Moebius circle to circle operators

Question

To map

$$ (x-h)^2+(y-k)^2 = r^2 $$ to $$ (x-H)^2+(y-K)^2 = R^2 $$

in the complex plane is there a easy/quick way to determine $ a,b,c,d $ in terms of $ A,B,C,D $

the Moebius transformation $ w =\dfrac{az+b}{cz+d} \,? $

EDIT1:

(Wlog if take $d=1)$ I hope to see column matrix /vector $ f,f^{-1} $ algebraically expressed:

$$ (H,K,R)= f(a,b,c) (h,k,r) \, ; \,(h,k,r)\,=f^{-1}(a,b,c)(H,K,R)$$

If $H, K , R$ can be$h , k ,r$ then identity $w = z$ works fine — IrbidMath, Mar 18 '17 at 05:55
Really? I thought it would be involved . Can you please explain it through an answer? — Narasimham, Mar 18 '17 at 05:58

IrbidMath · Answer 1 · 2017-03-18T19:59:23.730

0

I will solve an example hope it will help you to understand what Mobius transform do.

If we want to map the unit circle $x^2 + y^2 =1$ which can be written as $\mid z \mid = 1 $ into the circle $O: (x-2)^2 + (y-2)^2 = 4$ the circle with radius $2$ and center $(2,2)$ using Mobius maps. We can find many maps can do that.

To find one we will choose any three points choose any three points on the unit circle say $1, i , -1$ and any three points on the $O$ say $2 , 2i , 2+4i$. Our map will take ( I choose this ) $$1 \longmapsto 2 ,\;\;\;\ \ i\longmapsto 2i,\;\;\;\; -1 \longmapsto 2+4i$$

I will use the cross ratio (http://www.math.bas.bg/~rkovach/lectures/Moebius.pdf) $$(z,1,i,-1) = (w,2,2i,2+4i)$$ $$\frac{(z-1)(i--1)}{(z--1)(i-1)}= \frac{(w-2)(2i-(2+4i))}{(w-(2+4i))(2i-2)}$$ $$\frac{(z-1)(i+1)}{(z+1)(i-1)} = \frac{(w-2)(-2)(i+1)}{(w-2-4i)(2)(i-1)}$$ Solve it for $w$ $$w = 2 +2i -\frac{2i}{z}= \frac{(2+2i)z-2i}{z}$$ $a=2+2i , b = -2i , c = 1 , d = 0$

Mobius transform maps circles and lines into circles and lines and the map determined by three points.

The transform $$w= \frac{R}{r}\left(z-(h+i k)\right)+H+iK$$ Maps $(h, k-r)\longmapsto (H, K-R), \; \; (h, k+r)\longmapsto (H, K+R), \; \; (h+r, k)\longmapsto (H+R, K)$

edited Mar 18 '17 at 19:59

answered Mar 18 '17 at 17:45

IrbidMath

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Thanks, I know what it does, but wished to see the algebraic result. Please see my edit. – Narasimham Mar 18 '17 at 18:24
you want Mobius that transform $(x-h)^2 + (y-k)^2 = r^2$ into $(x-H)^2 + (y-K)^2 = R^2$ i.e giving $a,b,c,d$ values with respect to $h,k,r , H,K,R$ ? – IrbidMath Mar 18 '17 at 18:35
Yes, whatever it may be to see transformation complexity or simplicity.. – Narasimham Mar 18 '17 at 19:07
I edited my post check it please – IrbidMath Mar 18 '17 at 20:00
I choosed three points on the circle with small and maps them into the capitals – IrbidMath Mar 18 '17 at 20:01
For the inverse solve it for $z$ – IrbidMath Mar 18 '17 at 20:24
Is it possible to algebraically express $R$ in terms of $ r,a,b,c, .? $ along with another two equations? – Narasimham Mar 19 '17 at 11:38
$w= \frac{R}{r}\left(z-(h+i k)\right)+H+iK$ in this function $a= \frac{R}{r}, c=0, d=1$ and $b=\frac{R}{r}\left(-(h+i k)\right)+H+iK$ solve it for $R$ we get $R = \frac{r(H+iK -b)}{h+ik}$ or easily it is $R = ra$ – IrbidMath Mar 19 '17 at 13:57

Moebius circle to circle operators

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