Absolute value of a complex expression
0
$begingroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
amWhy
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
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|
show 3 more comments
0
$begingroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
amWhy
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
$endgroup$
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
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– Don Thousand
Jan 4 at 17:14
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Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
|
show 3 more comments
0
0
0
$begingroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
amWhy
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
$endgroup$
I've been stuck on this problem for a while:
$$vert 1-e^{-i2pi f}+.5e^{-i2pi fcdot 2}vert^2,$$
where $i =$ the imaginary unit, $(2pi f) =$ a real value, and $(2pi fcdot 2)=$ a real value.
I just don't know how to begin.
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
calculus algebra-precalculus exponentiation polar-coordinates absolute-value
edited Jan 4 at 17:58
amWhy
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
edited Jan 4 at 17:58
amWhy
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
edited Jan 4 at 17:58
amWhy
1
edited Jan 4 at 17:58
amWhy
1
edited Jan 4 at 17:58
amWhy
1
1
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
asked Jan 4 at 17:09
LowEnergyLowEnergy
11
11
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
|
show 3 more comments
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
2
2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32
|
show 3 more comments
1 Answer
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Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
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Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
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Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
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Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
$endgroup$
Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
answered Jan 4 at 17:46
EnnarEnnar
14.4k32443
14.4k32443
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2
$begingroup$
First off, what have you tried? Second, the notation is a little unclear. Some clarification would be helpful. And third, what exactly do you want out of this? This is not a polynomial, unlike what your title suggests.
$endgroup$
– Don Thousand
Jan 4 at 17:14
$begingroup$
Hello LowEnergy, why is the expression in latex so huge? :)
$endgroup$
– Ixion
Jan 4 at 17:24
$begingroup$
i = imaginary, (2*pi*f) = real value
$endgroup$
– LowEnergy
Jan 4 at 17:26
$begingroup$
So it's the absolute value of a complex number....
$endgroup$
– ÍgjøgnumMeg
Jan 4 at 17:28
$begingroup$
Yeah, I should've probably written complex value in the title.
$endgroup$
– LowEnergy
Jan 4 at 17:32