Mapping Real Axis and Im(z)=1 to Tangent Circles?

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Homework Statement


Find a Mobius transformation that maps the real axis to the circle |z-1|=1, and the line Im(z)=1 to the circle |z-2|=1

Homework Equations


A mobius transformation is one of the form z\rightarrow\frac{az+b}{cz+d} on the extended complex plane

The Attempt at a Solution



My attempt pretty much comes out to... well, mobius transformations are bijections, but the two lines I'm given in the pre-image only intersect at infinity. So how do I get the two intersection for the circles? I'm thinking maybe I'm really dumb, and they're just tangent, but it looks to me as if they meet at \frac{3}{2}\pm\frac{\sqrt{3}}{2}i
 
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You're right, there does seem to be a problem with the question.
 
I asked someone about it, and it's actually |z|=1 and |z-2|=1, so they're tangent at a point (I was reading off the circle descriptions from the question above previously... whoops).

I was having enough trouble with these stupid things without making them even more difficult for myself.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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