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Separating real and complex parts of a number

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello, I am supposed to express the and the phase part of expression:

    [itex]\displaystyle{S=\frac{k}{\sqrt{1+i\gamma_0}} \cdot exp\left(\frac{z}{1 + i\gamma_0}\right)}[/itex]

    2. Relevant equations
    The answer should be in the form:

    [itex]\displaystyle{S=a(\gamma_0) \cdot exp\left(i\varphi(\gamma_0)\right)}[/itex]


    3. The attempt at a solution
    Well, its clear(probably) that for the amplitude part I just have to multiply this equation by its complex conjugate and take a square root out of the result. This leaves me with the expression of:
    [itex]\displaystyle{ a(\gamma_0)=\frac{k}{\left(1+\gamma^2_0\right)^{1/4}} \cdot exp\left(\frac{z}{1+\gamma^2_0}\right) }[/itex]

    However, I dont quite understand how to get the complex(phase) part of the number? A hint where to start would be very gladly accepted :), thank you.
     
    Last edited: Oct 10, 2011
  2. jcsd
  3. Oct 10, 2011 #2

    ehild

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    It is the magnitude or absolute value of the complex number instead of the real part. So how do you get the real and imaginary parts?


    ehild
     
  4. Oct 10, 2011 #3
    Ah yes, sorry, its the amplitude, thanks for noticing. I'll reformulate the problem. I need to get the expressions for the amplitude and the phase.
     
  5. Oct 10, 2011 #4

    ehild

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    Is z in the exponent a complex number? Then write it out with its real and imaginary parts, and find the real and imaginary parts of z/(1+iγ0), so you have the exponential in the form exp(u+iv) .

    Write the fraction 1/(1+iγ0) in exponential form, too: exp(a+ib). Then your formula is equal to k*exp((a+u)+i(b+v)). k*exp(a+b) is the magnitude, and the phase is b+v.

    ehild
     
  6. Oct 11, 2011 #5
    z is real and negative and k is real and positive constant. Anyway, expressing the fraction sqrt(1/(1+iy0) can get me just as close as:

    [itex]\displaystyle{ k \cdot exp \left(\frac{1}{2}ln \left( \frac{1}{1+\gamma^2_0}-\frac{i\gamma_0}{1+\gamma^2_0}\right) \right) }[/itex]

    Which I dont understand how to simplify to form exp(a+ib). I have also tried expressing phase from the general form:

    [itex]\displaystyle{ \frac{S}{a(\gamma_0)}=exp \left(i \varphi \right) }[/itex]

    Which is kinnda closer to the answer with the expression:

    [itex]\displaystyle{ i\varphi=\frac{1}{4}ln\left(\frac{1-i\gamma_0}{1+i\gamma_0} \right) -
    \frac{i\gamma_0z}{1+\gamma^2_0} }[/itex]

    The answer should be:

    [itex]\displaystyle{\varphi = -\frac{1}{2}arctan\ \gamma_0 - \frac{\gamma_0z}{1+\gamma^2_0} }[/itex]
     
  7. Oct 11, 2011 #6

    ehild

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    Write all factors in the exponential form: r*e. Any complex number u+iv= r*e, where [itex]r=\sqrt{(u^2+v^2) }[/itex] and tan(φ)=v/u.
    This way, [itex]1+i \gamma_0=\sqrt{1+\gamma_0 ^2} e^ {i\arctan(\gamma_0)}[/itex]

    ehild
     
    Last edited: Oct 11, 2011
  8. Oct 11, 2011 #7
    ah yes, thank you very much:)
     
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