- #1

- 2,544

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[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.

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- Thread starter cragar
- Start date

- #1

- 2,544

- 3

[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

- 606

- 1

[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

- 2,544

- 3

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

- #4

- 798

- 34

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

- 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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