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Complex numbers equation/equality

  1. Jan 27, 2014 #1
    The problem statement, all variables and given/known data.

    Prove that, given constants ##A_1,A_2, \phi_1## and ##\phi_2##, there are constants ##A## and ##\phi## such that the following equality is satisfied:

    ##A_1\cos(kx+\phi_1)+A_2\cos(kx+\phi_2)=A\cos(kx+\phi)##

    The attempt at a solution.

    I've tried to use the identity ##\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)##, so

    ##A_1\cos(kx+\phi_1)+A_2\cos(kx+\phi_2)=A_1[\cos(kx)\cos(\phi_1)-\sin(kx)\sin(\phi_1)]+A_2[\cos(kx)\cos(\phi_2)-\sin(kx)\sin(\phi_2)]##, and then, the expresion on the right equals to

    ##[A_1\cos(\phi_1)+A_2\cos(\phi_2)]\cos(kx)-[A_1\sin(\phi_1)+A_2\sin(\phi_2)]\sin(kx)##.

    I am trying to write this last expression as something of the form ##A\cos(kx+\phi)## but I don't know how to, I would appreciate any suggestion to arrive to the desired expression.
     
  2. jcsd
  3. Jan 27, 2014 #2

    Dick

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    Here http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations is the sort of identity you need expressed for real functions. Can you figure out how to derive a similar sort of thing for complex functions? You don't need to actually find the complex magnitude or the phase, you just have to show they exist. I'm actually not sure I can now, it's getting kind of late. So I'm counting on you.
     
    Last edited: Jan 27, 2014
  4. Jan 28, 2014 #3

    Dick

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    Ok, here's a little more. You already have it in the form of a linear combination of ##sin(kx)## and ##cos(kx)##. Using the definitions of complex sin and cos you could equally well write that as ##C_1 e^{ikz}+C_2 e^{-ikz}##. Now let's introduce a complex phase ##\phi##, by computing it such that ##C_1 e^{-i \phi}=C_2 e^{i \phi}##. Can you see how it goes from here?
     
  5. Jan 28, 2014 #4
    As you've said, I could express ##[A_1\cos(\phi_1)+A_2\cos(\phi_2)]\cos(kx)-[A_1\sin(\phi_1)+A_2\sin(\phi_2)]\sin(kx)## as ##C1e^{ikx}+C_2e^{-ikx}##, if I call ##k_1=A_1\cos(\phi_1)+A_2\cos(\phi_2)## and ##k_2=A_1\sin(\phi_1)+A_2\sin(\phi_2)##, then ##C_1,C_2## must satisfy ##C_1=\dfrac{k_1-ik_2}{2}## and ##C_2=\dfrac{k_1+ik_2}{2}##.

    I didn't get your last suggestion, suppose I find ##\phi## such that ##C_1 e^{-i \phi}=C_2 e^{i \phi}##, and how would this help me to write the original expression in the form ##A\cos(kx+\phi)##? Sorry if it is too obvious.
     
  6. Jan 28, 2014 #5

    AlephZero

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    If you write ##A_1 \cos(kx + \phi_1) = A_1\cos\phi_1\cos kx - A_1\sin\phi_1\sin kx = P_1\cos kx + P_2\sin kx## say, and similarly for the other terms, you have to show that the result is true for any value of ##\cos kx## and ##\sin kx##. You can do that by equating the coefficients of ##\cos kx## and ##\sin kx##.

    Or, if you use ##e^{ikx} = \cos kx + i\sin kx##, this is the same as equating both the real and imaginary parts of ##e^{ikx}##.
     
  7. Jan 28, 2014 #6

    Dick

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    Ok, I'll tell you what the strategy is. Given I can write the expression as ##C_1e^{ikx}+C_2e^{-ikx}##. I would very much like to find a phase ##\phi## such that ##C_1e^{ikx}+C_2e^{-ikx}=De^{i(kx+\phi)}+De^{-i(kx+\phi)}##, what equation would I need to solve for ##\phi##? That actually probably would have been a better clue.
     
    Last edited: Jan 28, 2014
  8. Jan 28, 2014 #7

    Dick

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    All of the coefficients here are complex. So it's not quite the same as equating real and imaginary parts. It's equating coefficients of the linearly independent functions ##e^{ikx}## and ##e^{-ikx}##.
     
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