How To Do Integral Of Complex Conjugate?

**soma** · April 6th 2010, 06:41 PM

$\int\limits_{1+i}^{2+i}s\, dx$

Where s is the complex conjugate of x

I have the answer but I don't have the foggiest idea how it is obtained.

answer: [(2^2 -1 )/2] -i

**Prove It** · April 6th 2010, 06:53 PM

Originally Posted by soma

$\int\limits_{1+i}^{2+i}s\, dx$

Where s is the complex conjugate of x

I have the answer but I don't have the foggiest idea how it is obtained.

answer: [(2^2 -1 )/2] -i

Define $x = e^{i\theta}$ .

Then $\overline{x} = e^{-i\theta}$

and $dx = i\,e^{i\theta}$ .

Also note that $\theta = \frac{\log{x}}{i}$ .

So the integral becomes

$\int{\overline{x}\,dx} = \int{e^{-i\theta}\,i\,e^{i\theta}\,d\theta}$

$= \int{i\,d\theta}$

$= i\theta + C$

$= i\left(\frac{\log{x}}{i}\right) + C$

$= \log{x} + C$ .

Now you should be able to substitute the limits of integration.

Remember that $\log{x} = \ln{|x|} + i\,\textrm{arg}\,{x}$ .

Thread: How To Do Integral Of Complex Conjugate?

LinkBack

Thread Tools

Display

How To Do Integral Of Complex Conjugate?

Similar Math Help Forum Discussions

Complex Equations with a complex number and its conjugate

Complex conjugate?

complex conjugate

Complex Conjugate

conjugate complex

Search Tags