cragar said: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.
cragar said:where did you get the cos(x)^2 and what does arg mean
The purpose of extracting the real part from ln(e^{ix}+i) is to simplify the expression and make it easier to solve or manipulate in mathematical equations. By isolating the real part, it becomes easier to analyze and understand the behavior of the equation.
To extract the real part from ln(e^{ix}+i), we can use the properties of logarithms and Euler's formula. First, we can rewrite ln(e^{ix}+i) as ln(e^{ix}*(1+i)). Then, we can use the property ln(ab) = ln(a) + ln(b) to split the expression into ln(e^{ix}) + ln(1+i). Finally, we can use Euler's formula, e^{ix} = cos(x) + i*sin(x), to simplify ln(e^{ix}) to ix. Therefore, the real part of ln(e^{ix}+i) is just the imaginary part of ix, which is x.
No, extracting the real part from ln(e^{ix}+i) is not always necessary. It depends on the specific problem or equation you are trying to solve. Sometimes, it may be more useful to keep the expression in its original form or extract the imaginary part instead.
If x = 0, the real part of ln(e^{ix}+i) would be 0. This is because when x = 0, ln(e^{ix}+i) becomes ln(1+i), which can be simplified to ln(sqrt(2)) = 0.
Yes, the real part of ln(e^{ix}+i) can be negative. It depends on the value of x. For example, if x = -pi/2, the real part would be -pi/2. However, in most cases, the real part would be positive or 0.