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I need to find a map that takes the region between two circles $|z|=1$ and $|z-1/4| = 1/4$ to an annulus $a<|z|<1$.
Now I know that the bilinear transform $f(z) = \frac{z-\alpha}{1-\bar{\alpha}z}$ maps the unit disk to itself, so I've constructed 2 maps, one that takes the unit disk to itself, and another that takes the inside of the smaller circle to the disk $|z| < a$:

$f_1(z) = \frac{z-\alpha_1}{1-\bar{\alpha}_1z}$

$f_2(z) = a\big(\frac{4(z-1/4)-\alpha_2}{1-\bar{\alpha}_2(z-1/4)}\big)$

The idea is that if I can find a map that simultaneously does these two things, I'll have my answer. Unfortunately, I'm not sure how to 'combine' these maps. Obviously composition isn't the answer since I want simultaneous mapping, not sequential mappings.

Can someone help me equate these?

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Note that the map $f(z)={z-\alpha \over 1-\bar\alpha z}$ does not only map the unit disk to itself, it also maps circles to circles. If you let $\alpha=a$ with $0<a<1$, then $f$ will map the circle $|z|=a$ to a circle which passes through the origin, just as the circle $|z-{1\over 4}|={1\over 4}$ does. So try to choose $a$ so that the image circle has radius $1\over 4$, and follow up with a rotation if necessary, and you will have the inverse of the map you are looking for.

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