Extracting real part

  • Thread starter cragar
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  • #1
cragar
2,552
3
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.
 

Answers and Replies

  • #2
DonAntonio
606
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.



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

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

$$\frac{1}{2}\log\left(\cos^2x+(\sin x+1)^2\right)=\frac{1}{2}\log\left(2(1+\sin x)\right)$$
 
  • #3
cragar
2,552
3
where did you get the cos(x)^2 and what does arg mean
 
  • #4
JJacquelin
801
34
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))
 
  • #5
DonAntonio
606
1
where did you get the cos(x)^2 and what does arg mean

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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