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Prove that an entire function with $Imfle (Ref)^2$is constant
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This is an old question from Ph.D Qualifying Exam of Complex Analysis.
Without using Picard's theorem directly, prove that if $f$ is an entire function such that $textImf(z)le (textRef(z))^2$ for all $zin mathbbC$ then $f$ is constant.
My attempt: I'd like to apply Liouville's theorem, but I can't find the entire function to apply the theorem. First I tried for $e^f^2$, but its real part becomes $e^(Ref)^2-(Imf)^2ge e^Imf-(Imf)^2$ so I failed to find the bound.
How should I do?
Thanks in advance!
complex-analysis entire-functions
asked Jul 19 at 5:04
bellcircle
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up vote
1
down vote
favorite
This is an old question from Ph.D Qualifying Exam of Complex Analysis.
Without using Picard's theorem directly, prove that if $f$ is an entire function such that $textImf(z)le (textRef(z))^2$ for all $zin mathbbC$ then $f$ is constant.
My attempt: I'd like to apply Liouville's theorem, but I can't find the entire function to apply the theorem. First I tried for $e^f^2$, but its real part becomes $e^(Ref)^2-(Imf)^2ge e^Imf-(Imf)^2$ so I failed to find the bound.
How should I do?
Thanks in advance!
complex-analysis entire-functions
asked Jul 19 at 5:04
bellcircle
1,123311
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is an old question from Ph.D Qualifying Exam of Complex Analysis.
Without using Picard's theorem directly, prove that if $f$ is an entire function such that $textImf(z)le (textRef(z))^2$ for all $zin mathbbC$ then $f$ is constant.
My attempt: I'd like to apply Liouville's theorem, but I can't find the entire function to apply the theorem. First I tried for $e^f^2$, but its real part becomes $e^(Ref)^2-(Imf)^2ge e^Imf-(Imf)^2$ so I failed to find the bound.
How should I do?
Thanks in advance!
complex-analysis entire-functions
asked Jul 19 at 5:04
bellcircle
1,123311
This is an old question from Ph.D Qualifying Exam of Complex Analysis.
Without using Picard's theorem directly, prove that if $f$ is an entire function such that $textImf(z)le (textRef(z))^2$ for all $zin mathbbC$ then $f$ is constant.
My attempt: I'd like to apply Liouville's theorem, but I can't find the entire function to apply the theorem. First I tried for $e^f^2$, but its real part becomes $e^(Ref)^2-(Imf)^2ge e^Imf-(Imf)^2$ so I failed to find the bound.
How should I do?
Thanks in advance!
complex-analysis entire-functions
asked Jul 19 at 5:04
bellcircle
1,123311
asked Jul 19 at 5:04
bellcircle
1,123311
asked Jul 19 at 5:04
bellcircle
1,123311
asked Jul 19 at 5:04
bellcircle
1,123311
1,123311
add a comment |Â
add a comment |Â
2 Answers
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4
down vote
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The image of $f$ does not intersect a disc centred at $i$. Therefore
$zmapsto 1/(f(z)-i)$ is entire and bounded. Use Liouville.
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
add a comment |Â
up vote
0
down vote
One can prove more general result using Lord Shark the Unknown's argument as follows:
If $f$ is a nonconstant entire function, then $f(mathbbC)$ is dense in $mathbbC$.
Proof: Suppose $f(mathbbC)$ is not dense in $mathbbC$. Then there exist $z_0in mathbbC$ and $delta>0$ such that $D_delta(z_0)cap f(mathbbC)=emptyset$. Now consider the function $g(z)=dfrac1f(z)-z_0$, then $|g(z)|le dfrac1delta$ for all $zin mathbbC$ because $|f(z)-z_0|ge delta$. Therefore, $g$ is constant by Liouville's theorem, and it follows that $f$ is constant.
answered Jul 26 at 8:35
bellcircle
1,123311
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The image of $f$ does not intersect a disc centred at $i$. Therefore
$zmapsto 1/(f(z)-i)$ is entire and bounded. Use Liouville.
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
add a comment |Â
up vote
4
down vote
accepted
The image of $f$ does not intersect a disc centred at $i$. Therefore
$zmapsto 1/(f(z)-i)$ is entire and bounded. Use Liouville.
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The image of $f$ does not intersect a disc centred at $i$. Therefore
$zmapsto 1/(f(z)-i)$ is entire and bounded. Use Liouville.
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
The image of $f$ does not intersect a disc centred at $i$. Therefore
$zmapsto 1/(f(z)-i)$ is entire and bounded. Use Liouville.
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
answered Jul 19 at 5:14
Lord Shark the Unknown
85.5k951112
85.5k951112
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
add a comment |Â
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
Clever, generalizable trick. Do you know of any other good techniques for these kinds of problems?
– Nitin
Jul 19 at 6:07
add a comment |Â
up vote
0
down vote
One can prove more general result using Lord Shark the Unknown's argument as follows:
If $f$ is a nonconstant entire function, then $f(mathbbC)$ is dense in $mathbbC$.
Proof: Suppose $f(mathbbC)$ is not dense in $mathbbC$. Then there exist $z_0in mathbbC$ and $delta>0$ such that $D_delta(z_0)cap f(mathbbC)=emptyset$. Now consider the function $g(z)=dfrac1f(z)-z_0$, then $|g(z)|le dfrac1delta$ for all $zin mathbbC$ because $|f(z)-z_0|ge delta$. Therefore, $g$ is constant by Liouville's theorem, and it follows that $f$ is constant.
answered Jul 26 at 8:35
bellcircle
1,123311
add a comment |Â
up vote
0
down vote
One can prove more general result using Lord Shark the Unknown's argument as follows:
If $f$ is a nonconstant entire function, then $f(mathbbC)$ is dense in $mathbbC$.
Proof: Suppose $f(mathbbC)$ is not dense in $mathbbC$. Then there exist $z_0in mathbbC$ and $delta>0$ such that $D_delta(z_0)cap f(mathbbC)=emptyset$. Now consider the function $g(z)=dfrac1f(z)-z_0$, then $|g(z)|le dfrac1delta$ for all $zin mathbbC$ because $|f(z)-z_0|ge delta$. Therefore, $g$ is constant by Liouville's theorem, and it follows that $f$ is constant.
answered Jul 26 at 8:35
bellcircle
1,123311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One can prove more general result using Lord Shark the Unknown's argument as follows:
If $f$ is a nonconstant entire function, then $f(mathbbC)$ is dense in $mathbbC$.
Proof: Suppose $f(mathbbC)$ is not dense in $mathbbC$. Then there exist $z_0in mathbbC$ and $delta>0$ such that $D_delta(z_0)cap f(mathbbC)=emptyset$. Now consider the function $g(z)=dfrac1f(z)-z_0$, then $|g(z)|le dfrac1delta$ for all $zin mathbbC$ because $|f(z)-z_0|ge delta$. Therefore, $g$ is constant by Liouville's theorem, and it follows that $f$ is constant.
answered Jul 26 at 8:35
bellcircle
1,123311
One can prove more general result using Lord Shark the Unknown's argument as follows:
If $f$ is a nonconstant entire function, then $f(mathbbC)$ is dense in $mathbbC$.
Proof: Suppose $f(mathbbC)$ is not dense in $mathbbC$. Then there exist $z_0in mathbbC$ and $delta>0$ such that $D_delta(z_0)cap f(mathbbC)=emptyset$. Now consider the function $g(z)=dfrac1f(z)-z_0$, then $|g(z)|le dfrac1delta$ for all $zin mathbbC$ because $|f(z)-z_0|ge delta$. Therefore, $g$ is constant by Liouville's theorem, and it follows that $f$ is constant.
answered Jul 26 at 8:35
bellcircle
1,123311
answered Jul 26 at 8:35
bellcircle
1,123311
answered Jul 26 at 8:35
bellcircle
1,123311
answered Jul 26 at 8:35
bellcircle
1,123311
1,123311
add a comment |Â
add a comment |Â
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