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The problem is as follows:

Proove that $U(x,y) = x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ is the real part of an analytic function.

where $f(z)$ is analytic such that $u(x,y) = \mathrm{Re}[f(z)]$?

I've been playing around with the Cauchy-Riemann equations trying to find a harmonic conjugate, but I feel pretty stuck. Does anybody have a clue where to begin?

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  • $\begingroup$ Try to show $U$ is harmonic.... $\endgroup$
    – Empty
    Commented Oct 13, 2015 at 6:19
  • $\begingroup$ As S. Panja and guest said, do and write $\displaystyle\frac{\partial}{\partial x}=\frac{\partial}{\partial z}+\frac{\partial}{\partial\bar{z}}$, $\displaystyle\frac{\partial}{\partial y}=\frac{1}{i}(\frac{\partial}{\partial z}+\frac{\partial}{\partial\bar{z}})$, $x=\frac12(z+\bar{z})$ and $y=\frac{1}{2i}(z+\bar{z})$ so with substituting we have $\displaystyle U=z\frac{\partial u}{\partial z}+\bar{z}\frac{\partial u}{\partial\bar{z}}$ and it's harmonic simply. $\endgroup$
    – Nosrati
    Commented Oct 13, 2015 at 7:30

1 Answer 1

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Like S. Panja said, you do not have to find a harmonic conjugate, just verify the C-R equations using the C-R equations for $f=u+iv$ in the form of the Laplace equation for $u$: $u_{xx}+u_{yy}=0$.

If you want an explicit conjugate, think about what form it should have: it must somehow be "like" $U$ but with the $x$ and $y$ multipliers mixed up (to mimic the situation with the C-R equations).

If you are still having trouble, try:

$V = yu_x-xu_y$

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