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I have shown that the following function is harmonic and am attempting to find it's harmonic conjugate:

$u=e^{-2xy}\sin(x^2-y^2)$

I know that to find the harmonic conjugate I need to use the Cauchy–Riemann equations:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

Subtituting in $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$, I obtained the following equations:

(1) $\frac{\partial v}{\partial y} = 2e^{-2xy}\left(x\cos(x^2-y^2)-y\sin(x^2-y^2)\right)$

(2) $\frac{\partial v}{\partial x} = 2e^{-2xy}\left(y\cos(x^2-y^2)+x\sin(x^2-y^2)\right)$

And so we need to integrate (1) and (2) with respect to y and x, respectively. Initially I thought this would be possible using integration by parts, but it doesn't seem to simplify and just goes round in circles.

By inspection I have determined that the harmonic conjugate $v=-e^{-2xy}\cos(x^2-y^2)$, but I wish to show this by integration. Could anyone advise as to a better method of integrating that would allow this?

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    You can, in principle, integrate along the line segment joining $(0,0)$ and $(x,y)$. But it is perhaps easier to note that $u = -\Re\left[ i \exp (i z^2)\right]$ where $z = (x + iy)$.2012-10-17

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