Uncovering the Error in Finding the Mobius Transformation for a Circle Mapping

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I'm looking for the error in my understanding here, not help with the problem itself. I'm making some kind of mistake, so I've listed out everything I think I know, and I'm hoping someone can either tell me what I'm misunderstanding, or tell me there is an error in the problem statement:

Problem Statement: find the Mobius transformation taking the circle |z|=1 to |z+2|=1 such that T(-1)=-3 and T(i)=-1.

What I think:

*Both the circle and its image have radius 1, with the first centered at the origin and the second at -2.

*The additional points we are given mappings for are -1 and i, which are the endpoints to an arc of a quarter circle. These are mapping to -3 and -1 which are the bounds for the half circle. This implies the mapping is going around twice for once around the circle being mapped, which would imply it is not a bijection, and Mobius transformations are bijections.
 
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mimsy57 said:
I'm looking for the error in my understanding here, not help with the problem itself. I'm making some kind of mistake, so I've listed out everything I think I know, and I'm hoping someone can either tell me what I'm misunderstanding, or tell me there is an error in the problem statement:

Problem Statement: find the Mobius transformation taking the circle |z|=1 to |z+2|=1 such that T(-1)=-3 and T(i)=-1.

What I think:

*Both the circle and its image have radius 1, with the first centered at the origin and the second at -2.

*The additional points we are given mappings for are -1 and i, which are the endpoints to an arc of a quarter circle. These are mapping to -3 and -1 which are the bounds for the half circle. This implies the mapping is going around twice for once around the circle being mapped, which would imply it is not a bijection, and Mobius transformations are bijections.

Sure you can do it. You can map any three points in the complex plane to any other three points with a Mobius transformation, right? Just pick a third point on each of the two circles and that will fix a transformation. I haven't tried to visualize it in terms of dilations, rotations, etc.
 
Yes about the three points. My problem isn't so much with that, I'm okay calculating it, I'm just bothered by how it can work.

If I translate a circle, rotate a circle, or dilate a circle, or invert a circle, I don't see how points on a quarter arc could move to a half arc. Does my visualization problem make sense? I have to be understanding something incorrectly. Maybe inversion since the other three are so simple to visualize.

I tried finding examples, but in every example, the points stay in the same relative (for lack of a better word) location
 
mimsy57 said:
Yes about the three points. My problem isn't so much with that, I'm okay calculating it, I'm just bothered by how it can work.

If I translate a circle, rotate a circle, or dilate a circle, or invert a circle, I don't see how points on a quarter arc could move to a half arc. Does my visualization problem make sense? I have to be understanding something incorrectly. Maybe inversion since the other three are so simple to visualize.

I tried finding examples, but in every example, the points stay in the same relative (for lack of a better word) location

Yes, it's inversions. Replace |z+2|=1 with, say |z+100|=99. It should be visually clear that most of the points on the circle map to points near 0. z=(-1) and points nearby don't. It stretches angles on the circle. It's a good thing to visualize, keep it up!
 
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Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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