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Show $f$ holomorphic and $textimage(f)subseteq C[0,1] implies f$ is constant.
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.8
Suppose $f$ is holomorphic in region $G$, and $f(G) subseteq =1 $. Prove $f$ is constant.
I will now attempt to elaborate the following proof at a Winter 2017 course in Oregon State University.
What errors if any are there? Or are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations.
OSU Pf: Let $g(z)=frac1+z1-z$, and define $h(z)=g(f(z)), z in G setminus z : f(z) = 1$. Then $h$ is holomorphic on its domain, and $h$ is imaginary valued by Exer 3.7. By a variation of Exer 2.19, $h$ is constant. QED
My Pf: $because f(G) subseteq C[0,1]$, let's consider the Möbius transformation in the preceding Exer 3.7 $g: mathbb C setminus z = 1 to mathbb C$ s.t. $g(z) := frac1+z1-z$:
If we plug in $C[0,1] setminus 1$ in $g$, then we'll get the imaginary axis by Exer 3.7. Precisely: $$g(e^it_t in mathbb R setminus 0) = is_s in mathbb R. tag1$$ Now, define $G' := G setminus z in G $ and $h: G' to mathbb C$ s.t. $h := g circ f$ s.t. $h(z) = frac1+f(z)1-f(z)$. If we plug in $G'$ in $h$, then we'll get the imaginary axis. Precisely: $$h(G') := frac1+f(G')1-f(G') stackrel(1)= is_s in mathbb R. tag2$$
Now Exer 2.19 says that a real valued holomorphic function over a region is constant: $f(z)=u(z) implies u_x=0=u_y implies f'=0$ to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $u$ is constant. Actually, an imaginary valued holomorphic function over a region is constant too: $f(z)=iv(z) implies v_x=0=v_y implies f'=0$ again by Cauchy-Riemann Thm 2.13 to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $v$ is constant.
$(2)$ precisely says that $h$ is imaginary valued over $G'$. $therefore,$ if $G'$ is a region (A) and if $h$ is holomorphic on $G'$ (B), then $h$ is constant on $G'$ with value I'll denote $Hi, H in mathbb R$:
$forall z in G',$
$$Hi = frac1+f(z)1-f(z) implies f(z) = fracHi-1Hi+1, tag3$$
where $Hi+1 ne 0 forall H in mathbb R$.
$therefore, f$ is constant on $G'$ (Q4) with value given in $(3)$.
QED except possibly for (C)
(A) $G'$ is a region
I guess if $G setminus G'$ is finite, then G' is a region. I'm thinking $D[0,1]$ is a region and then $D[0,1] setminus 0$ is still a region.
(B) To show $h$ is holomorphic in $G'$:
Well $h(z)$ is differentiable $forall z in G'$ and $f(z) ne 1 forall z in G'$ and $f'(z)$ exists in $G' subseteq G$ because $f$ is differentiable in $G$ because $f$ is holomorphic in $G$.
$$h'(z) = g'(f(z)) f'(z) = frac2(1-w)^2|_w=f(z) f'(z) = frac2 f'(z)(1-f(z))^2 $$
Now, $f'(z)$ exists on an open disc $D[z,r_z] forall z in G$ where $r_z$ denotes the radius of the open disc s.t. $f(z)$ is holomorphic at $z$. So, I guess $frac2 f'(z)(1-f(z))^2 = h'(z)$ exists on an open disc with the same radius $D[z,r_z] forall z in G'$, and $therefore, h$ is holomorphic in $G'$.
(C) Possible flaw:
It seems that on $G'$, $f$ has value $fracHi-1Hi+1$ while on $G setminus G'$, $f$ has value $1$.
$$therefore, forall z in G, f(z) = fracHi-1Hi+1 1_G'(z) + 1_G setminus G'(z)$$
It seems then that we've actually show only that $f$ is constant on $G$ except for the subset of G where $f=1$.
general-topology complex-analysis holomorphic-functions mobius-transformation cauchy-riemann-equation
asked Jul 28 at 9:14
BCLC
6,98421973
 |Â
show 13 more comments
up vote
1
down vote
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.8
Suppose $f$ is holomorphic in region $G$, and $f(G) subseteq =1 $. Prove $f$ is constant.
I will now attempt to elaborate the following proof at a Winter 2017 course in Oregon State University.
What errors if any are there? Or are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations.
OSU Pf: Let $g(z)=frac1+z1-z$, and define $h(z)=g(f(z)), z in G setminus z : f(z) = 1$. Then $h$ is holomorphic on its domain, and $h$ is imaginary valued by Exer 3.7. By a variation of Exer 2.19, $h$ is constant. QED
My Pf: $because f(G) subseteq C[0,1]$, let's consider the Möbius transformation in the preceding Exer 3.7 $g: mathbb C setminus z = 1 to mathbb C$ s.t. $g(z) := frac1+z1-z$:
If we plug in $C[0,1] setminus 1$ in $g$, then we'll get the imaginary axis by Exer 3.7. Precisely: $$g(e^it_t in mathbb R setminus 0) = is_s in mathbb R. tag1$$ Now, define $G' := G setminus z in G $ and $h: G' to mathbb C$ s.t. $h := g circ f$ s.t. $h(z) = frac1+f(z)1-f(z)$. If we plug in $G'$ in $h$, then we'll get the imaginary axis. Precisely: $$h(G') := frac1+f(G')1-f(G') stackrel(1)= is_s in mathbb R. tag2$$
Now Exer 2.19 says that a real valued holomorphic function over a region is constant: $f(z)=u(z) implies u_x=0=u_y implies f'=0$ to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $u$ is constant. Actually, an imaginary valued holomorphic function over a region is constant too: $f(z)=iv(z) implies v_x=0=v_y implies f'=0$ again by Cauchy-Riemann Thm 2.13 to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $v$ is constant.
$(2)$ precisely says that $h$ is imaginary valued over $G'$. $therefore,$ if $G'$ is a region (A) and if $h$ is holomorphic on $G'$ (B), then $h$ is constant on $G'$ with value I'll denote $Hi, H in mathbb R$:
$forall z in G',$
$$Hi = frac1+f(z)1-f(z) implies f(z) = fracHi-1Hi+1, tag3$$
where $Hi+1 ne 0 forall H in mathbb R$.
$therefore, f$ is constant on $G'$ (Q4) with value given in $(3)$.
QED except possibly for (C)
(A) $G'$ is a region
I guess if $G setminus G'$ is finite, then G' is a region. I'm thinking $D[0,1]$ is a region and then $D[0,1] setminus 0$ is still a region.
(B) To show $h$ is holomorphic in $G'$:
Well $h(z)$ is differentiable $forall z in G'$ and $f(z) ne 1 forall z in G'$ and $f'(z)$ exists in $G' subseteq G$ because $f$ is differentiable in $G$ because $f$ is holomorphic in $G$.
$$h'(z) = g'(f(z)) f'(z) = frac2(1-w)^2|_w=f(z) f'(z) = frac2 f'(z)(1-f(z))^2 $$
Now, $f'(z)$ exists on an open disc $D[z,r_z] forall z in G$ where $r_z$ denotes the radius of the open disc s.t. $f(z)$ is holomorphic at $z$. So, I guess $frac2 f'(z)(1-f(z))^2 = h'(z)$ exists on an open disc with the same radius $D[z,r_z] forall z in G'$, and $therefore, h$ is holomorphic in $G'$.
(C) Possible flaw:
It seems that on $G'$, $f$ has value $fracHi-1Hi+1$ while on $G setminus G'$, $f$ has value $1$.
$$therefore, forall z in G, f(z) = fracHi-1Hi+1 1_G'(z) + 1_G setminus G'(z)$$
It seems then that we've actually show only that $f$ is constant on $G$ except for the subset of G where $f=1$.
general-topology complex-analysis holomorphic-functions mobius-transformation cauchy-riemann-equation
asked Jul 28 at 9:14
BCLC
6,98421973
2
This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic?
– Lord Shark the Unknown
Jul 28 at 9:18
2
If it's not, then so much the worse for that text.
– Lord Shark the Unknown
Jul 28 at 9:35
3
I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann.
– Lord Shark the Unknown
Jul 28 at 9:43
1
What do you mean by $f(G) subseteq z=1 $?
– zhw.
Aug 5 at 16:27
1
But you wrote $z=1$ not $|z|=1$
– zhw.
Aug 5 at 18:26
 |Â
show 13 more comments
up vote
1
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up vote
1
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.8
Suppose $f$ is holomorphic in region $G$, and $f(G) subseteq =1 $. Prove $f$ is constant.
I will now attempt to elaborate the following proof at a Winter 2017 course in Oregon State University.
What errors if any are there? Or are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations.
OSU Pf: Let $g(z)=frac1+z1-z$, and define $h(z)=g(f(z)), z in G setminus z : f(z) = 1$. Then $h$ is holomorphic on its domain, and $h$ is imaginary valued by Exer 3.7. By a variation of Exer 2.19, $h$ is constant. QED
My Pf: $because f(G) subseteq C[0,1]$, let's consider the Möbius transformation in the preceding Exer 3.7 $g: mathbb C setminus z = 1 to mathbb C$ s.t. $g(z) := frac1+z1-z$:
If we plug in $C[0,1] setminus 1$ in $g$, then we'll get the imaginary axis by Exer 3.7. Precisely: $$g(e^it_t in mathbb R setminus 0) = is_s in mathbb R. tag1$$ Now, define $G' := G setminus z in G $ and $h: G' to mathbb C$ s.t. $h := g circ f$ s.t. $h(z) = frac1+f(z)1-f(z)$. If we plug in $G'$ in $h$, then we'll get the imaginary axis. Precisely: $$h(G') := frac1+f(G')1-f(G') stackrel(1)= is_s in mathbb R. tag2$$
Now Exer 2.19 says that a real valued holomorphic function over a region is constant: $f(z)=u(z) implies u_x=0=u_y implies f'=0$ to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $u$ is constant. Actually, an imaginary valued holomorphic function over a region is constant too: $f(z)=iv(z) implies v_x=0=v_y implies f'=0$ again by Cauchy-Riemann Thm 2.13 to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $v$ is constant.
$(2)$ precisely says that $h$ is imaginary valued over $G'$. $therefore,$ if $G'$ is a region (A) and if $h$ is holomorphic on $G'$ (B), then $h$ is constant on $G'$ with value I'll denote $Hi, H in mathbb R$:
$forall z in G',$
$$Hi = frac1+f(z)1-f(z) implies f(z) = fracHi-1Hi+1, tag3$$
where $Hi+1 ne 0 forall H in mathbb R$.
$therefore, f$ is constant on $G'$ (Q4) with value given in $(3)$.
QED except possibly for (C)
(A) $G'$ is a region
I guess if $G setminus G'$ is finite, then G' is a region. I'm thinking $D[0,1]$ is a region and then $D[0,1] setminus 0$ is still a region.
(B) To show $h$ is holomorphic in $G'$:
Well $h(z)$ is differentiable $forall z in G'$ and $f(z) ne 1 forall z in G'$ and $f'(z)$ exists in $G' subseteq G$ because $f$ is differentiable in $G$ because $f$ is holomorphic in $G$.
$$h'(z) = g'(f(z)) f'(z) = frac2(1-w)^2|_w=f(z) f'(z) = frac2 f'(z)(1-f(z))^2 $$
Now, $f'(z)$ exists on an open disc $D[z,r_z] forall z in G$ where $r_z$ denotes the radius of the open disc s.t. $f(z)$ is holomorphic at $z$. So, I guess $frac2 f'(z)(1-f(z))^2 = h'(z)$ exists on an open disc with the same radius $D[z,r_z] forall z in G'$, and $therefore, h$ is holomorphic in $G'$.
(C) Possible flaw:
It seems that on $G'$, $f$ has value $fracHi-1Hi+1$ while on $G setminus G'$, $f$ has value $1$.
$$therefore, forall z in G, f(z) = fracHi-1Hi+1 1_G'(z) + 1_G setminus G'(z)$$
It seems then that we've actually show only that $f$ is constant on $G$ except for the subset of G where $f=1$.
general-topology complex-analysis holomorphic-functions mobius-transformation cauchy-riemann-equation
asked Jul 28 at 9:14
BCLC
6,98421973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.8
Suppose $f$ is holomorphic in region $G$, and $f(G) subseteq =1 $. Prove $f$ is constant.
I will now attempt to elaborate the following proof at a Winter 2017 course in Oregon State University.
What errors if any are there? Or are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations.
OSU Pf: Let $g(z)=frac1+z1-z$, and define $h(z)=g(f(z)), z in G setminus z : f(z) = 1$. Then $h$ is holomorphic on its domain, and $h$ is imaginary valued by Exer 3.7. By a variation of Exer 2.19, $h$ is constant. QED
My Pf: $because f(G) subseteq C[0,1]$, let's consider the Möbius transformation in the preceding Exer 3.7 $g: mathbb C setminus z = 1 to mathbb C$ s.t. $g(z) := frac1+z1-z$:
If we plug in $C[0,1] setminus 1$ in $g$, then we'll get the imaginary axis by Exer 3.7. Precisely: $$g(e^it_t in mathbb R setminus 0) = is_s in mathbb R. tag1$$ Now, define $G' := G setminus z in G $ and $h: G' to mathbb C$ s.t. $h := g circ f$ s.t. $h(z) = frac1+f(z)1-f(z)$. If we plug in $G'$ in $h$, then we'll get the imaginary axis. Precisely: $$h(G') := frac1+f(G')1-f(G') stackrel(1)= is_s in mathbb R. tag2$$
Now Exer 2.19 says that a real valued holomorphic function over a region is constant: $f(z)=u(z) implies u_x=0=u_y implies f'=0$ to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $u$ is constant. Actually, an imaginary valued holomorphic function over a region is constant too: $f(z)=iv(z) implies v_x=0=v_y implies f'=0$ again by Cauchy-Riemann Thm 2.13 to conclude by Thm 2.17 that $f$ is constant or simply by partial integration that $v$ is constant.
$(2)$ precisely says that $h$ is imaginary valued over $G'$. $therefore,$ if $G'$ is a region (A) and if $h$ is holomorphic on $G'$ (B), then $h$ is constant on $G'$ with value I'll denote $Hi, H in mathbb R$:
$forall z in G',$
$$Hi = frac1+f(z)1-f(z) implies f(z) = fracHi-1Hi+1, tag3$$
where $Hi+1 ne 0 forall H in mathbb R$.
$therefore, f$ is constant on $G'$ (Q4) with value given in $(3)$.
QED except possibly for (C)
(A) $G'$ is a region
I guess if $G setminus G'$ is finite, then G' is a region. I'm thinking $D[0,1]$ is a region and then $D[0,1] setminus 0$ is still a region.
(B) To show $h$ is holomorphic in $G'$:
Well $h(z)$ is differentiable $forall z in G'$ and $f(z) ne 1 forall z in G'$ and $f'(z)$ exists in $G' subseteq G$ because $f$ is differentiable in $G$ because $f$ is holomorphic in $G$.
$$h'(z) = g'(f(z)) f'(z) = frac2(1-w)^2|_w=f(z) f'(z) = frac2 f'(z)(1-f(z))^2 $$
Now, $f'(z)$ exists on an open disc $D[z,r_z] forall z in G$ where $r_z$ denotes the radius of the open disc s.t. $f(z)$ is holomorphic at $z$. So, I guess $frac2 f'(z)(1-f(z))^2 = h'(z)$ exists on an open disc with the same radius $D[z,r_z] forall z in G'$, and $therefore, h$ is holomorphic in $G'$.
(C) Possible flaw:
It seems that on $G'$, $f$ has value $fracHi-1Hi+1$ while on $G setminus G'$, $f$ has value $1$.
$$therefore, forall z in G, f(z) = fracHi-1Hi+1 1_G'(z) + 1_G setminus G'(z)$$
It seems then that we've actually show only that $f$ is constant on $G$ except for the subset of G where $f=1$.
general-topology complex-analysis holomorphic-functions mobius-transformation cauchy-riemann-equation
asked Jul 28 at 9:14
BCLC
6,98421973
edited Aug 6 at 15:59
asked Jul 28 at 9:14
BCLC
6,98421973
asked Jul 28 at 9:14
BCLC
6,98421973
asked Jul 28 at 9:14
BCLC
6,98421973
6,98421973
2
This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic?
– Lord Shark the Unknown
Jul 28 at 9:18
2
If it's not, then so much the worse for that text.
– Lord Shark the Unknown
Jul 28 at 9:35
3
I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann.
– Lord Shark the Unknown
Jul 28 at 9:43
1
What do you mean by $f(G) subseteq z=1 $?
– zhw.
Aug 5 at 16:27
1
But you wrote $z=1$ not $|z|=1$
– zhw.
Aug 5 at 18:26
 |Â
show 13 more comments
2
This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic?
– Lord Shark the Unknown
Jul 28 at 9:18
2
If it's not, then so much the worse for that text.
– Lord Shark the Unknown
Jul 28 at 9:35
3
I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann.
– Lord Shark the Unknown
Jul 28 at 9:43
1
What do you mean by $f(G) subseteq z=1 $?
– zhw.
Aug 5 at 16:27
1
But you wrote $z=1$ not $|z|=1$
– zhw.
Aug 5 at 18:26
2
2
This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic?
– Lord Shark the Unknown
Jul 28 at 9:18
This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic?
– Lord Shark the Unknown
Jul 28 at 9:18
2
2
If it's not, then so much the worse for that text.
– Lord Shark the Unknown
Jul 28 at 9:35
If it's not, then so much the worse for that text.
– Lord Shark the Unknown
Jul 28 at 9:35
3
3
I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann.
– Lord Shark the Unknown
Jul 28 at 9:43
I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann.
– Lord Shark the Unknown
Jul 28 at 9:43
1
1
What do you mean by $f(G) subseteq z=1 $?
– zhw.
Aug 5 at 16:27
What do you mean by $f(G) subseteq z=1 $?
– zhw.
Aug 5 at 16:27
1
1
But you wrote $z=1$ not $|z|=1$
– zhw.
Aug 5 at 18:26
But you wrote $z=1$ not $|z|=1$
– zhw.
Aug 5 at 18:26
 |Â
show 13 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
The Cauchy-Riemann equations have a geometric interpretation. Let $f$
be holomorphic at $a$ and let $f'(a)ne0$. Consider the horizontal
line through $a$ consisting of points $a+s$ for real $s$, and
also the vertical line through $a$, that is the points $a+it$ for
$t$ real.
Then these are mapped by $f$ into two curves $C_1$ and $C_2$ meeting
at $f(a)$. Cauchy-Riemann implies these meet at right angles there.
But if the image of $f$ were within a 1-dimensional subspace such as
the unit circle, then $C_1$ and $C_2$ would be restricted within
too, which means they cannot intersect orthogonally. The only
way out of this impasse is for $f'(a)=0$. This must happen for all $a$.
An introductory book that makes much of such geometric interpretations
is Needham's Visual Complex Analysis (OUP).
If you really don't like geometry, write $f(x+iy)=u+iv$ in the usual
way. If $f$ maps to the unit circle, then $u^2+v^2=1$. Differentiating
gives $uu_x+vv_x=uu_y+vv_y=0$. Cauchy-Riemann gives
$-uv_x+vu_x=0$. Then
$$u_x=u^2u_x+v^2u_x=u(uu_x+vv_x)+v(-uv_x+vu_x)=0$$
and similarly $v_x=0$. Therefore $f'=0$.
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
add a comment |Â
up vote
1
down vote
Note that $foverline f=1$ in $G.$ This implies $overline f =1/f$ in $G.$ Hence $overline f$ is holomorphic in $G.$ This implies both $f+overline f= 2text Re f$ and $f-overline f=2itext Im f$ are holomorphic in $G.$ By the remarks you made right after $(2)$ in your question, these functions are constant, which implies $f$ is constant.
answered Aug 6 at 15:55
zhw.
65.3k42870
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
1
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
add a comment |Â
up vote
0
down vote
In every point on a line in the plane the possible tangent vectors form a real monodimensional vector space. An holomorphic function has in every point of its domain a derivative map that is a complex linear, that is a roto-homothetic real transformation of the plane.
Turning attention to your problem: in every point of the region $G$ the tangent vectors form a bidimensional real vector space. When a tangent vector is transformed by the derivative map it is rotated and/or enlarged in the plane of the tangent vectors in the image of the point. But the only possible tangent vectors in any point of the image (subset of the unit circle, that is a set of arcs and/or points) are in a real monodimensional vector space (for points on an arc) or zero-dimensional vector space (for isolated points). So for that roto-homothetic transformation to be fitted, it must make any tangent vector in the region $G$ vanish, that is, the derivative map must be $0$.
answered Jul 28 at 9:50
trying
4,0461722
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The Cauchy-Riemann equations have a geometric interpretation. Let $f$
be holomorphic at $a$ and let $f'(a)ne0$. Consider the horizontal
line through $a$ consisting of points $a+s$ for real $s$, and
also the vertical line through $a$, that is the points $a+it$ for
$t$ real.
Then these are mapped by $f$ into two curves $C_1$ and $C_2$ meeting
at $f(a)$. Cauchy-Riemann implies these meet at right angles there.
But if the image of $f$ were within a 1-dimensional subspace such as
the unit circle, then $C_1$ and $C_2$ would be restricted within
too, which means they cannot intersect orthogonally. The only
way out of this impasse is for $f'(a)=0$. This must happen for all $a$.
An introductory book that makes much of such geometric interpretations
is Needham's Visual Complex Analysis (OUP).
If you really don't like geometry, write $f(x+iy)=u+iv$ in the usual
way. If $f$ maps to the unit circle, then $u^2+v^2=1$. Differentiating
gives $uu_x+vv_x=uu_y+vv_y=0$. Cauchy-Riemann gives
$-uv_x+vu_x=0$. Then
$$u_x=u^2u_x+v^2u_x=u(uu_x+vv_x)+v(-uv_x+vu_x)=0$$
and similarly $v_x=0$. Therefore $f'=0$.
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
add a comment |Â
up vote
4
down vote
accepted
The Cauchy-Riemann equations have a geometric interpretation. Let $f$
be holomorphic at $a$ and let $f'(a)ne0$. Consider the horizontal
line through $a$ consisting of points $a+s$ for real $s$, and
also the vertical line through $a$, that is the points $a+it$ for
$t$ real.
Then these are mapped by $f$ into two curves $C_1$ and $C_2$ meeting
at $f(a)$. Cauchy-Riemann implies these meet at right angles there.
But if the image of $f$ were within a 1-dimensional subspace such as
the unit circle, then $C_1$ and $C_2$ would be restricted within
too, which means they cannot intersect orthogonally. The only
way out of this impasse is for $f'(a)=0$. This must happen for all $a$.
An introductory book that makes much of such geometric interpretations
is Needham's Visual Complex Analysis (OUP).
If you really don't like geometry, write $f(x+iy)=u+iv$ in the usual
way. If $f$ maps to the unit circle, then $u^2+v^2=1$. Differentiating
gives $uu_x+vv_x=uu_y+vv_y=0$. Cauchy-Riemann gives
$-uv_x+vu_x=0$. Then
$$u_x=u^2u_x+v^2u_x=u(uu_x+vv_x)+v(-uv_x+vu_x)=0$$
and similarly $v_x=0$. Therefore $f'=0$.
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The Cauchy-Riemann equations have a geometric interpretation. Let $f$
be holomorphic at $a$ and let $f'(a)ne0$. Consider the horizontal
line through $a$ consisting of points $a+s$ for real $s$, and
also the vertical line through $a$, that is the points $a+it$ for
$t$ real.
Then these are mapped by $f$ into two curves $C_1$ and $C_2$ meeting
at $f(a)$. Cauchy-Riemann implies these meet at right angles there.
But if the image of $f$ were within a 1-dimensional subspace such as
the unit circle, then $C_1$ and $C_2$ would be restricted within
too, which means they cannot intersect orthogonally. The only
way out of this impasse is for $f'(a)=0$. This must happen for all $a$.
An introductory book that makes much of such geometric interpretations
is Needham's Visual Complex Analysis (OUP).
If you really don't like geometry, write $f(x+iy)=u+iv$ in the usual
way. If $f$ maps to the unit circle, then $u^2+v^2=1$. Differentiating
gives $uu_x+vv_x=uu_y+vv_y=0$. Cauchy-Riemann gives
$-uv_x+vu_x=0$. Then
$$u_x=u^2u_x+v^2u_x=u(uu_x+vv_x)+v(-uv_x+vu_x)=0$$
and similarly $v_x=0$. Therefore $f'=0$.
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
The Cauchy-Riemann equations have a geometric interpretation. Let $f$
be holomorphic at $a$ and let $f'(a)ne0$. Consider the horizontal
line through $a$ consisting of points $a+s$ for real $s$, and
also the vertical line through $a$, that is the points $a+it$ for
$t$ real.
Then these are mapped by $f$ into two curves $C_1$ and $C_2$ meeting
at $f(a)$. Cauchy-Riemann implies these meet at right angles there.
But if the image of $f$ were within a 1-dimensional subspace such as
the unit circle, then $C_1$ and $C_2$ would be restricted within
too, which means they cannot intersect orthogonally. The only
way out of this impasse is for $f'(a)=0$. This must happen for all $a$.
An introductory book that makes much of such geometric interpretations
is Needham's Visual Complex Analysis (OUP).
If you really don't like geometry, write $f(x+iy)=u+iv$ in the usual
way. If $f$ maps to the unit circle, then $u^2+v^2=1$. Differentiating
gives $uu_x+vv_x=uu_y+vv_y=0$. Cauchy-Riemann gives
$-uv_x+vu_x=0$. Then
$$u_x=u^2u_x+v^2u_x=u(uu_x+vv_x)+v(-uv_x+vu_x)=0$$
and similarly $v_x=0$. Therefore $f'=0$.
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
edited Aug 5 at 15:18
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
answered Aug 5 at 14:45
Lord Shark the Unknown
84.6k950111
84.6k950111
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
add a comment |Â
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
I just knew there had to be a way to do this with just Cauchy-Riemann. Suddenly, I no longer care about the kinda inelegant Möbius approach. Thanks!
– BCLC
Aug 6 at 5:43
add a comment |Â
up vote
1
down vote
Note that $foverline f=1$ in $G.$ This implies $overline f =1/f$ in $G.$ Hence $overline f$ is holomorphic in $G.$ This implies both $f+overline f= 2text Re f$ and $f-overline f=2itext Im f$ are holomorphic in $G.$ By the remarks you made right after $(2)$ in your question, these functions are constant, which implies $f$ is constant.
answered Aug 6 at 15:55
zhw.
65.3k42870
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
1
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
add a comment |Â
up vote
1
down vote
Note that $foverline f=1$ in $G.$ This implies $overline f =1/f$ in $G.$ Hence $overline f$ is holomorphic in $G.$ This implies both $f+overline f= 2text Re f$ and $f-overline f=2itext Im f$ are holomorphic in $G.$ By the remarks you made right after $(2)$ in your question, these functions are constant, which implies $f$ is constant.
answered Aug 6 at 15:55
zhw.
65.3k42870
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
1
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that $foverline f=1$ in $G.$ This implies $overline f =1/f$ in $G.$ Hence $overline f$ is holomorphic in $G.$ This implies both $f+overline f= 2text Re f$ and $f-overline f=2itext Im f$ are holomorphic in $G.$ By the remarks you made right after $(2)$ in your question, these functions are constant, which implies $f$ is constant.
answered Aug 6 at 15:55
zhw.
65.3k42870
Note that $foverline f=1$ in $G.$ This implies $overline f =1/f$ in $G.$ Hence $overline f$ is holomorphic in $G.$ This implies both $f+overline f= 2text Re f$ and $f-overline f=2itext Im f$ are holomorphic in $G.$ By the remarks you made right after $(2)$ in your question, these functions are constant, which implies $f$ is constant.
answered Aug 6 at 15:55
zhw.
65.3k42870
edited Aug 6 at 16:26
answered Aug 6 at 15:55
zhw.
65.3k42870
answered Aug 6 at 15:55
zhw.
65.3k42870
answered Aug 6 at 15:55
zhw.
65.3k42870
65.3k42870
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
1
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
add a comment |Â
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
1
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
I just knew there had to be a way to do this with just regular differentiation/holomorphic stuff and not Möbius. Thanks!
– BCLC
Aug 6 at 16:00
1
1
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
Thanks. I changed your edit as that page is not what I would refer a reader to. I was referring to the remarks in your question right after $(2);$ no need for anything else.
– zhw.
Aug 6 at 16:28
add a comment |Â
up vote
0
down vote
In every point on a line in the plane the possible tangent vectors form a real monodimensional vector space. An holomorphic function has in every point of its domain a derivative map that is a complex linear, that is a roto-homothetic real transformation of the plane.
Turning attention to your problem: in every point of the region $G$ the tangent vectors form a bidimensional real vector space. When a tangent vector is transformed by the derivative map it is rotated and/or enlarged in the plane of the tangent vectors in the image of the point. But the only possible tangent vectors in any point of the image (subset of the unit circle, that is a set of arcs and/or points) are in a real monodimensional vector space (for points on an arc) or zero-dimensional vector space (for isolated points). So for that roto-homothetic transformation to be fitted, it must make any tangent vector in the region $G$ vanish, that is, the derivative map must be $0$.
answered Jul 28 at 9:50
trying
4,0461722
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
add a comment |Â
up vote
0
down vote
In every point on a line in the plane the possible tangent vectors form a real monodimensional vector space. An holomorphic function has in every point of its domain a derivative map that is a complex linear, that is a roto-homothetic real transformation of the plane.
Turning attention to your problem: in every point of the region $G$ the tangent vectors form a bidimensional real vector space. When a tangent vector is transformed by the derivative map it is rotated and/or enlarged in the plane of the tangent vectors in the image of the point. But the only possible tangent vectors in any point of the image (subset of the unit circle, that is a set of arcs and/or points) are in a real monodimensional vector space (for points on an arc) or zero-dimensional vector space (for isolated points). So for that roto-homothetic transformation to be fitted, it must make any tangent vector in the region $G$ vanish, that is, the derivative map must be $0$.
answered Jul 28 at 9:50
trying
4,0461722
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In every point on a line in the plane the possible tangent vectors form a real monodimensional vector space. An holomorphic function has in every point of its domain a derivative map that is a complex linear, that is a roto-homothetic real transformation of the plane.
Turning attention to your problem: in every point of the region $G$ the tangent vectors form a bidimensional real vector space. When a tangent vector is transformed by the derivative map it is rotated and/or enlarged in the plane of the tangent vectors in the image of the point. But the only possible tangent vectors in any point of the image (subset of the unit circle, that is a set of arcs and/or points) are in a real monodimensional vector space (for points on an arc) or zero-dimensional vector space (for isolated points). So for that roto-homothetic transformation to be fitted, it must make any tangent vector in the region $G$ vanish, that is, the derivative map must be $0$.
answered Jul 28 at 9:50
trying
4,0461722
In every point on a line in the plane the possible tangent vectors form a real monodimensional vector space. An holomorphic function has in every point of its domain a derivative map that is a complex linear, that is a roto-homothetic real transformation of the plane.
Turning attention to your problem: in every point of the region $G$ the tangent vectors form a bidimensional real vector space. When a tangent vector is transformed by the derivative map it is rotated and/or enlarged in the plane of the tangent vectors in the image of the point. But the only possible tangent vectors in any point of the image (subset of the unit circle, that is a set of arcs and/or points) are in a real monodimensional vector space (for points on an arc) or zero-dimensional vector space (for isolated points). So for that roto-homothetic transformation to be fitted, it must make any tangent vector in the region $G$ vanish, that is, the derivative map must be $0$.
answered Jul 28 at 9:50
trying
4,0461722
answered Jul 28 at 9:50
trying
4,0461722
answered Jul 28 at 9:50
trying
4,0461722
answered Jul 28 at 9:50
trying
4,0461722
4,0461722
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
add a comment |Â
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
HWAT. At this point, the reader is only 3 chapters into complex analysis and is not expected to have a topology background apart from basic topology in elementary analysis. Anyhoo, my understanding of your answer is...that that's your best (as in most elementary proof you can think of) guess for Q5, and you're not answering Q1-4?
– BCLC
Jul 28 at 10:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
trying, edited question w/c is now reopened.
– BCLC
Aug 5 at 14:06
add a comment |Â
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2
This is all very complicated: "to show $h$ is holomorphic in $G'$?". Why not just appeal to the fact that the composite of two holomorphic functions is holomorphic?
– Lord Shark the Unknown
Jul 28 at 9:18
2
If it's not, then so much the worse for that text.
– Lord Shark the Unknown
Jul 28 at 9:35
3
I'd use the open mapping theorem.... A more naive way is to apply Cauchy-Riemann.
– Lord Shark the Unknown
Jul 28 at 9:43
1
What do you mean by $f(G) subseteq z=1 $?
– zhw.
Aug 5 at 16:27
1
But you wrote $z=1$ not $|z|=1$
– zhw.
Aug 5 at 18:26