1
$\begingroup$

I have the general recipe for finding the complex conjugate of a function down as follows:

Suppose I have $f(z)$:

  1. Separate $f(z)$ into a sum of real and imaginary functions such that $f(z)=u(x,y)+iv(x,y)$
  2. Negate the $i$ and you have your complex conjugate

I have tried the above recipe for the following function(which I obtained from another question on math.SE -- Complex conjugate of function): $\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$

(I basically didn't understand what they did over there, but I think my procedure is correct, and I tried to rework the problem of finding the complex conjugate using the above process.)

So I write $A=x_1+iy_1$ and $B=x_2+iy_2$, $\theta_1=kx-\omega t$ and $\theta_2=kx+\omega t$, and multiply out the following expression:$\psi(x,t)=Ae^{i(\theta_1)}+ Be^{-i(\theta_2)}$using the Euler relation, and I should get the answer, right? I tried this out on pen and paper and I get something like:$\psi^*=x_1e^{-\theta_1}+x_2e^{\theta_2}-y_1e^{\frac{\pi}{2}-\theta_1}+y_2e^{\frac{\pi}{2}-\theta_1}$

I am not 100% sure about my result as I could have made a mistake, but I'd like to know if the general procedure is correct. Thanks!

  • 0
    Thanks, I got it. It's much quicker to work with the exponential form.2012-11-19

0 Answers 0