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

Question

(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, $Tcirc f$ is entire and bounded, by Liouville theorem, $Tcirc f$ is  constant. If $f$ is not constant, by open mapping theorem, $f$ is open map therefore $Tcirc f$ is open map but $Tcirc f$ 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,
$f(z)=1-1/z, f:mathbb{C}-[0,1]to mathbb{C}-]-infty,0]$
$g(z)=sqrt{z}, g:mathbb{C}-]-infty,0]to left{z:Re(z)>0right}$
$h(z)=frac{z-1}{z+1}, h:left{z:Re(z)>0right}to B(0,1)$
Therefore $T=fcirc gcirc h$ holds (i)
Why $f:mathbb{C}-[0,1]to mathbb{C}-]-infty,0]?$ Exists a intuitive idea to proves this?

Martin R · Answer

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 ...

H. H. Rugh · Answer

Hint: Let $s(r e^{itheta}) = sqrt{r} e^{itheta/2}$, $|theta|<pi$, $r>0$ 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. $wmapsto (w-a)/(w+a)$, for some $a>0$.

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

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