Mixing
Does entire and v>0 implies f is constant? [duplicate]
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This question already has an answer here:
What might I use to show that an entire function with positive real parts is constant?
5 answers
Let $f$ : $C$ to $C$ be a function. Assume that $f$ is entire and Im(f(z)) > 0 for all $z$ belongs to $C$. Does that imply $f$ is constant?
I think f will be constant,but can't deduce it
Its easy when the imaginary part(v) is constant but does v>0 also implies that f is constant?
what if v is non-negative that is it can attain $0$ also, will the result remain same?
complex-analysis
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
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marked as duplicate by Martin R, Song, RRL, Lord Shark the Unknown, Leucippus Jan 17 at 6:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What might I use to show that an entire function with positive real parts is constant?
5 answers
Let $f$ : $C$ to $C$ be a function. Assume that $f$ is entire and Im(f(z)) > 0 for all $z$ belongs to $C$. Does that imply $f$ is constant?
I think f will be constant,but can't deduce it
Its easy when the imaginary part(v) is constant but does v>0 also implies that f is constant?
what if v is non-negative that is it can attain $0$ also, will the result remain same?
complex-analysis
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
$endgroup$
marked as duplicate by Martin R, Song, RRL, Lord Shark the Unknown, Leucippus Jan 17 at 6:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance?
$endgroup$
– Wojowu
Jan 15 at 12:06
$begingroup$
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad
$endgroup$
– Nilesh Khatri
Jan 15 at 13:49
add a comment |
$begingroup$
This question already has an answer here:
What might I use to show that an entire function with positive real parts is constant?
5 answers
Let $f$ : $C$ to $C$ be a function. Assume that $f$ is entire and Im(f(z)) > 0 for all $z$ belongs to $C$. Does that imply $f$ is constant?
I think f will be constant,but can't deduce it
Its easy when the imaginary part(v) is constant but does v>0 also implies that f is constant?
what if v is non-negative that is it can attain $0$ also, will the result remain same?
complex-analysis
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
$endgroup$
This question already has an answer here:
What might I use to show that an entire function with positive real parts is constant?
5 answers
Let $f$ : $C$ to $C$ be a function. Assume that $f$ is entire and Im(f(z)) > 0 for all $z$ belongs to $C$. Does that imply $f$ is constant?
I think f will be constant,but can't deduce it
Its easy when the imaginary part(v) is constant but does v>0 also implies that f is constant?
what if v is non-negative that is it can attain $0$ also, will the result remain same?
This question already has an answer here:
What might I use to show that an entire function with positive real parts is constant?
5 answers
complex-analysis
complex-analysis
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
edited Jan 15 at 13:50
Nilesh Khatri
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
asked Jan 15 at 12:04
Nilesh KhatriNilesh Khatri
195
195
marked as duplicate by Martin R, Song, RRL, Lord Shark the Unknown, Leucippus Jan 17 at 6:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Song, RRL, Lord Shark the Unknown, Leucippus Jan 17 at 6:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance?
$endgroup$
– Wojowu
Jan 15 at 12:06
$begingroup$
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad
$endgroup$
– Nilesh Khatri
Jan 15 at 13:49
add a comment |
$begingroup$
"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance?
$endgroup$
– Wojowu
Jan 15 at 12:06
$begingroup$
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad
$endgroup$
– Nilesh Khatri
Jan 15 at 13:49
$begingroup$
"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance?
$endgroup$
– Wojowu
Jan 15 at 12:06
$begingroup$
"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance?
$endgroup$
– Wojowu
Jan 15 at 12:06
$begingroup$
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad
$endgroup$
– Nilesh Khatri
Jan 15 at 13:49
$begingroup$
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad
$endgroup$
– Nilesh Khatri
Jan 15 at 13:49
add a comment |
1 Answer
1
active
oldest
votes
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If $f$ is a non-constant entire function then $f(Bbb C)$ is dense in the plane.
Proof: Suppose that $f(Bbb C)cap D(p,r)=emptyset$. Then $g=dots$ is bounded, hence $g$ is constant, so $f$ is constant.
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $f$ is a non-constant entire function then $f(Bbb C)$ is dense in the plane.
Proof: Suppose that $f(Bbb C)cap D(p,r)=emptyset$. Then $g=dots$ is bounded, hence $g$ is constant, so $f$ is constant.
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
add a comment |
$begingroup$
If $f$ is a non-constant entire function then $f(Bbb C)$ is dense in the plane.
Proof: Suppose that $f(Bbb C)cap D(p,r)=emptyset$. Then $g=dots$ is bounded, hence $g$ is constant, so $f$ is constant.
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
add a comment |
$begingroup$
If $f$ is a non-constant entire function then $f(Bbb C)$ is dense in the plane.
Proof: Suppose that $f(Bbb C)cap D(p,r)=emptyset$. Then $g=dots$ is bounded, hence $g$ is constant, so $f$ is constant.
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
If $f$ is a non-constant entire function then $f(Bbb C)$ is dense in the plane.
Proof: Suppose that $f(Bbb C)cap D(p,r)=emptyset$. Then $g=dots$ is bounded, hence $g$ is constant, so $f$ is constant.
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 15 at 13:35
David C. UllrichDavid C. Ullrich
60.9k43994
60.9k43994
add a comment |
add a comment |
$begingroup$
"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance?
$endgroup$
– Wojowu
Jan 15 at 12:06
$begingroup$
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad
$endgroup$
– Nilesh Khatri
Jan 15 at 13:49