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- Thread starter pantboio
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- Feb 7, 2012

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A linear fractional transformation is uniquely specified by three points and their images. You have already found the (unique) LFT taking $1$ to $1$, $-1$ to $-1$ and $i$ to $-i$, namely $f(z) = 1/z$. Since this does not take the unit disc to itself, you are right to conclude that $\mathscr{F}=\varnothing.$

- Feb 13, 2012

- 1,704

Why $f(z)=\frac{1}{z}$ don't map the unit disk in itself?... Setting $z= e^{i\ \theta}$ is $f(z)=e^{-i\ \theta}$... right?... and, of course, $f(z)= \frac{1}{z}$ when is a=0, b=1, c=1, d=0...

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$\chi$ $\sigma$

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- #4

- Feb 7, 2012

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In English, "disc" means "circle together with its interior". Maybe Italian distinguishes between these two concepts in a different way?Why $f(z)=\frac{1}{z}$ don't map the unit disk in itself?... Setting $z= e^{i\ \theta}$ is $f(z)=e^{-i\ \theta}$... right?...

Kind regards,

Opalg

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- #5

By unit disc i mean the set of complex numbers with modulus less than 1, not only the unit circleWhy $f(z)=\frac{1}{z}$ don't map the unit disk in itself?... Setting $z= e^{i\ \theta}$ is $f(z)=e^{-i\ \theta}$... right?...

Kind regards

$\chi$ $\sigma$

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- #6

i'm no longer sure if what i said is correct. We have seen that such a function doesn't exist. But suppose it exists, with this property:I'm trying to find the set $\mathscr{F}$ of all linear fractional transformations (l.f.t.) of the unit disc D in itself which map 1 in 1, -1 in -1 and i in -i. By l.f.t. i mean a function$$f(z)=\frac{az+b}{cz+d}$$with $a,b,c,d\in\mathbb C$, $ad-bc\neq0$.I know that this kind of maps sends lines to lines and circles to circles.In this particular case, $f$ fixes the real axis.

1) $f$ is a fractional linear transformation

2)$f$ map the unit disc onto itself

3)$f$ fixes 1 and -1, and maps i to -i

Can i conclude that $f$ fixes the real axis? Couldn't it happen that $f$ maps the real axis into a circle?

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- #7

- Feb 7, 2012

- 2,785

You have already seen that there is no function satisfying all three of those conditions. So forget (2) and part of (3), and suppose that $f$ is a fractional linear transformation that fixes 1 and -1. Then yes, it could happen that $f$ maps the real axis into a circle. If $L$ denotes the set of lines and $C$ denotes the set of circles, then a fractional linear transformation always maps $L\cup C$ to $L\cup C$. But it does not need to map $L$ to $L$, or $C$ to $C$.i'm no longer sure if what i said is correct. We have seen that such a function doesn't exist. But suppose it exists, with this property:

1) $f$ is a fractional linear transformation

2)$f$ map the unit disc onto itself

3)$f$ fixes 1 and -1, and maps i to -i

Can i conclude that $f$ fixes the real axis? Couldn't it happen that $f$ maps the real axis into a circle?

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tell me if i'm right: substitute condition 3) with the following:3')$f$ fixes 1. SoYou have already seen that there is no function satisfying all three of those conditions. So forget (2) and part of (3), and suppose that $f$ is a fractional linear transformation that fixes 1 and -1. Then yes, it could happen that $f$ maps the real axis into a circle. If $L$ denotes the set of lines and $C$ denotes the set of circles, then a fractional linear transformation always maps $L\cup C$ to $L\cup C$. But it does not need to map $L$ to $L$, or $C$ to $C$.

1)$f$ is fractional linear map

2)$f$ maps the unit disc in itself

3') $f$ fixes 1

I want to prove that, in this particular case, $f$ maps real axis in real axis (so a line into a line, non into a circle). I remember that all fractional linear of unit disc have the form (up to a rotation)

$$f(z)=\frac{z-a}{1-\overline a z}$$

for $|a|<1$. Condition 3') implies that $a$ must be real. Hence $f$ maps real numbers to real numbers

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- #9

- Feb 7, 2012

- 2,785

That argument is incorrect, because you are neglecting the rotation. The general form for an FLT is $$f(z)=\lambda\frac{z-a}{1-\overline a z},$$ where $|a|<1$ and $|\lambda|=1.$ It is quite possible for the fraction $\frac{z-a}{1-\overline a z}$ to shift the point 1 to some other point on the circle, and then for the rotation by $\lambda$ to shift it back to 1.tell me if i'm right: substitute condition 3) with the following:3')$f$ fixes 1. So

1)$f$ is fractional linear map

2)$f$ maps the unit disc in itself

3') $f$ fixes 1

I want to prove that, in this particular case, $f$ maps real axis in real axis (so a line into a line, non into a circle). I remember that all fractional linear of unit disc have the form (up to a rotation)

$$f(z)=\frac{z-a}{1-\overline a z}$$

for $|a|<1$. Condition 3') implies that $a$ must be real. Hence $f$ maps real numbers to real numbers

For example, if $a = \frac12(1+i)$ and $\lambda = i$ then the map becomes $$f(z)=\frac{i(2z-1-i)}{2-(1-i)z}.$$ You can check that this fixes $1$ but not $-1$, and it takes the real axis to a circle.

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- #10

Thank you very much. Let's see if i can state something correct. Let $f$ satisfie the following conditions:That argument is incorrect, because you are neglecting the rotation. The general form for an FLT is $$f(z)=\lambda\frac{z-a}{1-\overline a z},$$ where $|a|<1$ and $|\lambda|=1.$ It is quite possible for the fraction $\frac{z-a}{1-\overline a z}$ to shift the point 1 to some other point on the circle, and then for the rotation by $\lambda$ to shift it back to 1.

For example, if $a = \frac12(1+i)$ and $\lambda = i$ then the map becomes $$f(z)=\frac{2z-1-i}{2-(1-i)z}.$$ You can check that this fixes $1$ but not $-1$, and it takes the real axis to a circle.

1)$f$ is a linear fractional map;

2)$f$ maps the unit disc in itself;

3)$f$ fixes 1 and -1.

Hence $f$ is the identity.

proof: $f$ has the form

$$f(z)=\lambda\frac{z-a}{1-\overline a z}$$

for some $|\lambda|=1$. Condition 3) implies

$$\lambda=1$$

Thus

$$f(z)=\frac{z-a}{1-\overline a z}$$

and again by 3) i get $a$ real. Hence $f$ has infinite fixed points in the unit disc ( the real segment (-1,1) ). By identity principle for holomorphic function, i conclude $f=id$

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- #11

- Feb 7, 2012

- 2,785

Sorry to disappoint you, but this is still not right.Thank you very much. Let's see if i can state something correct. Let $f$ satisfie the following conditions:

1)$f$ is a linear fractional map;

2)$f$ maps the unit disc in itself;

3)$f$ fixes 1 and -1.

Hence $f$ is the identity.

proof: $f$ has the form

$$f(z)=\lambda\frac{z-a}{1-\overline a z}$$

for some $|\lambda|=1$. Condition 3) implies

$$\lambda=1$$

Thus

$$f(z)=\frac{z-a}{1-\overline a z}$$

and again by 3) i get $a$ real. Hence $f$ has infinite fixed points in the unit disc ( the real segment (-1,1) ). By identity principle for holomorphic function, i conclude $f=id$

You are correct that those conditions imply that $f(z)=\frac{z-a}{1-a z}$ for some real $a$ with $|a|<1$. But it does not follow that $f$ fixes the segment (-1,1), unless $a=0$. More precisely, it maps that segment to itself, but it does not fix each point in the segment. In fact, it stretches part of the segment and shrinks another part.

You can see for example that $f(0)=-a$. The only value of $a$ for which $f$ fixes the whole segment is $a=0$. In that case only, $f$ becomes the identity map.

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