True or False $Re f(z)>0$ [duplicate]
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This question already has an answer here:
Prove that a holomorphic function with postive real part is constant
1 answer
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the property:(Justify your answer)
$Re f(z)>0$ for all $zinBbb{C}$ and $f$ is not constant.
My attemp: I find a question in Conway that is discussed in
$f : D rightarrow mathbb{C} $ analytic. Show $operatorname{Re}(f(z)) geq 0$.
But my question is difference. Because we have not the hypothesis $Re f(z)geq 0$ and the domain is $Bbb{C}$ instead of $D$.
So we can not use the open mapping theorem and I think there is not such function!
complex-analysis
asked Jan 9 at 18:43
seyedseyed
566
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marked as duplicate by Martin R, Community♦ Jan 9 at 19:09
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:
Prove that a holomorphic function with postive real part is constant
1 answer
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the property:(Justify your answer)
$Re f(z)>0$ for all $zinBbb{C}$ and $f$ is not constant.
My attemp: I find a question in Conway that is discussed in
$f : D rightarrow mathbb{C} $ analytic. Show $operatorname{Re}(f(z)) geq 0$.
But my question is difference. Because we have not the hypothesis $Re f(z)geq 0$ and the domain is $Bbb{C}$ instead of $D$.
So we can not use the open mapping theorem and I think there is not such function!
complex-analysis
asked Jan 9 at 18:43
seyedseyed
566
$endgroup$
marked as duplicate by Martin R, Community♦ Jan 9 at 19:09
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.
$begingroup$
Also related: math.stackexchange.com/q/229312/42969.
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– Martin R
Jan 9 at 18:58
add a comment |
$begingroup$
This question already has an answer here:
Prove that a holomorphic function with postive real part is constant
1 answer
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the property:(Justify your answer)
$Re f(z)>0$ for all $zinBbb{C}$ and $f$ is not constant.
My attemp: I find a question in Conway that is discussed in
$f : D rightarrow mathbb{C} $ analytic. Show $operatorname{Re}(f(z)) geq 0$.
But my question is difference. Because we have not the hypothesis $Re f(z)geq 0$ and the domain is $Bbb{C}$ instead of $D$.
So we can not use the open mapping theorem and I think there is not such function!
complex-analysis
asked Jan 9 at 18:43
seyedseyed
566
$endgroup$
This question already has an answer here:
Prove that a holomorphic function with postive real part is constant
1 answer
Make an examples $f:Bbb{C}toBbb{C}$ holomorph (If it exists) with the property:(Justify your answer)
$Re f(z)>0$ for all $zinBbb{C}$ and $f$ is not constant.
My attemp: I find a question in Conway that is discussed in
$f : D rightarrow mathbb{C} $ analytic. Show $operatorname{Re}(f(z)) geq 0$.
But my question is difference. Because we have not the hypothesis $Re f(z)geq 0$ and the domain is $Bbb{C}$ instead of $D$.
So we can not use the open mapping theorem and I think there is not such function!
This question already has an answer here:
Prove that a holomorphic function with postive real part is constant
1 answer
complex-analysis
complex-analysis
asked Jan 9 at 18:43
seyedseyed
566
asked Jan 9 at 18:43
seyedseyed
566
asked Jan 9 at 18:43
seyedseyed
566
asked Jan 9 at 18:43
seyedseyed
566
asked Jan 9 at 18:43
seyedseyed
566
566
marked as duplicate by Martin R, Community♦ Jan 9 at 19:09
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, Community♦ Jan 9 at 19:09
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.
$begingroup$
Also related: math.stackexchange.com/q/229312/42969.
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– Martin R
Jan 9 at 18:58
add a comment |
$begingroup$
Also related: math.stackexchange.com/q/229312/42969.
$endgroup$
– Martin R
Jan 9 at 18:58
$begingroup$
Also related: math.stackexchange.com/q/229312/42969.
$endgroup$
– Martin R
Jan 9 at 18:58
$begingroup$
Also related: math.stackexchange.com/q/229312/42969.
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– Martin R
Jan 9 at 18:58
add a comment |
1 Answer
1
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No such $f$ exists. Suppose a non constant function $f(z)$ and let $f(z)=u+iv$ and consider the function $F(z)=e^{-f(z)}$. Then, $$|F(z)|=|e^{-u-iv}|=e^{-u}$$ Note that since $Re f(z)= u>0$, it follows that $|F(z)|<1$ and by Liouville's Theorem $F(z)$ is constant.
This implies that $F'(z)=-f'(z)e^{-f(z)}=0$ and this further implies that $f'(z)=0$. Hence, $f(z)$ must be a constant.
answered Jan 9 at 18:52
Ya GYa G
536211
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No such $f$ exists. Suppose a non constant function $f(z)$ and let $f(z)=u+iv$ and consider the function $F(z)=e^{-f(z)}$. Then, $$|F(z)|=|e^{-u-iv}|=e^{-u}$$ Note that since $Re f(z)= u>0$, it follows that $|F(z)|<1$ and by Liouville's Theorem $F(z)$ is constant.
This implies that $F'(z)=-f'(z)e^{-f(z)}=0$ and this further implies that $f'(z)=0$. Hence, $f(z)$ must be a constant.
answered Jan 9 at 18:52
Ya GYa G
536211
$endgroup$
add a comment |
$begingroup$
No such $f$ exists. Suppose a non constant function $f(z)$ and let $f(z)=u+iv$ and consider the function $F(z)=e^{-f(z)}$. Then, $$|F(z)|=|e^{-u-iv}|=e^{-u}$$ Note that since $Re f(z)= u>0$, it follows that $|F(z)|<1$ and by Liouville's Theorem $F(z)$ is constant.
This implies that $F'(z)=-f'(z)e^{-f(z)}=0$ and this further implies that $f'(z)=0$. Hence, $f(z)$ must be a constant.
answered Jan 9 at 18:52
Ya GYa G
536211
$endgroup$
add a comment |
$begingroup$
No such $f$ exists. Suppose a non constant function $f(z)$ and let $f(z)=u+iv$ and consider the function $F(z)=e^{-f(z)}$. Then, $$|F(z)|=|e^{-u-iv}|=e^{-u}$$ Note that since $Re f(z)= u>0$, it follows that $|F(z)|<1$ and by Liouville's Theorem $F(z)$ is constant.
This implies that $F'(z)=-f'(z)e^{-f(z)}=0$ and this further implies that $f'(z)=0$. Hence, $f(z)$ must be a constant.
answered Jan 9 at 18:52
Ya GYa G
536211
$endgroup$
No such $f$ exists. Suppose a non constant function $f(z)$ and let $f(z)=u+iv$ and consider the function $F(z)=e^{-f(z)}$. Then, $$|F(z)|=|e^{-u-iv}|=e^{-u}$$ Note that since $Re f(z)= u>0$, it follows that $|F(z)|<1$ and by Liouville's Theorem $F(z)$ is constant.
This implies that $F'(z)=-f'(z)e^{-f(z)}=0$ and this further implies that $f'(z)=0$. Hence, $f(z)$ must be a constant.
answered Jan 9 at 18:52
Ya GYa G
536211
edited Jan 9 at 19:04
answered Jan 9 at 18:52
Ya GYa G
536211
answered Jan 9 at 18:52
Ya GYa G
536211
answered Jan 9 at 18:52
Ya GYa G
536211
536211
add a comment |
add a comment |
$begingroup$
Also related: math.stackexchange.com/q/229312/42969.
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– Martin R
Jan 9 at 18:58