1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...