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Extracting real part

  1. Sep 30, 2012 #1
    How do I take the real part of this
    [itex] ln(e^{ix}+i) [/itex]
    I know you can write the arctan(x) in terms of logs with complex numbers should I do something like that? I came across this because I was trying to integrate sec(x) with Eulers formula.
     
  2. jcsd
  3. Oct 1, 2012 #2


    $$e^{ix}+i=\cos x+i\sin x+i=\cos x+(\sin x+1)i\Longrightarrow Log(e^{ix}+i)=\frac{1}{2}\log\left(\cos^2x+(\sin x+1)^2\right)+i\arg(e^{ix}+i)$$

    So never mind what branch of the complex logarithm we choose, the real part is:

    $$\frac{1}{2}\log\left(\cos^2x+(\sin x+1)^2\right)=\frac{1}{2}\log\left(2(1+\sin x)\right)$$
     
  4. Oct 3, 2012 #3
    where did you get the cos(x)^2 and what does arg mean
     
  5. Oct 3, 2012 #4
    Another way :
    2*Real part = ln(exp(ix)+i)+ln(exp(-ix)-i)
    (exp(ix)+i)*(exp(-ix)-i) = 1-i*exp(ix)+i*exp(-ix)+1 = 2+2sin(x)
    Real part =(1/2) ln(2+2sin(x))
     
  6. Oct 3, 2012 #5
    Cosine square is part of the (complex) module (or absolute value) of the complex number written there, and arg = argument.

    DonAntonio
     
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