is constant,

$f(z)=1-1/z, f:\mathbb{C}-[0,1]\to \mathbb{C}-]-\infty,0]$

$g(z)=\sqrt{z}, g:\mathbb{C}-]-\infty,0]\to \left\{z:Re(z)>0\right\}$

$h(z)=\frac{z-1}{z+1}, h:\left\{z:Re(z)>0\right\}\to B(0,1)$

Therefore $T=f\circ g\circ h$ holds (i)

Why $f:\mathbb{C}-[0,1]\to \mathbb{C}-]-\infty,0]?$ Exists a intuitive idea to proves this?

eraldcoil
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2 Answers2

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Hint: Let $s(r e^{i\theta}) = \sqrt{r} e^{i\theta/2}$, $|\theta|<\pi$, $r>0$ and show that $\frac{s(z)}{s(z-1)}$ may be extended to become a holomorphic function on $ {\Bbb C}\setminus [0.1]$ into the right halfplane. Then post-compose with e.g. $w\mapsto (w-a)/(w+a)$, for some $a>0$.

H. H. Rugh
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  • It seems that your $s$ is not defined on all of $\mathbb{C}-[0,1]$, which includes the negative real axis with $\theta = \pi$. (There is no holomophic square root in that domain.) – Or do you mean that $\frac{s(z)}{s(z-1)}$ can be continued analytically? – Martin R Jul 21 '20 at 18:19
  • Yes, it can be continued as a holomorphic function. I have clarified. Anyway, your solution is more elegant. – H. H. Rugh Jul 21 '20 at 18:34
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The Möbius transformation $\phi(z) = \frac{z}{z-1}$ maps the line segment $[0, 1]$ onto the negative real axis (plus the point at infinity). On the complement you can define a holomorphic branch $s$ of the square root, so that $s \circ \phi$ maps $\mathbb{C}-[0,1]$ into the right halfplane.

Now find a Möbius transformation which maps the right halfplane into the unit disk ...

Martin R
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