I am given the homogeneous polynomial of degree $l$: \begin{align} u(x,y,z)=\sum_{a,b}c_{ab}(x+iy)^a(x-iy)^bz^{l-a-b} \end{align} Where $0 \leq a,b \leq l$.
I have to show that given $l$ and $m=a-b$, $c_{ab}$ is determined up to one multiplicative constant, by $l$ and $m=a-b$, when the laplacian of $u(x,y,z)$ is $0$.
Im having difficulty taking the laplacian of $u$ (which is also an exercise), but i've reached:
\begin{align} \Delta u=\sum_{a,b} c_{ab}4ab(x+iy)^{a-1}(x-iy)^{b-1}z^{l-a-b}+c_{ab}(l-a-b)(l-a-b-1)(x+iy)^a(x-iy)^bz^{l-a-b-2} \end{align}
I've reached the following about $b$ using that $a=m+b$: \begin{align} 0 \leq b \leq l-m \quad \text{for} \ m \geq 0\\ -m \leq b \leq l \quad \text{for} \ m \leq 0 \end{align}
Simply rewriting $\Delta u$ has yielded nothing. I think it has something to do with Legendre Functions, but i'm not sure how.