Möbius Transformation: Have they argumentet correctly?

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There is an example in my book where they do something I am not sure why their argument work. First I will present the problem short in text, then I will present it again with pictures from the book:

Short describtion of the problem:

  1. The have a circle, and a fractional linear transformation. They want to see where the inside of the circle maps.
  2. They first look at the boundary of the circle. Since there is a pole here, they know that the circle is mapped to a line.
  3. They find out that the line is the imgainary axis.
  4. Then they use an argument presentet earlier. "A holomorphic function takes connected domains to connected domains". And because of this they say that our function have to map to either the left or the right half plane.

My problem is with 4. even though the region is connected, it doesn't have to be the entire left or right plane does it?

Description of problem with pictures:

Here is the example from the book: enter image description here

Here is what they refer to in 7.2:

enter image description here

Now theoretically speaking, how do they not know that we do not for example end up with the set:

$\{z: 50 < \Re(z)\le0 \}$. I mean this is also an open connected set. I guess that by putting in values you can show that it can't be this set either. But how can we exclude any open connected set on the left hand-side? Have they done it correctly, and there is something in the argument I am missing? Or do we need to put in values and argue in another way?

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The boundary of circle C maps to the boundary of the circle under the transform. We know that this circle maps to a line along the imaginary axis in the w-plane. Because $f(2+2i) = -\frac{i}{2}$ and $f(0) = 0$, as the when you traverse the circle C with the interior to the left, you travel from $2+2i$ to $0$, which corresponds to traveling from $-\frac{i}{2}$ to $0$ in the w-plane. Since the transform is conformal and preserves orientation, the interior will also be to the left. Because the function maps the circle C to a vertical line on the imaginary axis, the interior maps to all the points to the left of that line.