$f$ is an entire function s.t $|f(z)|=1$ $forall z in Bbb R$. prove that $f$ has no zeros in $Bbb C$
up vote
3
down vote
favorite
My attempt: I was trying to apply identity theorem that if the zeros of the function do have any limit point and it will be a zero function but setting $g(z)=f(z)-1$ will not help me. Can anyone help me in that!!
complex-analysis analysis holomorphic-functions analytic-functions entire-functions
asked Nov 23 at 2:47
Gimgim
565
|
show 1 more comment
up vote
3
down vote
favorite
My attempt: I was trying to apply identity theorem that if the zeros of the function do have any limit point and it will be a zero function but setting $g(z)=f(z)-1$ will not help me. Can anyone help me in that!!
complex-analysis analysis holomorphic-functions analytic-functions entire-functions
asked Nov 23 at 2:47
Gimgim
565
$f(z)$ must be of the form $f(z)=e^{iz}$ for all $z in R$, so if it is equal to that along a line by analytic continuation it must be equal to that everywhere, and $e^{iz}$ has no zeroes other than at $-infty$
– Seth
Nov 23 at 2:57
@seth Your assumption does not seem right. For one thing, $f$ may be $e^{i z^2}$, or $e^{i g(z)}$ for any entire function $g$ that sends $Bbb R$ to $Bbb R$.
– Saucy O'Path
Nov 23 at 3:05
Good point, I missed that. It should be more like $f(z)=e^{iRe(g(z))}$ or something, but still it would have to be a real number and its zeroes would be in the same places I think
– Seth
Nov 23 at 3:08
4
Perhaps Using maximum principle modulus to map upper (lower) half plane to unit disk, be helpful.
– Nosrati
Nov 23 at 3:11
So if $I$ is the map which takes the upper half plane into a disc analytically then $f circ I^{-1}$ takes the disc to the complex plane s.t it has modulus $1$ on the boundary of the disc. Now what?
– Gimgim
Nov 23 at 4:44
|
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
My attempt: I was trying to apply identity theorem that if the zeros of the function do have any limit point and it will be a zero function but setting $g(z)=f(z)-1$ will not help me. Can anyone help me in that!!
complex-analysis analysis holomorphic-functions analytic-functions entire-functions
asked Nov 23 at 2:47
Gimgim
565
My attempt: I was trying to apply identity theorem that if the zeros of the function do have any limit point and it will be a zero function but setting $g(z)=f(z)-1$ will not help me. Can anyone help me in that!!
complex-analysis analysis holomorphic-functions analytic-functions entire-functions
complex-analysis analysis holomorphic-functions analytic-functions entire-functions
asked Nov 23 at 2:47
Gimgim
565
asked Nov 23 at 2:47
Gimgim
565
asked Nov 23 at 2:47
Gimgim
565
asked Nov 23 at 2:47
Gimgim
565
asked Nov 23 at 2:47
Gimgim
565
565
$f(z)$ must be of the form $f(z)=e^{iz}$ for all $z in R$, so if it is equal to that along a line by analytic continuation it must be equal to that everywhere, and $e^{iz}$ has no zeroes other than at $-infty$
– Seth
Nov 23 at 2:57
@seth Your assumption does not seem right. For one thing, $f$ may be $e^{i z^2}$, or $e^{i g(z)}$ for any entire function $g$ that sends $Bbb R$ to $Bbb R$.
– Saucy O'Path
Nov 23 at 3:05
Good point, I missed that. It should be more like $f(z)=e^{iRe(g(z))}$ or something, but still it would have to be a real number and its zeroes would be in the same places I think
– Seth
Nov 23 at 3:08
4
Perhaps Using maximum principle modulus to map upper (lower) half plane to unit disk, be helpful.
– Nosrati
Nov 23 at 3:11
So if $I$ is the map which takes the upper half plane into a disc analytically then $f circ I^{-1}$ takes the disc to the complex plane s.t it has modulus $1$ on the boundary of the disc. Now what?
– Gimgim
Nov 23 at 4:44
|
show 1 more comment
$f(z)$ must be of the form $f(z)=e^{iz}$ for all $z in R$, so if it is equal to that along a line by analytic continuation it must be equal to that everywhere, and $e^{iz}$ has no zeroes other than at $-infty$
– Seth
Nov 23 at 2:57
@seth Your assumption does not seem right. For one thing, $f$ may be $e^{i z^2}$, or $e^{i g(z)}$ for any entire function $g$ that sends $Bbb R$ to $Bbb R$.
– Saucy O'Path
Nov 23 at 3:05
Good point, I missed that. It should be more like $f(z)=e^{iRe(g(z))}$ or something, but still it would have to be a real number and its zeroes would be in the same places I think
– Seth
Nov 23 at 3:08
4
Perhaps Using maximum principle modulus to map upper (lower) half plane to unit disk, be helpful.
– Nosrati
Nov 23 at 3:11
So if $I$ is the map which takes the upper half plane into a disc analytically then $f circ I^{-1}$ takes the disc to the complex plane s.t it has modulus $1$ on the boundary of the disc. Now what?
– Gimgim
Nov 23 at 4:44
$f(z)$ must be of the form $f(z)=e^{iz}$ for all $z in R$, so if it is equal to that along a line by analytic continuation it must be equal to that everywhere, and $e^{iz}$ has no zeroes other than at $-infty$
– Seth
Nov 23 at 2:57
$f(z)$ must be of the form $f(z)=e^{iz}$ for all $z in R$, so if it is equal to that along a line by analytic continuation it must be equal to that everywhere, and $e^{iz}$ has no zeroes other than at $-infty$
– Seth
Nov 23 at 2:57
@seth Your assumption does not seem right. For one thing, $f$ may be $e^{i z^2}$, or $e^{i g(z)}$ for any entire function $g$ that sends $Bbb R$ to $Bbb R$.
– Saucy O'Path
Nov 23 at 3:05
@seth Your assumption does not seem right. For one thing, $f$ may be $e^{i z^2}$, or $e^{i g(z)}$ for any entire function $g$ that sends $Bbb R$ to $Bbb R$.
– Saucy O'Path
Nov 23 at 3:05
Good point, I missed that. It should be more like $f(z)=e^{iRe(g(z))}$ or something, but still it would have to be a real number and its zeroes would be in the same places I think
– Seth
Nov 23 at 3:08
Good point, I missed that. It should be more like $f(z)=e^{iRe(g(z))}$ or something, but still it would have to be a real number and its zeroes would be in the same places I think
– Seth
Nov 23 at 3:08
4
4
Perhaps Using maximum principle modulus to map upper (lower) half plane to unit disk, be helpful.
– Nosrati
Nov 23 at 3:11
Perhaps Using maximum principle modulus to map upper (lower) half plane to unit disk, be helpful.
– Nosrati
Nov 23 at 3:11
So if $I$ is the map which takes the upper half plane into a disc analytically then $f circ I^{-1}$ takes the disc to the complex plane s.t it has modulus $1$ on the boundary of the disc. Now what?
– Gimgim
Nov 23 at 4:44
So if $I$ is the map which takes the upper half plane into a disc analytically then $f circ I^{-1}$ takes the disc to the complex plane s.t it has modulus $1$ on the boundary of the disc. Now what?
– Gimgim
Nov 23 at 4:44
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Since $f$ is entire, so is the function $g$ defined by $g(z)=overline{f(overline z)}.$ The hypothesis gives $g(z)=1/f(z)$ for $zinmathbb R,$ which forces $g=1/f$ by the identity principle. So $f$ is invertible.
answered Nov 23 at 11:09
Dap
14.1k533
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $f$ is entire, so is the function $g$ defined by $g(z)=overline{f(overline z)}.$ The hypothesis gives $g(z)=1/f(z)$ for $zinmathbb R,$ which forces $g=1/f$ by the identity principle. So $f$ is invertible.
answered Nov 23 at 11:09
Dap
14.1k533
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
add a comment |
up vote
2
down vote
accepted
Since $f$ is entire, so is the function $g$ defined by $g(z)=overline{f(overline z)}.$ The hypothesis gives $g(z)=1/f(z)$ for $zinmathbb R,$ which forces $g=1/f$ by the identity principle. So $f$ is invertible.
answered Nov 23 at 11:09
Dap
14.1k533
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $f$ is entire, so is the function $g$ defined by $g(z)=overline{f(overline z)}.$ The hypothesis gives $g(z)=1/f(z)$ for $zinmathbb R,$ which forces $g=1/f$ by the identity principle. So $f$ is invertible.
answered Nov 23 at 11:09
Dap
14.1k533
Since $f$ is entire, so is the function $g$ defined by $g(z)=overline{f(overline z)}.$ The hypothesis gives $g(z)=1/f(z)$ for $zinmathbb R,$ which forces $g=1/f$ by the identity principle. So $f$ is invertible.
answered Nov 23 at 11:09
Dap
14.1k533
answered Nov 23 at 11:09
Dap
14.1k533
answered Nov 23 at 11:09
Dap
14.1k533
answered Nov 23 at 11:09
Dap
14.1k533
14.1k533
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
add a comment |
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
+1. just a small nitpick though: it seems that since you are applying the identity principle, you are assuming $frac{1}{f}$ is entire, which you can't. I think a better way to phrase things would be $f(z)g(z)$ is entire and is $1$ on the real line. So, it's $1$ everywhere.
– mathworker21
Nov 25 at 5:14
add a comment |
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$f(z)$ must be of the form $f(z)=e^{iz}$ for all $z in R$, so if it is equal to that along a line by analytic continuation it must be equal to that everywhere, and $e^{iz}$ has no zeroes other than at $-infty$
– Seth
Nov 23 at 2:57
@seth Your assumption does not seem right. For one thing, $f$ may be $e^{i z^2}$, or $e^{i g(z)}$ for any entire function $g$ that sends $Bbb R$ to $Bbb R$.
– Saucy O'Path
Nov 23 at 3:05
Good point, I missed that. It should be more like $f(z)=e^{iRe(g(z))}$ or something, but still it would have to be a real number and its zeroes would be in the same places I think
– Seth
Nov 23 at 3:08
4
Perhaps Using maximum principle modulus to map upper (lower) half plane to unit disk, be helpful.
– Nosrati
Nov 23 at 3:11
So if $I$ is the map which takes the upper half plane into a disc analytically then $f circ I^{-1}$ takes the disc to the complex plane s.t it has modulus $1$ on the boundary of the disc. Now what?
– Gimgim
Nov 23 at 4:44