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Let $u : \mathbb{C} \rightarrow \mathbb{R}$ be a harmonic function. Is there a $k \in \mathbb{R}$ such that $v(z) = u(x^2-y^2, kxy)$ is harmonic $\forall z = x+iy \in \mathbb{ C}$ ?

$v_x = 2x u_a + ky u_b \leadsto v_{x,x } = 2u_a + 4 x^2 u_{a,a} + k^2 y^2 u_{ b,b} $.
$v_y = -2y u_a + kx u_b \leadsto v_{ y,y} = -2u_a + 4y^2 u_{ a,a} + k^2 x^2 u_{b,b } $.
$\Delta v = v_{ x,x}+ u_{ y,y} = 4u_{ a,a} (x^2+y^2) + k^2 u_{ b,b}(x^2+y^2) = (x^2+y^2)(4 u_{ a,a} + k^2 u_{ b,b}) $.
So if $k^2 = - \dfrac{ 4 u_{ a,a} }{u_{ b,b} } \Rightarrow \Delta v = 0 $. This is true in particular if $ k = \pm 2$, since $\Delta u = 0$. Is this the only solution?

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    Your calculation seems to be ok up to the point where you write 'So if $a^2$....'. What you need is $a^2 = 4$ since $u_{aa} = -u_{bb}$ -- I get $ a=2$ or $a=-2$ from this.2012-06-10

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Please don't use the same letter $a$ to mean two different things in the same formula...

Otherwise, your solution (with correction by @Thomas) is fine. You may want to use a concrete example such as $u(\xi,\eta)=\xi^2-\eta^2$ to show that $a\ne \pm 2$ does not work: otherwise the reader might have a suspicion that $4u_{\xi \xi}+a^2 u_{\eta \eta}$ could vanish for some mysterious reason.