Solving for the Real part of a complex number
0
$begingroup$
I have the following two complex values:
$A = a e^{j omega t + theta }$
$B = b e^{j omega t + theta}$
where $j = sqrt{-1}$.
I am trying to find the real part of the product AB.
So far I've done this: $A B = a b e^{2 j omega t+ theta}$
I am not sure where to go from here with the complex variables...
complex-analysis
edited Jan 20 at 2:45
David G. Stork
11k41432
asked Jan 20 at 0:41
articatarticat
63
$endgroup$
add a comment |
0
$begingroup$
I have the following two complex values:
$A = a e^{j omega t + theta }$
$B = b e^{j omega t + theta}$
where $j = sqrt{-1}$.
I am trying to find the real part of the product AB.
So far I've done this: $A B = a b e^{2 j omega t+ theta}$
I am not sure where to go from here with the complex variables...
complex-analysis
edited Jan 20 at 2:45
David G. Stork
11k41432
asked Jan 20 at 0:41
articatarticat
63
$endgroup$
$begingroup$
Please use MathJax to format your question
$endgroup$
– saulspatz
Jan 20 at 0:44
$begingroup$
The exponent should be $2jomega t + 2theta$
$endgroup$
– Ross Millikan
Jan 20 at 2:48
add a comment |
0
0
0
$begingroup$
I have the following two complex values:
$A = a e^{j omega t + theta }$
$B = b e^{j omega t + theta}$
where $j = sqrt{-1}$.
I am trying to find the real part of the product AB.
So far I've done this: $A B = a b e^{2 j omega t+ theta}$
I am not sure where to go from here with the complex variables...
complex-analysis
edited Jan 20 at 2:45
David G. Stork
11k41432
asked Jan 20 at 0:41
articatarticat
63
$endgroup$
I have the following two complex values:
$A = a e^{j omega t + theta }$
$B = b e^{j omega t + theta}$
where $j = sqrt{-1}$.
I am trying to find the real part of the product AB.
So far I've done this: $A B = a b e^{2 j omega t+ theta}$
I am not sure where to go from here with the complex variables...
complex-analysis
complex-analysis
edited Jan 20 at 2:45
David G. Stork
11k41432
asked Jan 20 at 0:41
articatarticat
63
edited Jan 20 at 2:45
David G. Stork
11k41432
asked Jan 20 at 0:41
articatarticat
63
edited Jan 20 at 2:45
David G. Stork
11k41432
edited Jan 20 at 2:45
David G. Stork
11k41432
edited Jan 20 at 2:45
David G. Stork
11k41432
11k41432
asked Jan 20 at 0:41
articatarticat
63
asked Jan 20 at 0:41
articatarticat
63
asked Jan 20 at 0:41
articatarticat
63
63
$begingroup$
Please use MathJax to format your question
$endgroup$
– saulspatz
Jan 20 at 0:44
$begingroup$
The exponent should be $2jomega t + 2theta$
$endgroup$
– Ross Millikan
Jan 20 at 2:48
add a comment |
$begingroup$
Please use MathJax to format your question
$endgroup$
– saulspatz
Jan 20 at 0:44
$begingroup$
The exponent should be $2jomega t + 2theta$
$endgroup$
– Ross Millikan
Jan 20 at 2:48
$begingroup$
Please use MathJax to format your question
$endgroup$
– saulspatz
Jan 20 at 0:44
$begingroup$
Please use MathJax to format your question
$endgroup$
– saulspatz
Jan 20 at 0:44
$begingroup$
The exponent should be $2jomega t + 2theta$
$endgroup$
– Ross Millikan
Jan 20 at 2:48
$begingroup$
The exponent should be $2jomega t + 2theta$
$endgroup$
– Ross Millikan
Jan 20 at 2:48
add a comment |
1 Answer
1
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oldest
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1
$begingroup$
$$A B = a b e^{2 (j omega t + theta)} = underbrace{a b [ cos (2 (j omega t + theta))}_{real} + i sin (2 (j omega t + theta))]$$
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
1
$begingroup$
$$A B = a b e^{2 (j omega t + theta)} = underbrace{a b [ cos (2 (j omega t + theta))}_{real} + i sin (2 (j omega t + theta))]$$
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
add a comment |
1
$begingroup$
$$A B = a b e^{2 (j omega t + theta)} = underbrace{a b [ cos (2 (j omega t + theta))}_{real} + i sin (2 (j omega t + theta))]$$
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
add a comment |
1
1
1
$begingroup$
$$A B = a b e^{2 (j omega t + theta)} = underbrace{a b [ cos (2 (j omega t + theta))}_{real} + i sin (2 (j omega t + theta))]$$
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
$endgroup$
$$A B = a b e^{2 (j omega t + theta)} = underbrace{a b [ cos (2 (j omega t + theta))}_{real} + i sin (2 (j omega t + theta))]$$
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
answered Jan 20 at 0:47
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
add a comment |
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
I forgot to mention that for engineering notation, j = sqrt(-1)
$endgroup$
– articat
Jan 20 at 1:31
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
A recommendation: Always update your problem with such information (rather than make a comment), so that helpers will better see it. (See my edit to your problem.)
$endgroup$
– David G. Stork
Jan 20 at 2:45
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
Thank you for the recommendation. In regards to the solution, it is still imaginary because j is in the real part right? Would I then have to take the magnitude of the solution?
$endgroup$
– articat
Jan 20 at 17:00
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
$begingroup$
@articat: No. A complex number $z$ is made of a real part and an imaginary part. $z = {rm real~part} + {rm imaginary~part} = underbrace{a}_{real} + underbrace{i b}_{imaginary}$. You are thinking of the magnitude.
$endgroup$
– David G. Stork
Jan 20 at 19:59
add a comment |
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$begingroup$
Please use MathJax to format your question
$endgroup$
– saulspatz
Jan 20 at 0:44
$begingroup$
The exponent should be $2jomega t + 2theta$
$endgroup$
– Ross Millikan
Jan 20 at 2:48