Describe those polynomials $$P=a + bx + cy + dx^2 + exy + fy^2$$ with real coefficients that are the real parts of analytic functions on $\mathbb{C}$.
Idea (1): We are given that $P$ is the real part of a holomorphic function, so it is harmonic. Computing its Lapacian we have $\Delta(P)=2d+2f=0,$ so $d=-f$.
Question: Is there more that can be said about $P$?
Other ideas:
(2) Mean Value Property: Given a harmonic function an an open set in $\mathbb{C}$, such that the closed disk $\bar D(Q,r)\subset U$, we have $$u(Q)=\frac{1}{2\pi}\int_0^{2\pi}u(Q+re^{i\theta})$$ I attempted this formula at the point $0$ on the unit disc. $$P(0)=\frac{1}{2\pi}\int_0^{2\pi} P(e^{i\theta})\iff a=a+\frac{d}{2}+\frac{f}{2}$$ But this gives no new information as we have already shown $d=-f$. Using values of $r$ different than $1$ doesn't give new information. I could try this formula at other points, but then the integrals become tedious.
Naive Attempt (3): The set of local extremes is discrete. Hence given a local extreme $(x,y)$, I can consider a disc $D((x,y),\varepsilon)$ such that $P$ has only one local extreme in the closed disk. Since $P$ attains its sup or inf in this disk, it is constant by the Maximum Principle for Harmonic Functions (http://en.wikipedia.org/wiki/Maximum_principle#The_classical_example). Hence $b=c=d=e=f=0$ and $P=a$. I am not sure if this reasoning is correct or that if $P$ even has local extremes.
On a simply connected domain (in your case the whole complex plane), every harmonic function is the real part of a holomorphic function.
In other words, you can say no more than what you have already.