Proving a harmonic polynomial in $x$ and $y$ is a linear combination of $\Re(x+\mathrm iy)^n$ and $\Im(x+\mathrm iy)^n$ 
Prove that a harmonic polynomial in $x$ and $y$ is a linear combination of $\Re(x+\mathrm iy)^n$ and $\Im(x+\mathrm iy)^n$.

My train of thought is as follows: 
Prove that a harmonic polynomial is a linear combination of $(x + iy)^n$ and $(x - iy)^n$, denoted by $z$ and $\bar{z}$, and show that $\Delta p = 0$ if and only if $p$ consists of only monomials that purely involve $z$ or $\bar{z}$.

An easy proof of this uses the fact that the Laplacian $\Delta=4\partial_z \partial_{\bar z}$. Then using $2x=z+\bar z, 2iy =z-\bar z$ write the polynomial as $P(x,y)=P(z,\partial z)=\sum {c_{m,n}z^n\bar z^m}$ and think of $z, \bar z$ as independent (real) variables too. 
Then $\Delta P=4\sum_{n,m \ge 1}c_{n,m}z^{n-1}\bar z^{m-1}$, so $\Delta P=0$ iff $c_{n,m}=0, n \ge 1, m \ge 1$ by the usual reasoning ($c_{1,1}=0$ by taking $z=\bar z=0$ and then one can divide by |z| and take again $z, \bar z \to 0$ to show that $c_{1,2}=c_{2,1}=0$ etc - the only difference from usual polynomials as that here $z, \bar z$ go simulatenously to zero, so one divides by $|z|^k$ to eliminate higher powers and show inductively that all coefficients of degree $k$ are zero using the arguments)
