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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.15

(Exer 3.15) Using the cross ratio, with different choices of $z_k$, find two different Möbius transformations that transform $C[1 + i, 1]$ onto the real axis plus $\infty$. In each case, find the image of the center of the circle.

Note: $C[i+1,1] = \{|z-(i+1)|=1\}$ is the circle with centre (1,1) and radius 1


(Part I) I'm going to try $g=[z,(2,1),(1,2),(0,1)], -g=[z,(2,1),(1,0),(0,1)]$. Please point out and explain any errors.

I observe that that $\pm g$ map $C[i+1,1] \to \mathbb R$. By plugging in $z=(1+i)+e^{i \phi}$, we can show that $\pm g((1+i)+e^{i \phi}) \in \mathbb R$. Then we extend $\pm g$ to be defined for the singularity $z=z_3=i=(0,1)$.

Finally, the images are (WA: plus, minus)

$$\pm g(1+i) = \pm i$$

Did I go wrong anywhere, and why?

(Part II) Can I actually pick any 3 points on $C[i+1,1]$ to form a required cross-ratio?

By inputting in the 3 points used to make up a cross ratio, into the cross ratio, it seems that the cross ratio will output $0,1,\infty$, soooo it looks like all cross ratios will map 3 distinct points on a circle to the real line.

$\therefore,$ although the question asks for 2 Möbius transformations, there are uncountably $\infty$ly many Möbius transformations/cross-ratios that work, so it seems?

So how will we compute the images $[i+1,z_1,z_2,z_3]$? Is it simply that with no particular pattern?

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    $\begingroup$ Why do you think you have gone wrong? $\endgroup$ – Somos Jul 31 '18 at 13:39
  • $\begingroup$ @Somos I think my ±1 was wrong. Edited. If it's $\pm 1$, then I think I did something wrong because I'm getting something from somewhere inside the circle rather than on the circle and then it gets mapped to the real line...? $\endgroup$ – BCLC Aug 9 '18 at 5:19
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For Part I, you are almost there. The instructions ask you to try

Using the cross ratio, with different choices of $\ z_i$, find two different Möbius transformations

which you tried to do using $\ g=[z,(2,1),(1,2),(0,1)], \ -g=[z,(2,1),(1,0),(0,1)] \ $ which produces $\ g(z) = (-i z -1 +2i)/(z-i). \ $ Even though $\ -g(z) \ $ is a Möbius transformation, you did not get it using a different choice of $\ z_i$. Notice that, by the characteristic property of cross ratio, we immediately know that $\ g(2+i) = 0, \ g(1+2i) = 1, \ g(i) = \infty. \ $ and also $\ g(1) = -1, \ g(1+i) = i. \ $ It follows that the Möbius transformation $\ h=[z,(2,1),(1,0),(0,1)] \ $ is the same as $\ -g. \ $ Now $\ g \ $ maps the center of the circle to $\ i \ $, and $\ -g = h \ $ maps it to $\ -i. \ $

For part II, you can pick any three distinct points on the circle and each such choice will produce different Möbius transformations which share the common property that they map the circle one-to-one onto the real line. Depending on the order of the three points, the interior of the circle, including the center, will get mapped one-to-one onto either the upper or lower half plane.

You can think of all Möbius transformations as being conjugate to two cases depending on the number of fixed points. Case 1 with one fixed point is a parabolic transformation $\ f(z) = z + a \ $ for some complex constant $\ a $ and the fixed point is at infinity. Case 2 with two fixed points is $\ f(z) = b\ z \ $ for some complex constant $\ b $ and the two fixed points are zero and infinity.

There is much more information about all this in the Wikipedia article Möbius transformation.

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  • $\begingroup$ Thanks Somos! For Part I, are you interpreting that that each of the 3 points for $h$ should have been different from each of the 3 points of $g$, and mine doesn't work because I changed only 1 point? $\endgroup$ – BCLC Aug 10 '18 at 10:11
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    $\begingroup$ I think you have t o take the instructions literally. Given three distinct points, then $f =[z,z_1,z_2,z_3]$ is the map determined by them. Conversely, given $f$, then the three points are the preimages of $(0,1,\infty).$ It is accidental that $-g$ is also a tranformation, but for a different triple of points, and since $-0 = 0$ and $-\infty = \infty,$ two of those points are the same as the original triple. I think that the question requires you to use the different triple of points. $\endgroup$ – Somos Aug 10 '18 at 11:25
  • $\begingroup$ I'm too tired to analyse that, but ayt thanks! ^-^ $\endgroup$ – BCLC Aug 10 '18 at 11:41
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Write the identity stating that the cross-ratio is preserved: $$\frac {z - z_1} {z - z_3} \frac {z_2 - z_3} {z_2 - z_1} = \frac {w - w_1} {w - w_3} \frac {w_2 - w_3} {w_2 - w_1}.$$ Take $z_1, z_2, z_3$ on the circle and $w_1, w_2, w_3$ on the real line. If one of the $w_i$ is infinity, cancel the two factors containing this $w_i$. The function $w(z)$ is the answer.

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    $\begingroup$ The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$. $\endgroup$ – Maxim Aug 9 '18 at 4:03
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    $\begingroup$ Take $w_1 = 0, w_2 = \pm 1, w_3 = \infty$. The rhs becomes $\pm w$, therefore $w = \pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice. $\endgroup$ – Maxim Aug 9 '18 at 4:15
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    $\begingroup$ Strictly speaking, we are not making any assumptions about $w$, we've constructed the function $w = w(z)$. A linear-fractional transform maps circles to circles, and a circle is fixed by three points. $\endgroup$ – Maxim Aug 9 '18 at 4:22
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    $\begingroup$ Yes, there are uncountably many choices. Consider that we can add any real number to $w(z)$. $\endgroup$ – Maxim Aug 9 '18 at 4:49
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    $\begingroup$ You know that $w(z_i) = w_i$. You know that a circle is mapped to a circle (a straight line can be treated as a circle going through $\infty$). Why do you think that some choice of $z_i$ might not work? $\endgroup$ – Maxim Aug 9 '18 at 4:56
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Part I is right.

Part II No! They have to be distinct. Otherwise, we don't have that they are Möbius transformations. Other than that, yes. Also, I don't see any particular pattern with the images.

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