I'm searching for harmonic functions inside a set of HOMOGENIOUS polynomials in two variables. Let's say that $$P_n =\left\{\sum_{i+j = n} a_{ij} x^i y^j \quad|\quad a_{ij} \in \Bbb R\right\}.$$
Let $H_n$ denote the set of all harmonic functions from $P_n$.
So, obviously $H_n \subset P_n$.
Show that $\dim(H_n) = 2$ for all $n$.
I tried it using induction on $n$, but I couldn't prove it. Then I tried it by separating polynomials like this: $$ p(x, y) \in P_n \\ p(x, y) = x a(x, y) + y b(x, y)$$ But again, couldn't come far, because now $ a(x, y), b(x, y)$ aren't necessarily homogenous.
Write $x+iy=z$ so $2x=z+\bar z, 2iy=z-\bar z$ and express the polynomial as $P(z,\bar z)=\sum_{j+k=n} c_{jk}z^k \bar z^j, c_{jk} \in \mathbb C$
Since a polynomial in $z, \bar z$ is harmonic iff all the mixed terms are zero, we get $P(x,y)=P(z, \bar z)=cz^n+d\bar z^n=c(x+iy)^n+d(x-iy)^n, c, d \in \mathbb C$
But the original coefficients are real so $P$ is real hence $P(z, \bar z)=\bar P(z, \bar z)$ or $d=\bar c$ so the complex dimension is $1$ hence the real dimension is $2$ since clearly the polynomials for $c=1,i$ or if you want $\Re z^n, \Im z^n$ are harmonic, homogeneous of degree $n$ and independent over the reals. Done!