Mixing
What Algebra law is this called: $|alpha beta| le |alpha| |beta|$ for complex numbers
0
$begingroup$
I'm trying to remember the name of this law in complex number algebra:
$$|alpha beta| le |alpha| |beta|$$
Where:
$$alpha = Re{alpha}+ i Im{alpha}$$
$$beta = Re{beta}+ i Im{beta}$$
complex-analysis complex-numbers
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
$endgroup$
add a comment |
0
$begingroup$
I'm trying to remember the name of this law in complex number algebra:
$$|alpha beta| le |alpha| |beta|$$
Where:
$$alpha = Re{alpha}+ i Im{alpha}$$
$$beta = Re{beta}+ i Im{beta}$$
complex-analysis complex-numbers
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
$endgroup$
1
$begingroup$
It is not an Algebra law, but just an equation: $|zw|=|z||w|$ for complex $z,w$. It is sometimes called "norm", and it is multiplicative. A proof is given here.
$endgroup$
– Dietrich Burde
Feb 2 at 20:51
$begingroup$
its a law because I started with the magnitude formed by the product of two complex numbers and then inferred that its less than or equal to the magnitude of each complex number individually and then multiplied together.... and this equation always holds.... its never less than...always greater-than or equal-to
$endgroup$
– MrCasuality
Feb 2 at 20:58
$begingroup$
could be a linear algebra property as well... I was just trying to avoid dealing with dot products...so I was looking for the complex algebra law name to get around thinking about dot products...
$endgroup$
– MrCasuality
Feb 2 at 21:03
$begingroup$
There are no dot products here anyway. Just complex numbers and their absolute value.
$endgroup$
– Dietrich Burde
Feb 2 at 21:04
add a comment |
0
0
0
$begingroup$
I'm trying to remember the name of this law in complex number algebra:
$$|alpha beta| le |alpha| |beta|$$
Where:
$$alpha = Re{alpha}+ i Im{alpha}$$
$$beta = Re{beta}+ i Im{beta}$$
complex-analysis complex-numbers
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
$endgroup$
I'm trying to remember the name of this law in complex number algebra:
$$|alpha beta| le |alpha| |beta|$$
Where:
$$alpha = Re{alpha}+ i Im{alpha}$$
$$beta = Re{beta}+ i Im{beta}$$
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
asked Feb 2 at 20:48
MrCasualityMrCasuality
82
82
1
$begingroup$
It is not an Algebra law, but just an equation: $|zw|=|z||w|$ for complex $z,w$. It is sometimes called "norm", and it is multiplicative. A proof is given here.
$endgroup$
– Dietrich Burde
Feb 2 at 20:51
$begingroup$
its a law because I started with the magnitude formed by the product of two complex numbers and then inferred that its less than or equal to the magnitude of each complex number individually and then multiplied together.... and this equation always holds.... its never less than...always greater-than or equal-to
$endgroup$
– MrCasuality
Feb 2 at 20:58
$begingroup$
could be a linear algebra property as well... I was just trying to avoid dealing with dot products...so I was looking for the complex algebra law name to get around thinking about dot products...
$endgroup$
– MrCasuality
Feb 2 at 21:03
$begingroup$
There are no dot products here anyway. Just complex numbers and their absolute value.
$endgroup$
– Dietrich Burde
Feb 2 at 21:04
add a comment |
1
$begingroup$
It is not an Algebra law, but just an equation: $|zw|=|z||w|$ for complex $z,w$. It is sometimes called "norm", and it is multiplicative. A proof is given here.
$endgroup$
– Dietrich Burde
Feb 2 at 20:51
$begingroup$
its a law because I started with the magnitude formed by the product of two complex numbers and then inferred that its less than or equal to the magnitude of each complex number individually and then multiplied together.... and this equation always holds.... its never less than...always greater-than or equal-to
$endgroup$
– MrCasuality
Feb 2 at 20:58
$begingroup$
could be a linear algebra property as well... I was just trying to avoid dealing with dot products...so I was looking for the complex algebra law name to get around thinking about dot products...
$endgroup$
– MrCasuality
Feb 2 at 21:03
$begingroup$
There are no dot products here anyway. Just complex numbers and their absolute value.
$endgroup$
– Dietrich Burde
Feb 2 at 21:04
1
1
$begingroup$
It is not an Algebra law, but just an equation: $|zw|=|z||w|$ for complex $z,w$. It is sometimes called "norm", and it is multiplicative. A proof is given here.
$endgroup$
– Dietrich Burde
Feb 2 at 20:51
$begingroup$
It is not an Algebra law, but just an equation: $|zw|=|z||w|$ for complex $z,w$. It is sometimes called "norm", and it is multiplicative. A proof is given here.
$endgroup$
– Dietrich Burde
Feb 2 at 20:51
$begingroup$
its a law because I started with the magnitude formed by the product of two complex numbers and then inferred that its less than or equal to the magnitude of each complex number individually and then multiplied together.... and this equation always holds.... its never less than...always greater-than or equal-to
$endgroup$
– MrCasuality
Feb 2 at 20:58
$begingroup$
its a law because I started with the magnitude formed by the product of two complex numbers and then inferred that its less than or equal to the magnitude of each complex number individually and then multiplied together.... and this equation always holds.... its never less than...always greater-than or equal-to
$endgroup$
– MrCasuality
Feb 2 at 20:58
$begingroup$
could be a linear algebra property as well... I was just trying to avoid dealing with dot products...so I was looking for the complex algebra law name to get around thinking about dot products...
$endgroup$
– MrCasuality
Feb 2 at 21:03
$begingroup$
could be a linear algebra property as well... I was just trying to avoid dealing with dot products...so I was looking for the complex algebra law name to get around thinking about dot products...
$endgroup$
– MrCasuality
Feb 2 at 21:03
$begingroup$
There are no dot products here anyway. Just complex numbers and their absolute value.
$endgroup$
– Dietrich Burde
Feb 2 at 21:04
$begingroup$
There are no dot products here anyway. Just complex numbers and their absolute value.
$endgroup$
– Dietrich Burde
Feb 2 at 21:04
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
For complex this is an immediate consequence of $|z|^2=zbar z$ and associative property of multiplication.
Indeed $|ab|^2=(ab)(overline{ab})=abbar abar b=(abar a)(bbar b)=|a|^2|b|^2$
And since $|cdot|$ is positive then we can get rid of squares.
For a more general result, you may be interested in this post:
A name for the property $ | x star y | = | x | | y | $.
answered Feb 2 at 21:02
zwimzwim
12.7k832
$endgroup$
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
add a comment |
1
$begingroup$
You not only have $vert alpha beta vert le vert alpha vert vert beta vert$ but even $vert alpha beta vert = vert alpha vert vert beta vert$.
The complex modulus is multiplicative
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
1
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
For complex this is an immediate consequence of $|z|^2=zbar z$ and associative property of multiplication.
Indeed $|ab|^2=(ab)(overline{ab})=abbar abar b=(abar a)(bbar b)=|a|^2|b|^2$
And since $|cdot|$ is positive then we can get rid of squares.
For a more general result, you may be interested in this post:
A name for the property $ | x star y | = | x | | y | $.
answered Feb 2 at 21:02
zwimzwim
12.7k832
$endgroup$
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
add a comment |
2
$begingroup$
For complex this is an immediate consequence of $|z|^2=zbar z$ and associative property of multiplication.
Indeed $|ab|^2=(ab)(overline{ab})=abbar abar b=(abar a)(bbar b)=|a|^2|b|^2$
And since $|cdot|$ is positive then we can get rid of squares.
For a more general result, you may be interested in this post:
A name for the property $ | x star y | = | x | | y | $.
answered Feb 2 at 21:02
zwimzwim
12.7k832
$endgroup$
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
add a comment |
2
2
2
$begingroup$
For complex this is an immediate consequence of $|z|^2=zbar z$ and associative property of multiplication.
Indeed $|ab|^2=(ab)(overline{ab})=abbar abar b=(abar a)(bbar b)=|a|^2|b|^2$
And since $|cdot|$ is positive then we can get rid of squares.
For a more general result, you may be interested in this post:
A name for the property $ | x star y | = | x | | y | $.
answered Feb 2 at 21:02
zwimzwim
12.7k832
$endgroup$
For complex this is an immediate consequence of $|z|^2=zbar z$ and associative property of multiplication.
Indeed $|ab|^2=(ab)(overline{ab})=abbar abar b=(abar a)(bbar b)=|a|^2|b|^2$
And since $|cdot|$ is positive then we can get rid of squares.
For a more general result, you may be interested in this post:
A name for the property $ | x star y | = | x | | y | $.
answered Feb 2 at 21:02
zwimzwim
12.7k832
answered Feb 2 at 21:02
zwimzwim
12.7k832
answered Feb 2 at 21:02
zwimzwim
12.7k832
answered Feb 2 at 21:02
zwimzwim
12.7k832
12.7k832
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
add a comment |
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
what's the meaning of the "double absolute value braces" is that just a fancy absolute value brace or does it have special meaning? also, is the "star" implicit if you don't write it in a complex algebra equation?
$endgroup$
– MrCasuality
Feb 2 at 21:07
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
This is used for a norm en.wikipedia.org/wiki/Norm_(mathematics), this is a multidimensional extension of absolute value. The star here, just represent any binary operation similar to a dot product.
$endgroup$
– zwim
Feb 2 at 21:09
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
thanks... i guess my math book is just lax about calling norms as absolute values...
$endgroup$
– Bill Moore
Feb 2 at 21:14
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
$begingroup$
That's interesting... so you only need the inequality if we are dealing with vectors? instead of scalar complex number? its actually exactly equal when dealing with scalar numbers. that's actually, a little bit confusing because the book was saying that complex numbers are vectors of real and imaginary numbers
$endgroup$
– MrCasuality
Feb 2 at 21:28
add a comment |
1
$begingroup$
You not only have $vert alpha beta vert le vert alpha vert vert beta vert$ but even $vert alpha beta vert = vert alpha vert vert beta vert$.
The complex modulus is multiplicative
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
1
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
add a comment |
1
$begingroup$
You not only have $vert alpha beta vert le vert alpha vert vert beta vert$ but even $vert alpha beta vert = vert alpha vert vert beta vert$.
The complex modulus is multiplicative
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
1
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
add a comment |
1
1
1
$begingroup$
You not only have $vert alpha beta vert le vert alpha vert vert beta vert$ but even $vert alpha beta vert = vert alpha vert vert beta vert$.
The complex modulus is multiplicative
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
You not only have $vert alpha beta vert le vert alpha vert vert beta vert$ but even $vert alpha beta vert = vert alpha vert vert beta vert$.
The complex modulus is multiplicative
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 2 at 20:53
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
1
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
add a comment |
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
1
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
$begingroup$
it only equals if $alpha$ and $beta$ are scalars, or they have the same complex phase angle.
$endgroup$
– MrCasuality
Feb 2 at 21:01
1
1
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
$begingroup$
@MrCasuality You’re confusing sum and multiplication.
$endgroup$
– mathcounterexamples.net
Feb 2 at 21:04
add a comment |
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1
$begingroup$
It is not an Algebra law, but just an equation: $|zw|=|z||w|$ for complex $z,w$. It is sometimes called "norm", and it is multiplicative. A proof is given here.
$endgroup$
– Dietrich Burde
Feb 2 at 20:51
$begingroup$
its a law because I started with the magnitude formed by the product of two complex numbers and then inferred that its less than or equal to the magnitude of each complex number individually and then multiplied together.... and this equation always holds.... its never less than...always greater-than or equal-to
$endgroup$
– MrCasuality
Feb 2 at 20:58
$begingroup$
could be a linear algebra property as well... I was just trying to avoid dealing with dot products...so I was looking for the complex algebra law name to get around thinking about dot products...
$endgroup$
– MrCasuality
Feb 2 at 21:03
$begingroup$
There are no dot products here anyway. Just complex numbers and their absolute value.
$endgroup$
– Dietrich Burde
Feb 2 at 21:04