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Prove this Möbius function maps unit disc to itself bijectively.
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.9
I got $(a)$ and $(b)$. My attempt for $(c)$:
First, I interpret that $(c)$ is equivalent to $f_a(D[0,1]), f_-a(D[0,1]) subseteq D[0,1]$.
Next, for $f_a(D[0,1])$, my approach is to let $z=x+iy in D[0,1]$, i.e. $|z|^2 = x^2+y^2 < 1$ and then plug it in $f_a$:
$$f_a(x+iy) = frac(x+iy)-(Re(a)+i Im(a))1-(Re(a)+i Im(a))(x+iy) = fracAC+BDC^2+D^2 + ifracBC-ADC^2+D^2$$
where
$$A := x - Re(a)$$
$$B := y - Im(a)$$
$$C := 1- (xRe(a)+yIm(a))$$
$$D := xIm(a)-yRe(a)$$
Now $$f_a(x+iy) in D[0,1] iff |f_a(x+iy)| < 1 iff |f_a(x+iy)|^2 = fracA^2+B^2C^2+D^2 < 1$$
$$iff 0 < (1-|a|^2)(1-|z|^2).$$
Finally, for $f_-a(D[0,1])$, I hope that we will similarly have that
$$f_-a(x+iy) = frac(x+iy)+(Re(a)+i Im(a))1+(Re(a)-i Im(a))(x+iy) in D[0,1]$$
- Where have I gone wrong, and why?
- How could I have more efficiently shown that $f_a(x+iy), f_-a(x+iy) in D[0,1]$? Perhaps polar? Or is this exercise indeed meant to be gory?
complex-analysis mobius-transformation
asked Jul 28 at 10:02
BCLC
6,98621973
add a comment |Â
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.9
I got $(a)$ and $(b)$. My attempt for $(c)$:
First, I interpret that $(c)$ is equivalent to $f_a(D[0,1]), f_-a(D[0,1]) subseteq D[0,1]$.
Next, for $f_a(D[0,1])$, my approach is to let $z=x+iy in D[0,1]$, i.e. $|z|^2 = x^2+y^2 < 1$ and then plug it in $f_a$:
$$f_a(x+iy) = frac(x+iy)-(Re(a)+i Im(a))1-(Re(a)+i Im(a))(x+iy) = fracAC+BDC^2+D^2 + ifracBC-ADC^2+D^2$$
where
$$A := x - Re(a)$$
$$B := y - Im(a)$$
$$C := 1- (xRe(a)+yIm(a))$$
$$D := xIm(a)-yRe(a)$$
Now $$f_a(x+iy) in D[0,1] iff |f_a(x+iy)| < 1 iff |f_a(x+iy)|^2 = fracA^2+B^2C^2+D^2 < 1$$
$$iff 0 < (1-|a|^2)(1-|z|^2).$$
Finally, for $f_-a(D[0,1])$, I hope that we will similarly have that
$$f_-a(x+iy) = frac(x+iy)+(Re(a)+i Im(a))1+(Re(a)-i Im(a))(x+iy) in D[0,1]$$
- Where have I gone wrong, and why?
- How could I have more efficiently shown that $f_a(x+iy), f_-a(x+iy) in D[0,1]$? Perhaps polar? Or is this exercise indeed meant to be gory?
complex-analysis mobius-transformation
asked Jul 28 at 10:02
BCLC
6,98621973
See these very similar questions: math.stackexchange.com/questions/1227914/… and math.stackexchange.com/questions/343982/…
– Robert Z
Jul 28 at 10:27
2
Never write a complex number $z$ as $x+iy$ unless you really have to.
– Lord Shark the Unknown
Jul 28 at 10:33
@LordSharktheUnknown Thanks. Any consolation prize for my asking #2? :|
– BCLC
Jul 28 at 10:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.9
I got $(a)$ and $(b)$. My attempt for $(c)$:
First, I interpret that $(c)$ is equivalent to $f_a(D[0,1]), f_-a(D[0,1]) subseteq D[0,1]$.
Next, for $f_a(D[0,1])$, my approach is to let $z=x+iy in D[0,1]$, i.e. $|z|^2 = x^2+y^2 < 1$ and then plug it in $f_a$:
$$f_a(x+iy) = frac(x+iy)-(Re(a)+i Im(a))1-(Re(a)+i Im(a))(x+iy) = fracAC+BDC^2+D^2 + ifracBC-ADC^2+D^2$$
where
$$A := x - Re(a)$$
$$B := y - Im(a)$$
$$C := 1- (xRe(a)+yIm(a))$$
$$D := xIm(a)-yRe(a)$$
Now $$f_a(x+iy) in D[0,1] iff |f_a(x+iy)| < 1 iff |f_a(x+iy)|^2 = fracA^2+B^2C^2+D^2 < 1$$
$$iff 0 < (1-|a|^2)(1-|z|^2).$$
Finally, for $f_-a(D[0,1])$, I hope that we will similarly have that
$$f_-a(x+iy) = frac(x+iy)+(Re(a)+i Im(a))1+(Re(a)-i Im(a))(x+iy) in D[0,1]$$
- Where have I gone wrong, and why?
- How could I have more efficiently shown that $f_a(x+iy), f_-a(x+iy) in D[0,1]$? Perhaps polar? Or is this exercise indeed meant to be gory?
complex-analysis mobius-transformation
asked Jul 28 at 10:02
BCLC
6,98621973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.9
I got $(a)$ and $(b)$. My attempt for $(c)$:
First, I interpret that $(c)$ is equivalent to $f_a(D[0,1]), f_-a(D[0,1]) subseteq D[0,1]$.
Next, for $f_a(D[0,1])$, my approach is to let $z=x+iy in D[0,1]$, i.e. $|z|^2 = x^2+y^2 < 1$ and then plug it in $f_a$:
$$f_a(x+iy) = frac(x+iy)-(Re(a)+i Im(a))1-(Re(a)+i Im(a))(x+iy) = fracAC+BDC^2+D^2 + ifracBC-ADC^2+D^2$$
where
$$A := x - Re(a)$$
$$B := y - Im(a)$$
$$C := 1- (xRe(a)+yIm(a))$$
$$D := xIm(a)-yRe(a)$$
Now $$f_a(x+iy) in D[0,1] iff |f_a(x+iy)| < 1 iff |f_a(x+iy)|^2 = fracA^2+B^2C^2+D^2 < 1$$
$$iff 0 < (1-|a|^2)(1-|z|^2).$$
Finally, for $f_-a(D[0,1])$, I hope that we will similarly have that
$$f_-a(x+iy) = frac(x+iy)+(Re(a)+i Im(a))1+(Re(a)-i Im(a))(x+iy) in D[0,1]$$
- Where have I gone wrong, and why?
- How could I have more efficiently shown that $f_a(x+iy), f_-a(x+iy) in D[0,1]$? Perhaps polar? Or is this exercise indeed meant to be gory?
complex-analysis mobius-transformation
asked Jul 28 at 10:02
BCLC
6,98621973
edited Aug 5 at 11:53
asked Jul 28 at 10:02
BCLC
6,98621973
asked Jul 28 at 10:02
BCLC
6,98621973
asked Jul 28 at 10:02
BCLC
6,98621973
6,98621973
See these very similar questions: math.stackexchange.com/questions/1227914/… and math.stackexchange.com/questions/343982/…
– Robert Z
Jul 28 at 10:27
2
Never write a complex number $z$ as $x+iy$ unless you really have to.
– Lord Shark the Unknown
Jul 28 at 10:33
@LordSharktheUnknown Thanks. Any consolation prize for my asking #2? :|
– BCLC
Jul 28 at 10:54
add a comment |Â
See these very similar questions: math.stackexchange.com/questions/1227914/… and math.stackexchange.com/questions/343982/…
– Robert Z
Jul 28 at 10:27
2
Never write a complex number $z$ as $x+iy$ unless you really have to.
– Lord Shark the Unknown
Jul 28 at 10:33
@LordSharktheUnknown Thanks. Any consolation prize for my asking #2? :|
– BCLC
Jul 28 at 10:54
See these very similar questions: math.stackexchange.com/questions/1227914/… and math.stackexchange.com/questions/343982/…
– Robert Z
Jul 28 at 10:27
See these very similar questions: math.stackexchange.com/questions/1227914/… and math.stackexchange.com/questions/343982/…
– Robert Z
Jul 28 at 10:27
2
2
Never write a complex number $z$ as $x+iy$ unless you really have to.
– Lord Shark the Unknown
Jul 28 at 10:33
Never write a complex number $z$ as $x+iy$ unless you really have to.
– Lord Shark the Unknown
Jul 28 at 10:33
@LordSharktheUnknown Thanks. Any consolation prize for my asking #2? :|
– BCLC
Jul 28 at 10:54
@LordSharktheUnknown Thanks. Any consolation prize for my asking #2? :|
– BCLC
Jul 28 at 10:54
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
What you are supposed to prove is that $f_a$ is a bijection from the open unit disk $mathbb D$ onto itself. The first thing to check is whether $f_a(mathbbD)subsetmathbb D$. This is true, because, if $|z|<1$, thenbeginalignleft|fracz-a1-overline azright|^2&=frac(z-a)overline(z-a)left(1-overline azright)overlineleft(1-overline azright)\&=fracz^2\&=frac^2-2operatornameReleft(overline azright)+aendalignand thereforebeginalignleft|fracz-a1-overline azright|^2<1&iff|z|^2-2operatornameReleft(overline azright)+|a|^2<1-2operatornameReleft(overline azright)+|a|^2|z|^2\&iff|z|^2+|a|^2<1+|a|^2|z|^2\&iffbigl(1-|z|^2bigr)bigl(1-|a|^2bigr)>0,endalignwhich is true.
So, $f_a$ is indeed a map from $mathbb D$ into itself. But you have already checked that$$f_acirc f_-a=operatornameId_mathbb D=f_-acirc f_a.$$The first of these equalities implies that $f_a$ is surjective, whereas the second ont implies that it is injective. So, $f_a$ is a bijection.
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
I would prove that $|f_a(z)|<1 Rightarrow |z|<1$.
Assume $|f_a(x)|<1$. Since
$$ |f_a(z)| = left| fracz-a1-bar az right|
= frac1-bar az = frac1/bar a-z $$
we have
$$ |z-a| < |a| cdot |z-1/bar a| $$
This is now a geometric inequality about lengths in the complex plane. Furthermore the points $a$ and $1/bar a$ lie on the same ray from the origin, so it makes sense to declare that ray to be the $x$-axis of a new $xy$-coordinate system that we will calculate the lengths in! Then $a$ has coordinates $(A,0)$ for some $Ain(0,1)$, and $1/bar a$ is $(frac 1A, 0)$.
Our assumption is now
$$ sqrt(x-A)^2+y^2 < A sqrt(x-tfrac1A)^2+y^2. $$
Square both sides and rearrange, and we get
$$ (1-A^2)x^2 + (1-A^2)y^2 < 1-A^2 $$
so $x^2+y^2<1$, which is to say $|z|<1$, as desired.
answered Jul 28 at 10:50
Henning Makholm
225k16290516
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
1
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
First note that $f_a$ is defined for all points in $overline D[0,1]$. We will show that $|z|<1to|f_a(z)|<1$, equivalently;$$|z-a|<|1-bar az|$$$$|z-a|^2<|1-bar az|^2$$
$$(z-a)(bar z-bar a)<(1-bar az)(1-abar z)$$
$$zbar z-zbar a-abar z+abar a<1-abar z-bar az + bar azabar z$$
$$|z|^2+|a|^2<1+|z|^2|a|^2$$$$|a|^2(1-|z|^2)<1-|z|^2$$$$|a|^2<1,$$
where in the last step, we can divide by $1-|z|^2$ because $|z|<1$. Thus we can consider the restriction $f_a^star:D[0,1]to D[0,1]$ of $f_a$. $f_a^star$ is one-to-one because $f_a$ is one-to-one. Since the inverse $f_-a$ of $f_a$ also maps points of $D[0,1]$ to $D[0,1]$, $f_a^star$ is onto. Thus $f_a^star$ is a bijection, and $f_a$ maps the unit disc to itself bijectively.
answered Jul 29 at 10:56
Ludvig Lindström
585
1
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
What you are supposed to prove is that $f_a$ is a bijection from the open unit disk $mathbb D$ onto itself. The first thing to check is whether $f_a(mathbbD)subsetmathbb D$. This is true, because, if $|z|<1$, thenbeginalignleft|fracz-a1-overline azright|^2&=frac(z-a)overline(z-a)left(1-overline azright)overlineleft(1-overline azright)\&=fracz^2\&=frac^2-2operatornameReleft(overline azright)+aendalignand thereforebeginalignleft|fracz-a1-overline azright|^2<1&iff|z|^2-2operatornameReleft(overline azright)+|a|^2<1-2operatornameReleft(overline azright)+|a|^2|z|^2\&iff|z|^2+|a|^2<1+|a|^2|z|^2\&iffbigl(1-|z|^2bigr)bigl(1-|a|^2bigr)>0,endalignwhich is true.
So, $f_a$ is indeed a map from $mathbb D$ into itself. But you have already checked that$$f_acirc f_-a=operatornameId_mathbb D=f_-acirc f_a.$$The first of these equalities implies that $f_a$ is surjective, whereas the second ont implies that it is injective. So, $f_a$ is a bijection.
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
What you are supposed to prove is that $f_a$ is a bijection from the open unit disk $mathbb D$ onto itself. The first thing to check is whether $f_a(mathbbD)subsetmathbb D$. This is true, because, if $|z|<1$, thenbeginalignleft|fracz-a1-overline azright|^2&=frac(z-a)overline(z-a)left(1-overline azright)overlineleft(1-overline azright)\&=fracz^2\&=frac^2-2operatornameReleft(overline azright)+aendalignand thereforebeginalignleft|fracz-a1-overline azright|^2<1&iff|z|^2-2operatornameReleft(overline azright)+|a|^2<1-2operatornameReleft(overline azright)+|a|^2|z|^2\&iff|z|^2+|a|^2<1+|a|^2|z|^2\&iffbigl(1-|z|^2bigr)bigl(1-|a|^2bigr)>0,endalignwhich is true.
So, $f_a$ is indeed a map from $mathbb D$ into itself. But you have already checked that$$f_acirc f_-a=operatornameId_mathbb D=f_-acirc f_a.$$The first of these equalities implies that $f_a$ is surjective, whereas the second ont implies that it is injective. So, $f_a$ is a bijection.
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What you are supposed to prove is that $f_a$ is a bijection from the open unit disk $mathbb D$ onto itself. The first thing to check is whether $f_a(mathbbD)subsetmathbb D$. This is true, because, if $|z|<1$, thenbeginalignleft|fracz-a1-overline azright|^2&=frac(z-a)overline(z-a)left(1-overline azright)overlineleft(1-overline azright)\&=fracz^2\&=frac^2-2operatornameReleft(overline azright)+aendalignand thereforebeginalignleft|fracz-a1-overline azright|^2<1&iff|z|^2-2operatornameReleft(overline azright)+|a|^2<1-2operatornameReleft(overline azright)+|a|^2|z|^2\&iff|z|^2+|a|^2<1+|a|^2|z|^2\&iffbigl(1-|z|^2bigr)bigl(1-|a|^2bigr)>0,endalignwhich is true.
So, $f_a$ is indeed a map from $mathbb D$ into itself. But you have already checked that$$f_acirc f_-a=operatornameId_mathbb D=f_-acirc f_a.$$The first of these equalities implies that $f_a$ is surjective, whereas the second ont implies that it is injective. So, $f_a$ is a bijection.
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
What you are supposed to prove is that $f_a$ is a bijection from the open unit disk $mathbb D$ onto itself. The first thing to check is whether $f_a(mathbbD)subsetmathbb D$. This is true, because, if $|z|<1$, thenbeginalignleft|fracz-a1-overline azright|^2&=frac(z-a)overline(z-a)left(1-overline azright)overlineleft(1-overline azright)\&=fracz^2\&=frac^2-2operatornameReleft(overline azright)+aendalignand thereforebeginalignleft|fracz-a1-overline azright|^2<1&iff|z|^2-2operatornameReleft(overline azright)+|a|^2<1-2operatornameReleft(overline azright)+|a|^2|z|^2\&iff|z|^2+|a|^2<1+|a|^2|z|^2\&iffbigl(1-|z|^2bigr)bigl(1-|a|^2bigr)>0,endalignwhich is true.
So, $f_a$ is indeed a map from $mathbb D$ into itself. But you have already checked that$$f_acirc f_-a=operatornameId_mathbb D=f_-acirc f_a.$$The first of these equalities implies that $f_a$ is surjective, whereas the second ont implies that it is injective. So, $f_a$ is a bijection.
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
edited Jul 28 at 10:36
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
answered Jul 28 at 10:21
José Carlos Santos
112k1696173
112k1696173
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
I would prove that $|f_a(z)|<1 Rightarrow |z|<1$.
Assume $|f_a(x)|<1$. Since
$$ |f_a(z)| = left| fracz-a1-bar az right|
= frac1-bar az = frac1/bar a-z $$
we have
$$ |z-a| < |a| cdot |z-1/bar a| $$
This is now a geometric inequality about lengths in the complex plane. Furthermore the points $a$ and $1/bar a$ lie on the same ray from the origin, so it makes sense to declare that ray to be the $x$-axis of a new $xy$-coordinate system that we will calculate the lengths in! Then $a$ has coordinates $(A,0)$ for some $Ain(0,1)$, and $1/bar a$ is $(frac 1A, 0)$.
Our assumption is now
$$ sqrt(x-A)^2+y^2 < A sqrt(x-tfrac1A)^2+y^2. $$
Square both sides and rearrange, and we get
$$ (1-A^2)x^2 + (1-A^2)y^2 < 1-A^2 $$
so $x^2+y^2<1$, which is to say $|z|<1$, as desired.
answered Jul 28 at 10:50
Henning Makholm
225k16290516
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
1
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
I would prove that $|f_a(z)|<1 Rightarrow |z|<1$.
Assume $|f_a(x)|<1$. Since
$$ |f_a(z)| = left| fracz-a1-bar az right|
= frac1-bar az = frac1/bar a-z $$
we have
$$ |z-a| < |a| cdot |z-1/bar a| $$
This is now a geometric inequality about lengths in the complex plane. Furthermore the points $a$ and $1/bar a$ lie on the same ray from the origin, so it makes sense to declare that ray to be the $x$-axis of a new $xy$-coordinate system that we will calculate the lengths in! Then $a$ has coordinates $(A,0)$ for some $Ain(0,1)$, and $1/bar a$ is $(frac 1A, 0)$.
Our assumption is now
$$ sqrt(x-A)^2+y^2 < A sqrt(x-tfrac1A)^2+y^2. $$
Square both sides and rearrange, and we get
$$ (1-A^2)x^2 + (1-A^2)y^2 < 1-A^2 $$
so $x^2+y^2<1$, which is to say $|z|<1$, as desired.
answered Jul 28 at 10:50
Henning Makholm
225k16290516
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
1
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I would prove that $|f_a(z)|<1 Rightarrow |z|<1$.
Assume $|f_a(x)|<1$. Since
$$ |f_a(z)| = left| fracz-a1-bar az right|
= frac1-bar az = frac1/bar a-z $$
we have
$$ |z-a| < |a| cdot |z-1/bar a| $$
This is now a geometric inequality about lengths in the complex plane. Furthermore the points $a$ and $1/bar a$ lie on the same ray from the origin, so it makes sense to declare that ray to be the $x$-axis of a new $xy$-coordinate system that we will calculate the lengths in! Then $a$ has coordinates $(A,0)$ for some $Ain(0,1)$, and $1/bar a$ is $(frac 1A, 0)$.
Our assumption is now
$$ sqrt(x-A)^2+y^2 < A sqrt(x-tfrac1A)^2+y^2. $$
Square both sides and rearrange, and we get
$$ (1-A^2)x^2 + (1-A^2)y^2 < 1-A^2 $$
so $x^2+y^2<1$, which is to say $|z|<1$, as desired.
answered Jul 28 at 10:50
Henning Makholm
225k16290516
I would prove that $|f_a(z)|<1 Rightarrow |z|<1$.
Assume $|f_a(x)|<1$. Since
$$ |f_a(z)| = left| fracz-a1-bar az right|
= frac1-bar az = frac1/bar a-z $$
we have
$$ |z-a| < |a| cdot |z-1/bar a| $$
This is now a geometric inequality about lengths in the complex plane. Furthermore the points $a$ and $1/bar a$ lie on the same ray from the origin, so it makes sense to declare that ray to be the $x$-axis of a new $xy$-coordinate system that we will calculate the lengths in! Then $a$ has coordinates $(A,0)$ for some $Ain(0,1)$, and $1/bar a$ is $(frac 1A, 0)$.
Our assumption is now
$$ sqrt(x-A)^2+y^2 < A sqrt(x-tfrac1A)^2+y^2. $$
Square both sides and rearrange, and we get
$$ (1-A^2)x^2 + (1-A^2)y^2 < 1-A^2 $$
so $x^2+y^2<1$, which is to say $|z|<1$, as desired.
answered Jul 28 at 10:50
Henning Makholm
225k16290516
answered Jul 28 at 10:50
Henning Makholm
225k16290516
answered Jul 28 at 10:50
Henning Makholm
225k16290516
answered Jul 28 at 10:50
Henning Makholm
225k16290516
225k16290516
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
1
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
1
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28 at 17:22
1
1
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
@BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^-1=f_bar a$". My point in the first line is just that this direction seems to be easiest to do directly.
– Henning Makholm
Jul 28 at 17:44
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
Thanks! I'll analyse later.
– BCLC
Jul 29 at 1:25
add a comment |Â
up vote
1
down vote
First note that $f_a$ is defined for all points in $overline D[0,1]$. We will show that $|z|<1to|f_a(z)|<1$, equivalently;$$|z-a|<|1-bar az|$$$$|z-a|^2<|1-bar az|^2$$
$$(z-a)(bar z-bar a)<(1-bar az)(1-abar z)$$
$$zbar z-zbar a-abar z+abar a<1-abar z-bar az + bar azabar z$$
$$|z|^2+|a|^2<1+|z|^2|a|^2$$$$|a|^2(1-|z|^2)<1-|z|^2$$$$|a|^2<1,$$
where in the last step, we can divide by $1-|z|^2$ because $|z|<1$. Thus we can consider the restriction $f_a^star:D[0,1]to D[0,1]$ of $f_a$. $f_a^star$ is one-to-one because $f_a$ is one-to-one. Since the inverse $f_-a$ of $f_a$ also maps points of $D[0,1]$ to $D[0,1]$, $f_a^star$ is onto. Thus $f_a^star$ is a bijection, and $f_a$ maps the unit disc to itself bijectively.
answered Jul 29 at 10:56
Ludvig Lindström
585
1
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
add a comment |Â
up vote
1
down vote
First note that $f_a$ is defined for all points in $overline D[0,1]$. We will show that $|z|<1to|f_a(z)|<1$, equivalently;$$|z-a|<|1-bar az|$$$$|z-a|^2<|1-bar az|^2$$
$$(z-a)(bar z-bar a)<(1-bar az)(1-abar z)$$
$$zbar z-zbar a-abar z+abar a<1-abar z-bar az + bar azabar z$$
$$|z|^2+|a|^2<1+|z|^2|a|^2$$$$|a|^2(1-|z|^2)<1-|z|^2$$$$|a|^2<1,$$
where in the last step, we can divide by $1-|z|^2$ because $|z|<1$. Thus we can consider the restriction $f_a^star:D[0,1]to D[0,1]$ of $f_a$. $f_a^star$ is one-to-one because $f_a$ is one-to-one. Since the inverse $f_-a$ of $f_a$ also maps points of $D[0,1]$ to $D[0,1]$, $f_a^star$ is onto. Thus $f_a^star$ is a bijection, and $f_a$ maps the unit disc to itself bijectively.
answered Jul 29 at 10:56
Ludvig Lindström
585
1
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First note that $f_a$ is defined for all points in $overline D[0,1]$. We will show that $|z|<1to|f_a(z)|<1$, equivalently;$$|z-a|<|1-bar az|$$$$|z-a|^2<|1-bar az|^2$$
$$(z-a)(bar z-bar a)<(1-bar az)(1-abar z)$$
$$zbar z-zbar a-abar z+abar a<1-abar z-bar az + bar azabar z$$
$$|z|^2+|a|^2<1+|z|^2|a|^2$$$$|a|^2(1-|z|^2)<1-|z|^2$$$$|a|^2<1,$$
where in the last step, we can divide by $1-|z|^2$ because $|z|<1$. Thus we can consider the restriction $f_a^star:D[0,1]to D[0,1]$ of $f_a$. $f_a^star$ is one-to-one because $f_a$ is one-to-one. Since the inverse $f_-a$ of $f_a$ also maps points of $D[0,1]$ to $D[0,1]$, $f_a^star$ is onto. Thus $f_a^star$ is a bijection, and $f_a$ maps the unit disc to itself bijectively.
answered Jul 29 at 10:56
Ludvig Lindström
585
First note that $f_a$ is defined for all points in $overline D[0,1]$. We will show that $|z|<1to|f_a(z)|<1$, equivalently;$$|z-a|<|1-bar az|$$$$|z-a|^2<|1-bar az|^2$$
$$(z-a)(bar z-bar a)<(1-bar az)(1-abar z)$$
$$zbar z-zbar a-abar z+abar a<1-abar z-bar az + bar azabar z$$
$$|z|^2+|a|^2<1+|z|^2|a|^2$$$$|a|^2(1-|z|^2)<1-|z|^2$$$$|a|^2<1,$$
where in the last step, we can divide by $1-|z|^2$ because $|z|<1$. Thus we can consider the restriction $f_a^star:D[0,1]to D[0,1]$ of $f_a$. $f_a^star$ is one-to-one because $f_a$ is one-to-one. Since the inverse $f_-a$ of $f_a$ also maps points of $D[0,1]$ to $D[0,1]$, $f_a^star$ is onto. Thus $f_a^star$ is a bijection, and $f_a$ maps the unit disc to itself bijectively.
answered Jul 29 at 10:56
Ludvig Lindström
585
edited Jul 29 at 17:29
answered Jul 29 at 10:56
Ludvig Lindström
585
answered Jul 29 at 10:56
Ludvig Lindström
585
answered Jul 29 at 10:56
Ludvig Lindström
585
585
1
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
add a comment |Â
1
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
1
1
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29 at 12:01
add a comment |Â
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See these very similar questions: math.stackexchange.com/questions/1227914/… and math.stackexchange.com/questions/343982/…
– Robert Z
Jul 28 at 10:27
2
Never write a complex number $z$ as $x+iy$ unless you really have to.
– Lord Shark the Unknown
Jul 28 at 10:33
@LordSharktheUnknown Thanks. Any consolation prize for my asking #2? :|
– BCLC
Jul 28 at 10:54