Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.