# Extracting real part

1. Sep 30, 2012

### cragar

How do I take the real part of this
$ln(e^{ix}+i)$
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. Oct 1, 2012

### DonAntonio

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

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. Oct 3, 2012

### cragar

where did you get the cos(x)^2 and what does arg mean

4. Oct 3, 2012

### JJacquelin

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. Oct 3, 2012

### DonAntonio

Cosine square is part of the (complex) module (or absolute value) of the complex number written there, and arg = argument.

DonAntonio