An exercise from Conway (Phragmen-Lindelöf Theorem)
1
The exercise is from Chapter 6.4 Complex Analysis by Conway. I am new in this field and have no idea how to start or process. Any help would be really appreciated.
complex-analysis
edited Dec 5 '18 at 18:36
Bernard
118k639112
asked Dec 5 '18 at 17:31
syedasyeda
164
add a comment |
1
The exercise is from Chapter 6.4 Complex Analysis by Conway. I am new in this field and have no idea how to start or process. Any help would be really appreciated.
complex-analysis
edited Dec 5 '18 at 18:36
Bernard
118k639112
asked Dec 5 '18 at 17:31
syedasyeda
164
The $a<b<1$ is to ensure that $|arg(e^{bz})| = b |Im(z)|$ stays $le b pi/2$ so that $e^{A e^{Re(z)}}e^{−epsilon e^{bz}}to 0$ when $Re(z) to infty$
– reuns
Dec 5 '18 at 21:03
Still, I don't understand things. I actually saw the Wikipedia before. So if you could give me more detail.
– syeda
Dec 6 '18 at 18:16
add a comment |
1
1
1
The exercise is from Chapter 6.4 Complex Analysis by Conway. I am new in this field and have no idea how to start or process. Any help would be really appreciated.
complex-analysis
edited Dec 5 '18 at 18:36
Bernard
118k639112
asked Dec 5 '18 at 17:31
syedasyeda
164
The exercise is from Chapter 6.4 Complex Analysis by Conway. I am new in this field and have no idea how to start or process. Any help would be really appreciated.
complex-analysis
complex-analysis
edited Dec 5 '18 at 18:36
Bernard
118k639112
asked Dec 5 '18 at 17:31
syedasyeda
164
edited Dec 5 '18 at 18:36
Bernard
118k639112
asked Dec 5 '18 at 17:31
syedasyeda
164
edited Dec 5 '18 at 18:36
Bernard
118k639112
edited Dec 5 '18 at 18:36
Bernard
118k639112
edited Dec 5 '18 at 18:36
Bernard
118k639112
118k639112
asked Dec 5 '18 at 17:31
syedasyeda
164
asked Dec 5 '18 at 17:31
syedasyeda
164
asked Dec 5 '18 at 17:31
syedasyeda
164
164
The $a<b<1$ is to ensure that $|arg(e^{bz})| = b |Im(z)|$ stays $le b pi/2$ so that $e^{A e^{Re(z)}}e^{−epsilon e^{bz}}to 0$ when $Re(z) to infty$
– reuns
Dec 5 '18 at 21:03
Still, I don't understand things. I actually saw the Wikipedia before. So if you could give me more detail.
– syeda
Dec 6 '18 at 18:16
add a comment |
The $a<b<1$ is to ensure that $|arg(e^{bz})| = b |Im(z)|$ stays $le b pi/2$ so that $e^{A e^{Re(z)}}e^{−epsilon e^{bz}}to 0$ when $Re(z) to infty$
– reuns
Dec 5 '18 at 21:03
Still, I don't understand things. I actually saw the Wikipedia before. So if you could give me more detail.
– syeda
Dec 6 '18 at 18:16
The $a<b<1$ is to ensure that $|arg(e^{bz})| = b |Im(z)|$ stays $le b pi/2$ so that $e^{A e^{Re(z)}}e^{−epsilon e^{bz}}to 0$ when $Re(z) to infty$
– reuns
Dec 5 '18 at 21:03
The $a<b<1$ is to ensure that $|arg(e^{bz})| = b |Im(z)|$ stays $le b pi/2$ so that $e^{A e^{Re(z)}}e^{−epsilon e^{bz}}to 0$ when $Re(z) to infty$
– reuns
Dec 5 '18 at 21:03
Still, I don't understand things. I actually saw the Wikipedia before. So if you could give me more detail.
– syeda
Dec 6 '18 at 18:16
Still, I don't understand things. I actually saw the Wikipedia before. So if you could give me more detail.
– syeda
Dec 6 '18 at 18:16
add a comment |
1 Answer
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The complete answer resides on Wikipedia.
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
add a comment |
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1 Answer
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1 Answer
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The complete answer resides on Wikipedia.
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
add a comment |
1
The complete answer resides on Wikipedia.
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
add a comment |
1
1
1
The complete answer resides on Wikipedia.
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
The complete answer resides on Wikipedia.
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
answered Dec 5 '18 at 17:37
FedericoFederico
4,859514
4,859514
add a comment |
add a comment |
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The $a<b<1$ is to ensure that $|arg(e^{bz})| = b |Im(z)|$ stays $le b pi/2$ so that $e^{A e^{Re(z)}}e^{−epsilon e^{bz}}to 0$ when $Re(z) to infty$
– reuns
Dec 5 '18 at 21:03
Still, I don't understand things. I actually saw the Wikipedia before. So if you could give me more detail.
– syeda
Dec 6 '18 at 18:16