Complex Analysis: Harmonic Conjugates

1. Aug 22, 2010

AUGTRON

1. The problem statement, all variables and given/known data
For $$u(x,y)=e^{-y}(x\sin(x)+y\cos(x))$$ find a harmonic conjugate $$v(x,y)$$ and express the analytic function $$f=u +iv$$ as a function of z alone (where $$z=x+iy$$0

2. Relevant equations
The Cauchy Riemann equations $$u_x=v_y$$ and $$u_y=-v_x$$
and possibly:
$$sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$
$$cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$

3. The attempt at a solution
I've been toying around with algebra and I've been able to reduce up to $$u=\frac {e^{-y+ix}(y-ix)+e^{-y-ix}(y+ix)}{2}$$ but I can't get past here. I have also noted that if I divide by i, u(x,y) becomes u(y,-x) so I can replace these variables and I have $$u/i = e^{x}(y\sin(y)-x\cos(y)$$ which is, as someone on another forum pointed out, the real part of $$-ze^{z}$$. So I suppose I could write $$f=Re(-ze^{iz})i + iv$$ no?
But how should I go about finding v? I am familiar with the straightforward procedure of finding u_x, integrating w.r.t v, then finding v_x and setting it equal to u_y... but I can't seem to simplify u to the point of easy differentiation. I don't know if I should just try to brute force it, or if that's even possible. It seems like u should simplify more.

2. Aug 22, 2010

HallsofIvy

I would not change the sine and cosine to exponentials, although you certainly can do so.

With $v(x,y)= e^{-y}(x sin(x)- y cos(x))$, $v_x= e^{-y}(sin(x)+ y cos(x))+ y sin(x))$ and $v_y= -e^{-y}(x sin(x)- y cos(x))+ e^{-y}(-cos(x))$

Now you must find u(x,y) such that
$$u_x= v_y= -e^{-y}(x sin(x)- y cos(x))+ e^{-y}(-cos(x))$$
and
$$u_y= -v_x= -e^{-y}(sin(x)+ y cos(x))- y sin(x))[/itex] Integrate either of those equations with respect to the appropriate variable, remembering that the "constant" of integration may be a function of the other variable, then put that into the other equation to find that function. 3. Aug 22, 2010 AUGTRON Great. I guess the integration wasn't as bad as I thought. Your algebra is a bit off because you wrote down a different u(x,y) than I had. Using [tex]u(x,y)=e^{-y}(x\sin(x)+y\cos(x)$$ I came up with
v(x,y) = e^{-y}(ysin(x)-xcos(x)).

Now my question is how to put this in terms of z only. I think I have a THM I can use: http://planetmath.org/encyclopedia/IdentityTheoremOfHolomorphicFunctions.html" [Broken]. So if I have f(u,v) = e^{-y}(xsin(x)+ycos(x) +i(e^{-y}(ysin(x)-xcos(x)) can I say that some function g(z) is equal to f(z) along the iy axis of the complex plane (i.e. where x=o). Then f(z)=g(z) for all complex numbers. So f(z)=g(0+iy) ---> f(z)=ye^{-y}?

Sorry about the formatting, Latex is giving me trouble.

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