Absolute value of a complex expression












0














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







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  • 2




    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.
    – Don Thousand
    Jan 4 at 17:14










  • Hello LowEnergy, why is the expression in latex so huge? :)
    – Ixion
    Jan 4 at 17:24










  • i = imaginary, (2*pi*f) = real value
    – LowEnergy
    Jan 4 at 17:26










  • So it's the absolute value of a complex number....
    – ÍgjøgnumMeg
    Jan 4 at 17:28










  • Yeah, I should've probably written complex value in the title.
    – LowEnergy
    Jan 4 at 17:32
















0














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







share|cite|improve this question









New contributor




LowEnergy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    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.
    – Don Thousand
    Jan 4 at 17:14










  • Hello LowEnergy, why is the expression in latex so huge? :)
    – Ixion
    Jan 4 at 17:24










  • i = imaginary, (2*pi*f) = real value
    – LowEnergy
    Jan 4 at 17:26










  • So it's the absolute value of a complex number....
    – ÍgjøgnumMeg
    Jan 4 at 17:28










  • Yeah, I should've probably written complex value in the title.
    – LowEnergy
    Jan 4 at 17:32














0












0








0







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







share|cite|improve this question









New contributor




LowEnergy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






share|cite|improve this question









New contributor




LowEnergy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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LowEnergy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited Jan 4 at 17:58









amWhy

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192k28225439






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asked Jan 4 at 17:09









LowEnergyLowEnergy

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11




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LowEnergy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 2




    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.
    – Don Thousand
    Jan 4 at 17:14










  • Hello LowEnergy, why is the expression in latex so huge? :)
    – Ixion
    Jan 4 at 17:24










  • i = imaginary, (2*pi*f) = real value
    – LowEnergy
    Jan 4 at 17:26










  • So it's the absolute value of a complex number....
    – ÍgjøgnumMeg
    Jan 4 at 17:28










  • Yeah, I should've probably written complex value in the title.
    – LowEnergy
    Jan 4 at 17:32














  • 2




    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.
    – Don Thousand
    Jan 4 at 17:14










  • Hello LowEnergy, why is the expression in latex so huge? :)
    – Ixion
    Jan 4 at 17:24










  • i = imaginary, (2*pi*f) = real value
    – LowEnergy
    Jan 4 at 17:26










  • So it's the absolute value of a complex number....
    – ÍgjøgnumMeg
    Jan 4 at 17:28










  • Yeah, I should've probably written complex value in the title.
    – LowEnergy
    Jan 4 at 17:32








2




2




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.
– Don Thousand
Jan 4 at 17:14




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.
– Don Thousand
Jan 4 at 17:14












Hello LowEnergy, why is the expression in latex so huge? :)
– Ixion
Jan 4 at 17:24




Hello LowEnergy, why is the expression in latex so huge? :)
– Ixion
Jan 4 at 17:24












i = imaginary, (2*pi*f) = real value
– LowEnergy
Jan 4 at 17:26




i = imaginary, (2*pi*f) = real value
– LowEnergy
Jan 4 at 17:26












So it's the absolute value of a complex number....
– ÍgjøgnumMeg
Jan 4 at 17:28




So it's the absolute value of a complex number....
– ÍgjøgnumMeg
Jan 4 at 17:28












Yeah, I should've probably written complex value in the title.
– LowEnergy
Jan 4 at 17:32




Yeah, I should've probably written complex value in the title.
– LowEnergy
Jan 4 at 17:32










1 Answer
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Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$






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    1 Answer
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    Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$






    share|cite|improve this answer


























      0














      Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$






      share|cite|improve this answer
























        0












        0








        0






        Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$






        share|cite|improve this answer












        Hint: $|z|^2 = zbar z$ and $overline{e^{it}} = e^{-it}, tinmathbb R.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 17:46









        EnnarEnnar

        14.4k32343




        14.4k32343






















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