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I'm trying to find the most general harmonic polynomial of form $ax^3+bx^2y+cxy^2+dy^3$. I write this polynomial as $u(x,y)$.

I calculate $$ \frac{\partial^2 u}{\partial x^2}=6ax+2by,\qquad \frac{\partial^2 u}{\partial y^2}=2cx+6dy $$ and conclude $3a+c=0=b+3d$. Does this just mean the most general harmonic polynomial has form $ax^3-3dx^2y-3axy^2+dy^3$ with the above condition on the coefficients? "Most general" is what my book states, and I'm not quite sure what it means.

Also, I want to find the conjugate harmonic function, say $v$. I set $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$ and find $$ v=dx^3+3ax^2y+bxy^2-ay^3+K $$ for some constant $K$. By integrating $\frac{\partial v}{\partial x}$, finding $v$ as some polynomial in $x$ and $y$ plus some function in $y$, and then differentiating with respect to $y$ to determine that function in $y$ up to a constant. Is this the right approach?

Finally, the question asks for the corresponding analytic function. Is that just $u(x,y)+iv(x,y)$? Or something else? Thanks for taking the time to read this.

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Yes, you got everything right. A minor remark: you can write $v$ as $v=dx^3+3ax^2y-3dxy^2-ay^3+K$ since $b=-3d$. –  Paul Jan 21 '12 at 7:43
    
Thanks Paul. I can't figure out how to delete this post, but if someone else can, feel free to do so, since there isn't really a question here. Sorry! –  NivLac Jan 21 '12 at 7:45
1  
No. You don't need to delete it. In my opinion, you ask a good question. It's strongly encouraged that people post what they didn't understand and what they have tried when they post a question, so that others can help them more precisely. In your question, you did a good job by saying exactly what you have tried and what you were not sure. –  Paul Jan 21 '12 at 8:00

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