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I am trying to solve the following question:

Suppose $C_{1}$ and $C_{2}$ are two tangent circles with $C_{2}$ in the interior of $C_{1}$: Show that an infinite number of circles can be placed in the region between $C_{1}$ and $C_{2}$ , each tangent to $C_{1}$ and $C_{2}$ each tangent to the next. Show also that the points of tangency of these circles each with the next lie on a circle.

There was a hint for this question that we need to show that $f(z)=\frac{4}{z}$ maps the region $A=\{z: |z-1|>1 \,\text {and} \, |z-2|<2 \}$ one-to-one and onto the strip $B=\{z: 1<Rez <2\}$ which I did. The images of the smaller circle in the hint is the line $x=1$ and $x=2$ is the image of the other circle. So I guess the images of the tangent circles above are the vertical lines in the region so there are infinitely many of them and the image of the circle described in the last sentence of the question is the circle which is tangent to the lines $x=2$ and $x=4$ with the center $(3,0)$ if we placed the two circles in the question as the two circles in the hint. But I don't know how to prove these. Any help would be great. Thanks

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You've got it nearly right: the "tangent" circles in the original map to circles in the transformed plane that that stack up in the strip (like pennies in a coin-slot), like this:

|O|
|O| 
|O| 
|O| 
|O| 
|O| 
|O| 

where each of the "O"s is supposed to be tangent to the ones above and below it, and to the two sides of the strip.

Your idea, that the tangent circles are also vertical lines -- can't be right, for those vertical lines are not tangent to the two other vertical lines, nor to each other.

I think you're a little confused about the last sentence, although I might just be misreading what you wrote. The centers of these extra circles line up along the mid-line of the strip, which is transformed (as you can show) to a circle. So that "circle" (before untransforming with $z = \frac{1}{4w}$) is the line $x = 3$.

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  • $\begingroup$ I got the idea, thanks a lot. $\endgroup$ – user135582 Jan 5 '15 at 18:50

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