$|f(z)|=cinBbb Rforall zinOmega$ for $OmegasubsetBbb C$ open and connected $implies f$ is constant. Why must...
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Let $f$ be holomorphic on $OmegasubsetBbb C$, and $Omega$ is open and connected. If $|f(z)|$ is constant it is known that $f$ is constant on $Omega$
The proof of this fact says that if $f$ is not constant then $f(Omega)$ is open as the image of an open by a holomorphic function. But a circle of radius $cinBbb R_+$ is not open and $f(Omega)$ wouldn't be open either as a subset of that circle, which contradicts the first sentence.
Question 1
Where is the connectedness of $Omega$ used in the above proof?
Question 2
Why should $Omega$ have any of the two constraints? Is the following proof invalid?
writing $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$ since $f$ is holomorphic the Cauchy-Riemann equations hold: $$frac{partial u}{partial x}=frac{partial v}{partial y}~~text{ and }~~frac{partial u}{partial y}=-frac{partial v}{partial x}$$ $forall (x+iy)inOmega ~|f(x+iy)|=sqrt{u^2(x,y)+v^2(x,y)}=cinBbb R_+$
$impliesfrac{partial }{partial x}[sqrt{u^2(x,y)+v^2(x,y)}]=frac{partial }{partial y}[sqrt{u^2(x,y)+v^2(x,y)}]=0$
$iff (u_x+v_x)(u^2+v^2)^{-{1over2}}=(u_y+v_y)(u^2+v^2)^{-{1over2}}=0$
If $u^2+v^2=|f|^2=0forall zinOmega$ then $f=0$ and we're done
Let's suppose $u^2+v^2ne0$ then we can divide by $(u^2+v^2)^{-{1over2}}$ it on both sides to get: $$u_x+v_x=u_y+v_y=0$$ by the C-R equations $$u_x-u_y=u_y+u_ximplies u_y=-v_x=0$$ substituting $v_x=0$ into $u_x+v_x=0$ $$u_x=0implies v_y=0$$
All the partial derivatives of $u$ and $v$ are zero and since $f'(z)=u_x(x,y)+iv_x(x,y)=0$ we get that $f$ is constant.
complex-analysis proof-verification examples-counterexamples connectedness
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
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show 5 more comments
$begingroup$
Let $f$ be holomorphic on $OmegasubsetBbb C$, and $Omega$ is open and connected. If $|f(z)|$ is constant it is known that $f$ is constant on $Omega$
The proof of this fact says that if $f$ is not constant then $f(Omega)$ is open as the image of an open by a holomorphic function. But a circle of radius $cinBbb R_+$ is not open and $f(Omega)$ wouldn't be open either as a subset of that circle, which contradicts the first sentence.
Question 1
Where is the connectedness of $Omega$ used in the above proof?
Question 2
Why should $Omega$ have any of the two constraints? Is the following proof invalid?
writing $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$ since $f$ is holomorphic the Cauchy-Riemann equations hold: $$frac{partial u}{partial x}=frac{partial v}{partial y}~~text{ and }~~frac{partial u}{partial y}=-frac{partial v}{partial x}$$ $forall (x+iy)inOmega ~|f(x+iy)|=sqrt{u^2(x,y)+v^2(x,y)}=cinBbb R_+$
$impliesfrac{partial }{partial x}[sqrt{u^2(x,y)+v^2(x,y)}]=frac{partial }{partial y}[sqrt{u^2(x,y)+v^2(x,y)}]=0$
$iff (u_x+v_x)(u^2+v^2)^{-{1over2}}=(u_y+v_y)(u^2+v^2)^{-{1over2}}=0$
If $u^2+v^2=|f|^2=0forall zinOmega$ then $f=0$ and we're done
Let's suppose $u^2+v^2ne0$ then we can divide by $(u^2+v^2)^{-{1over2}}$ it on both sides to get: $$u_x+v_x=u_y+v_y=0$$ by the C-R equations $$u_x-u_y=u_y+u_ximplies u_y=-v_x=0$$ substituting $v_x=0$ into $u_x+v_x=0$ $$u_x=0implies v_y=0$$
All the partial derivatives of $u$ and $v$ are zero and since $f'(z)=u_x(x,y)+iv_x(x,y)=0$ we get that $f$ is constant.
complex-analysis proof-verification examples-counterexamples connectedness
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
$endgroup$
$begingroup$
In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain
$endgroup$
– vidyarthi
Dec 18 '18 at 17:09
$begingroup$
Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same.
$endgroup$
– copper.hat
Dec 18 '18 at 17:12
$begingroup$
The conditions are sufficient, not necessary. $f$ may be constant without $Omega$ being either open or connected.
$endgroup$
– copper.hat
Dec 18 '18 at 17:13
$begingroup$
But the connectedness is necessary, no?
$endgroup$
– John Cataldo
Dec 18 '18 at 17:14
1
$begingroup$
You were asking in Question 1 where connectedness is used, I am giving an example where $Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used.
$endgroup$
– copper.hat
Dec 18 '18 at 17:26
|
show 5 more comments
$begingroup$
Let $f$ be holomorphic on $OmegasubsetBbb C$, and $Omega$ is open and connected. If $|f(z)|$ is constant it is known that $f$ is constant on $Omega$
The proof of this fact says that if $f$ is not constant then $f(Omega)$ is open as the image of an open by a holomorphic function. But a circle of radius $cinBbb R_+$ is not open and $f(Omega)$ wouldn't be open either as a subset of that circle, which contradicts the first sentence.
Question 1
Where is the connectedness of $Omega$ used in the above proof?
Question 2
Why should $Omega$ have any of the two constraints? Is the following proof invalid?
writing $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$ since $f$ is holomorphic the Cauchy-Riemann equations hold: $$frac{partial u}{partial x}=frac{partial v}{partial y}~~text{ and }~~frac{partial u}{partial y}=-frac{partial v}{partial x}$$ $forall (x+iy)inOmega ~|f(x+iy)|=sqrt{u^2(x,y)+v^2(x,y)}=cinBbb R_+$
$impliesfrac{partial }{partial x}[sqrt{u^2(x,y)+v^2(x,y)}]=frac{partial }{partial y}[sqrt{u^2(x,y)+v^2(x,y)}]=0$
$iff (u_x+v_x)(u^2+v^2)^{-{1over2}}=(u_y+v_y)(u^2+v^2)^{-{1over2}}=0$
If $u^2+v^2=|f|^2=0forall zinOmega$ then $f=0$ and we're done
Let's suppose $u^2+v^2ne0$ then we can divide by $(u^2+v^2)^{-{1over2}}$ it on both sides to get: $$u_x+v_x=u_y+v_y=0$$ by the C-R equations $$u_x-u_y=u_y+u_ximplies u_y=-v_x=0$$ substituting $v_x=0$ into $u_x+v_x=0$ $$u_x=0implies v_y=0$$
All the partial derivatives of $u$ and $v$ are zero and since $f'(z)=u_x(x,y)+iv_x(x,y)=0$ we get that $f$ is constant.
complex-analysis proof-verification examples-counterexamples connectedness
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
$endgroup$
Let $f$ be holomorphic on $OmegasubsetBbb C$, and $Omega$ is open and connected. If $|f(z)|$ is constant it is known that $f$ is constant on $Omega$
The proof of this fact says that if $f$ is not constant then $f(Omega)$ is open as the image of an open by a holomorphic function. But a circle of radius $cinBbb R_+$ is not open and $f(Omega)$ wouldn't be open either as a subset of that circle, which contradicts the first sentence.
Question 1
Where is the connectedness of $Omega$ used in the above proof?
Question 2
Why should $Omega$ have any of the two constraints? Is the following proof invalid?
writing $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$ since $f$ is holomorphic the Cauchy-Riemann equations hold: $$frac{partial u}{partial x}=frac{partial v}{partial y}~~text{ and }~~frac{partial u}{partial y}=-frac{partial v}{partial x}$$ $forall (x+iy)inOmega ~|f(x+iy)|=sqrt{u^2(x,y)+v^2(x,y)}=cinBbb R_+$
$impliesfrac{partial }{partial x}[sqrt{u^2(x,y)+v^2(x,y)}]=frac{partial }{partial y}[sqrt{u^2(x,y)+v^2(x,y)}]=0$
$iff (u_x+v_x)(u^2+v^2)^{-{1over2}}=(u_y+v_y)(u^2+v^2)^{-{1over2}}=0$
If $u^2+v^2=|f|^2=0forall zinOmega$ then $f=0$ and we're done
Let's suppose $u^2+v^2ne0$ then we can divide by $(u^2+v^2)^{-{1over2}}$ it on both sides to get: $$u_x+v_x=u_y+v_y=0$$ by the C-R equations $$u_x-u_y=u_y+u_ximplies u_y=-v_x=0$$ substituting $v_x=0$ into $u_x+v_x=0$ $$u_x=0implies v_y=0$$
All the partial derivatives of $u$ and $v$ are zero and since $f'(z)=u_x(x,y)+iv_x(x,y)=0$ we get that $f$ is constant.
complex-analysis proof-verification examples-counterexamples connectedness
complex-analysis proof-verification examples-counterexamples connectedness
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
edited Dec 18 '18 at 17:03
John Cataldo
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
asked Dec 18 '18 at 16:58
John CataldoJohn Cataldo
1,1931316
1,1931316
$begingroup$
In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain
$endgroup$
– vidyarthi
Dec 18 '18 at 17:09
$begingroup$
Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same.
$endgroup$
– copper.hat
Dec 18 '18 at 17:12
$begingroup$
The conditions are sufficient, not necessary. $f$ may be constant without $Omega$ being either open or connected.
$endgroup$
– copper.hat
Dec 18 '18 at 17:13
$begingroup$
But the connectedness is necessary, no?
$endgroup$
– John Cataldo
Dec 18 '18 at 17:14
1
$begingroup$
You were asking in Question 1 where connectedness is used, I am giving an example where $Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used.
$endgroup$
– copper.hat
Dec 18 '18 at 17:26
|
show 5 more comments
$begingroup$
In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain
$endgroup$
– vidyarthi
Dec 18 '18 at 17:09
$begingroup$
Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same.
$endgroup$
– copper.hat
Dec 18 '18 at 17:12
$begingroup$
The conditions are sufficient, not necessary. $f$ may be constant without $Omega$ being either open or connected.
$endgroup$
– copper.hat
Dec 18 '18 at 17:13
$begingroup$
But the connectedness is necessary, no?
$endgroup$
– John Cataldo
Dec 18 '18 at 17:14
1
$begingroup$
You were asking in Question 1 where connectedness is used, I am giving an example where $Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used.
$endgroup$
– copper.hat
Dec 18 '18 at 17:26
$begingroup$
In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain
$endgroup$
– vidyarthi
Dec 18 '18 at 17:09
$begingroup$
In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain
$endgroup$
– vidyarthi
Dec 18 '18 at 17:09
$begingroup$
Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same.
$endgroup$
– copper.hat
Dec 18 '18 at 17:12
$begingroup$
Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same.
$endgroup$
– copper.hat
Dec 18 '18 at 17:12
$begingroup$
The conditions are sufficient, not necessary. $f$ may be constant without $Omega$ being either open or connected.
$endgroup$
– copper.hat
Dec 18 '18 at 17:13
$begingroup$
The conditions are sufficient, not necessary. $f$ may be constant without $Omega$ being either open or connected.
$endgroup$
– copper.hat
Dec 18 '18 at 17:13
$begingroup$
But the connectedness is necessary, no?
$endgroup$
– John Cataldo
Dec 18 '18 at 17:14
$begingroup$
But the connectedness is necessary, no?
$endgroup$
– John Cataldo
Dec 18 '18 at 17:14
1
1
$begingroup$
You were asking in Question 1 where connectedness is used, I am giving an example where $Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used.
$endgroup$
– copper.hat
Dec 18 '18 at 17:26
$begingroup$
You were asking in Question 1 where connectedness is used, I am giving an example where $Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used.
$endgroup$
– copper.hat
Dec 18 '18 at 17:26
|
show 5 more comments
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In your proof, you use the fact of connectedness of domain: If a multivariable function has its total derivative zero in a connected domain, then the function is constant on that domain
$endgroup$
– vidyarthi
Dec 18 '18 at 17:09
$begingroup$
Let $f=1$ on $B(0,1)$ and $f=-1$ on $B(2,1)$ then $|f|$ is constant but $f$ is not. You need connectedness so the constant is the same.
$endgroup$
– copper.hat
Dec 18 '18 at 17:12
$begingroup$
The conditions are sufficient, not necessary. $f$ may be constant without $Omega$ being either open or connected.
$endgroup$
– copper.hat
Dec 18 '18 at 17:13
$begingroup$
But the connectedness is necessary, no?
$endgroup$
– John Cataldo
Dec 18 '18 at 17:14
1
$begingroup$
You were asking in Question 1 where connectedness is used, I am giving an example where $Omega$ is not connected, but is open and $|f|$ is constant but $f$ is not. This should indicate where connectedness is used.
$endgroup$
– copper.hat
Dec 18 '18 at 17:26