Mixing
Bilinear transformations in one variable complex analysis.
0
$begingroup$
I am self studying complex analysis 1st time from complex variables and applications from ponnusamy and silverman.
I got struck on this question:
find all bilinear transformations mapping imaginary axis onto unit circle.
I tried assuming a general bilinear transformation w = (az+b )/(cz+d) and equating its modulus to 1. But couldn't get anything.
It's on ch-3 page 84 ex no -19
Please help.
complex-analysis
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
asked Jan 18 at 14:54
AMDEAMDE
42
$endgroup$
add a comment |
0
$begingroup$
I am self studying complex analysis 1st time from complex variables and applications from ponnusamy and silverman.
I got struck on this question:
find all bilinear transformations mapping imaginary axis onto unit circle.
I tried assuming a general bilinear transformation w = (az+b )/(cz+d) and equating its modulus to 1. But couldn't get anything.
It's on ch-3 page 84 ex no -19
Please help.
complex-analysis
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
asked Jan 18 at 14:54
AMDEAMDE
42
$endgroup$
$begingroup$
Do you mean linear fractional transformations? The term "bilinear" is usually used for functions with two arguments that are linear in both of them.
$endgroup$
– 0x539
Jan 18 at 16:39
$begingroup$
A fact you might use is that a complex fractional linear transformation is already determined by the images of any three points.
$endgroup$
– 0x539
Jan 18 at 16:46
$begingroup$
Yes, I mean linear fractional transformations
$endgroup$
– AMDE
Jan 18 at 18:13
$begingroup$
0x539 - could you please elaborately tell steps to reach the answer. I could not do much despite trying. I knew that it has to do something with three points , but could not apply it. Please help.
$endgroup$
– AMDE
Jan 18 at 18:15
$begingroup$
Linear fractional transformations are completely determined by how they act on three inputs (one might think 4 inputs, since there are 4 parameters, but you can divide one of them out without changing the transformation). I suggest that you find the transformation that takes sends $0 to 1, i to i, infty to -1$.
$endgroup$
– Paul Sinclair
Jan 18 at 22:03
add a comment |
0
0
0
1
$begingroup$
I am self studying complex analysis 1st time from complex variables and applications from ponnusamy and silverman.
I got struck on this question:
find all bilinear transformations mapping imaginary axis onto unit circle.
I tried assuming a general bilinear transformation w = (az+b )/(cz+d) and equating its modulus to 1. But couldn't get anything.
It's on ch-3 page 84 ex no -19
Please help.
complex-analysis
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
asked Jan 18 at 14:54
AMDEAMDE
42
$endgroup$
I am self studying complex analysis 1st time from complex variables and applications from ponnusamy and silverman.
I got struck on this question:
find all bilinear transformations mapping imaginary axis onto unit circle.
I tried assuming a general bilinear transformation w = (az+b )/(cz+d) and equating its modulus to 1. But couldn't get anything.
It's on ch-3 page 84 ex no -19
Please help.
complex-analysis
complex-analysis
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
asked Jan 18 at 14:54
AMDEAMDE
42
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
asked Jan 18 at 14:54
AMDEAMDE
42
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
edited Jan 18 at 15:43
J. W. Tanner
2,5821217
2,5821217
asked Jan 18 at 14:54
AMDEAMDE
42
asked Jan 18 at 14:54
AMDEAMDE
42
asked Jan 18 at 14:54
AMDEAMDE
42
42
$begingroup$
Do you mean linear fractional transformations? The term "bilinear" is usually used for functions with two arguments that are linear in both of them.
$endgroup$
– 0x539
Jan 18 at 16:39
$begingroup$
A fact you might use is that a complex fractional linear transformation is already determined by the images of any three points.
$endgroup$
– 0x539
Jan 18 at 16:46
$begingroup$
Yes, I mean linear fractional transformations
$endgroup$
– AMDE
Jan 18 at 18:13
$begingroup$
0x539 - could you please elaborately tell steps to reach the answer. I could not do much despite trying. I knew that it has to do something with three points , but could not apply it. Please help.
$endgroup$
– AMDE
Jan 18 at 18:15
$begingroup$
Linear fractional transformations are completely determined by how they act on three inputs (one might think 4 inputs, since there are 4 parameters, but you can divide one of them out without changing the transformation). I suggest that you find the transformation that takes sends $0 to 1, i to i, infty to -1$.
$endgroup$
– Paul Sinclair
Jan 18 at 22:03
add a comment |
$begingroup$
Do you mean linear fractional transformations? The term "bilinear" is usually used for functions with two arguments that are linear in both of them.
$endgroup$
– 0x539
Jan 18 at 16:39
$begingroup$
A fact you might use is that a complex fractional linear transformation is already determined by the images of any three points.
$endgroup$
– 0x539
Jan 18 at 16:46
$begingroup$
Yes, I mean linear fractional transformations
$endgroup$
– AMDE
Jan 18 at 18:13
$begingroup$
0x539 - could you please elaborately tell steps to reach the answer. I could not do much despite trying. I knew that it has to do something with three points , but could not apply it. Please help.
$endgroup$
– AMDE
Jan 18 at 18:15
$begingroup$
Linear fractional transformations are completely determined by how they act on three inputs (one might think 4 inputs, since there are 4 parameters, but you can divide one of them out without changing the transformation). I suggest that you find the transformation that takes sends $0 to 1, i to i, infty to -1$.
$endgroup$
– Paul Sinclair
Jan 18 at 22:03
$begingroup$
Do you mean linear fractional transformations? The term "bilinear" is usually used for functions with two arguments that are linear in both of them.
$endgroup$
– 0x539
Jan 18 at 16:39
$begingroup$
Do you mean linear fractional transformations? The term "bilinear" is usually used for functions with two arguments that are linear in both of them.
$endgroup$
– 0x539
Jan 18 at 16:39
$begingroup$
A fact you might use is that a complex fractional linear transformation is already determined by the images of any three points.
$endgroup$
– 0x539
Jan 18 at 16:46
$begingroup$
A fact you might use is that a complex fractional linear transformation is already determined by the images of any three points.
$endgroup$
– 0x539
Jan 18 at 16:46
$begingroup$
Yes, I mean linear fractional transformations
$endgroup$
– AMDE
Jan 18 at 18:13
$begingroup$
Yes, I mean linear fractional transformations
$endgroup$
– AMDE
Jan 18 at 18:13
$begingroup$
0x539 - could you please elaborately tell steps to reach the answer. I could not do much despite trying. I knew that it has to do something with three points , but could not apply it. Please help.
$endgroup$
– AMDE
Jan 18 at 18:15
$begingroup$
0x539 - could you please elaborately tell steps to reach the answer. I could not do much despite trying. I knew that it has to do something with three points , but could not apply it. Please help.
$endgroup$
– AMDE
Jan 18 at 18:15
$begingroup$
Linear fractional transformations are completely determined by how they act on three inputs (one might think 4 inputs, since there are 4 parameters, but you can divide one of them out without changing the transformation). I suggest that you find the transformation that takes sends $0 to 1, i to i, infty to -1$.
$endgroup$
– Paul Sinclair
Jan 18 at 22:03
$begingroup$
Linear fractional transformations are completely determined by how they act on three inputs (one might think 4 inputs, since there are 4 parameters, but you can divide one of them out without changing the transformation). I suggest that you find the transformation that takes sends $0 to 1, i to i, infty to -1$.
$endgroup$
– Paul Sinclair
Jan 18 at 22:03
add a comment |
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$begingroup$
Do you mean linear fractional transformations? The term "bilinear" is usually used for functions with two arguments that are linear in both of them.
$endgroup$
– 0x539
Jan 18 at 16:39
$begingroup$
A fact you might use is that a complex fractional linear transformation is already determined by the images of any three points.
$endgroup$
– 0x539
Jan 18 at 16:46
$begingroup$
Yes, I mean linear fractional transformations
$endgroup$
– AMDE
Jan 18 at 18:13
$begingroup$
0x539 - could you please elaborately tell steps to reach the answer. I could not do much despite trying. I knew that it has to do something with three points , but could not apply it. Please help.
$endgroup$
– AMDE
Jan 18 at 18:15
$begingroup$
Linear fractional transformations are completely determined by how they act on three inputs (one might think 4 inputs, since there are 4 parameters, but you can divide one of them out without changing the transformation). I suggest that you find the transformation that takes sends $0 to 1, i to i, infty to -1$.
$endgroup$
– Paul Sinclair
Jan 18 at 22:03