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Easy Peasy CA problem

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    I have a very simple complex arithmetic question.

    How do I express the quantity cos(1+i) in Cartesian (a+ib) and Euler(re^i*theta)

    Is this the right track?:

    [tex]cos(1+i)={e^{i(1+i)}\over2}+{e^{-i(1+i)}\over2}[/tex]

    I know that:

    [tex]cos(1+i)=cos(1)cos(i)-sin(1)sin(i)[/tex]

    but I am not sure how to get the "i's" out of the cosine and sine.
     
  2. jcsd
  3. Feb 2, 2010 #2

    Dick

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    Well, sure you are on the right track. The same formula you used before. cos(i)=(e^(i*i)+e^(i*(-i))/2=(e^(-1)+e^1)/2. That's cosh(1). Use similar expression for sin(i).
     
  4. Feb 2, 2010 #3

    So then as far as cartesian form goes:


    [tex]cos(1+i)=cos(1)cosh(1)-i(sin(1)sinh(1))[/tex]

    If that is correct, what is the relationship between the hyperbolic functions and the notation:

    [tex]re^{i\theta}[/tex]
     
  5. Feb 2, 2010 #4

    Dick

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    Now you've got cos(1+i) written in the form a+ib where a and b are real. Now it's the usual relation with the polar form. r=sqrt(a^2+b^2), theta=arctan(b/a), right? There's nothing terribly spooky or mysterious going on here.
     
  6. Feb 2, 2010 #5
    Thanks Dick. I don't know why this stuff cooks my noodle so much. I feel as though I cannot wrap my head around it. I always think there is some Identity that I am not taking in to account or that my answers are not nearly simplified enough.

    One last question. Would you consider it correct to write the equation in Euler form this way:


    [tex]\sqrt{(cos(1)cosh(1))^2+(sin(1)sinh(1))^2}e^{i(arctan{(sin(1)sinh(1)\over(cos(1)cosh(1)})}[/tex]
     
  7. Feb 2, 2010 #6

    Dick

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    That's about as good as I could come up with. But you are missing a sign in the argument. You could also write tan(1)*tanh(1) in the argument. But that's not all that much of a real simplification, is it?
     
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