If $f$ is analytic and $f(z)^2$ = $bar f(z)$ then $f$ is constant
$begingroup$
I'm currently stuck on the following problem.
Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.
I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.
real-analysis complex-analysis
asked Jan 4 at 2:24
user628226user628226
575
$endgroup$
add a comment |
$begingroup$
I'm currently stuck on the following problem.
Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.
I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.
real-analysis complex-analysis
asked Jan 4 at 2:24
user628226user628226
575
$endgroup$
4
$begingroup$
Maybe use $f^3 = f bar f$ is a pure real analytic function?
$endgroup$
– JonathanZ
Jan 4 at 2:29
1
$begingroup$
The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 5:40
add a comment |
$begingroup$
I'm currently stuck on the following problem.
Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.
I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.
real-analysis complex-analysis
asked Jan 4 at 2:24
user628226user628226
575
$endgroup$
I'm currently stuck on the following problem.
Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.
I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.
real-analysis complex-analysis
real-analysis complex-analysis
asked Jan 4 at 2:24
user628226user628226
575
asked Jan 4 at 2:24
user628226user628226
575
asked Jan 4 at 2:24
user628226user628226
575
asked Jan 4 at 2:24
user628226user628226
575
asked Jan 4 at 2:24
user628226user628226
575
575
4
$begingroup$
Maybe use $f^3 = f bar f$ is a pure real analytic function?
$endgroup$
– JonathanZ
Jan 4 at 2:29
1
$begingroup$
The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 5:40
add a comment |
4
$begingroup$
Maybe use $f^3 = f bar f$ is a pure real analytic function?
$endgroup$
– JonathanZ
Jan 4 at 2:29
1
$begingroup$
The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 5:40
4
4
$begingroup$
Maybe use $f^3 = f bar f$ is a pure real analytic function?
$endgroup$
– JonathanZ
Jan 4 at 2:29
$begingroup$
Maybe use $f^3 = f bar f$ is a pure real analytic function?
$endgroup$
– JonathanZ
Jan 4 at 2:29
1
1
$begingroup$
The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 5:40
$begingroup$
The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 5:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No need. First solve
$$
Z^2=bar{Z}qquad (1)
$$
Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
$$
j=e^{frac{2ipi}{3}}
$$
then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
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add a comment |
$begingroup$
$$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No need. First solve
$$
Z^2=bar{Z}qquad (1)
$$
Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
$$
j=e^{frac{2ipi}{3}}
$$
then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
add a comment |
$begingroup$
No need. First solve
$$
Z^2=bar{Z}qquad (1)
$$
Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
$$
j=e^{frac{2ipi}{3}}
$$
then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
add a comment |
$begingroup$
No need. First solve
$$
Z^2=bar{Z}qquad (1)
$$
Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
$$
j=e^{frac{2ipi}{3}}
$$
then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
No need. First solve
$$
Z^2=bar{Z}qquad (1)
$$
Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
$$
j=e^{frac{2ipi}{3}}
$$
then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
edited Jan 5 at 9:04
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 4 at 2:33
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
2,644919
add a comment |
add a comment |
$begingroup$
$$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
add a comment |
$begingroup$
$$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
add a comment |
$begingroup$
$$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
answered Jan 4 at 2:51
Jack D'AurizioJack D'Aurizio
291k33284667
291k33284667
add a comment |
add a comment |
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4
$begingroup$
Maybe use $f^3 = f bar f$ is a pure real analytic function?
$endgroup$
– JonathanZ
Jan 4 at 2:29
1
$begingroup$
The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 5:40