Holomorphic functions with polynomial real part $f:\mathbb{C}\rightarrow \mathbb{C}$, $f(x+iy)=u(x,y)+iv(x,y)$ is a holomorphic function, its real part $u$ is a harmonic polynomial, i.e. $u\in \mathbb{R}[x,y]$ and $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$. We are to describe all possible $v$.
I'll be very grateful if anybody notice any mistake in my reasonings. Furthermore, I'd like to know if the following description is full or it can be improved.
We prove that $v$ is defined uniquely up to an additive constant. First of all, let's prove that also $v\in \mathbb{R}[x,y]$. Really, $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ (Cauchy-Riemann equations), hence partial derivatives of $v$ are polynomial functions. It means that $v$ is a polynomial function. 
Further, let $u(x,y)=\sum a_{mn}x^my^n$, $v(x,y)=\sum b_{mn}x^my^n$ (where $m,n$ are not greater than $\max\{\deg u,\deg v\}$, some of $a_{nm}$ or $b_{mn}$ can be zeros). $u$ and $v$ are smooth functions, hence $f$ is holomorphic if and only if $u$ and $v$ satisfy Cauchy-Riemann equations everywhere in $\mathbb{R}^2$. After differentiating we have the following conditions: $\sum ma_{mn}x^{m-1}y^n=\sum nb_{mn}x^my^{n-1}$ and $\sum mb_{mn}x^{m-1}y^n=-\sum na_{mn}x^my^{n-1}$ (here $m^2+n^2>0$). So, $(m+1)a_{m+1,n}=(n+1)b_{m,n+1}$ and $(m+1)b_{m+1,n}=-(n+1)a_{m,n+1}$ $\forall$ $m$ and $n$. These results promise that if $u$ is given, all coefficients of $v$ can be evaluated explicitly and uniquely (except of $v(0,0)$ that can be chosen at will).
Can you improve the given description of imaginary part?
 A: Writing $f(z)=\sum_{n=0}^\infty c_n z^n$, we see that $$u=\frac12(f+\bar f) = \frac12\sum_{n=0}^\infty (c_n z^n+\bar c_n\bar z^n)\tag1$$ 
The $n$th term in the series in (1) is a homogeneous polynomial in $x,y$ of degree $n$. Therefore, there is no cancellation of monomials $x^ky^m$ between the terms with different $n$. 
Conclusion: for $d\ge 1$, $u$ is a harmonic polynomial of degree $d$ if and only if $f$ is a complex polynomial of degree $d$. 
(The case $d=0$ is a bit special, since zero constant and nonzero constants have different degrees.)
The coefficient of $x^n$ in (1) is $\operatorname{Re}c_n$ while the coefficient of $x^{n-1}y$ is $i\operatorname{Im}c_n$. This gives you $c_n$ for every $n\ge 1$. (Of course, for $n=0$ you don't know $\operatorname{Im}c_0$ for the lack of $x^{n-1}y$.) Having $c_n$, you can write down 
$$v=\frac12(f-\bar f) = \frac12\sum_{n=0}^\infty (c_n z^n-\bar c_n\bar z^n)\tag2$$ 
and expand it into $x$ and $y$ monomials if you wish.
(Personally, I never find much need to expand harmonic functions into $x^ky^m$ monomials; the expansion into powers of $z$ and of $\bar z$ is more efficient because it encodes the harmonicity of the function.)
A: You are correct. 
A more straight way to prove this is: 
$$\text{Let  }u(x,y)=\sum a_{mn}x^my^n$$
Then use the $1$st of the C-R equations : $$\dfrac{\partial v}{\partial y}=\dfrac{\partial u}{\partial x}=\sum a_{mn}\cdot m\cdot  x^{m-1}y^{n}\Rightarrow $$
$$\dfrac{\partial v}{\partial y}=\sum a_{mn}\cdot m\cdot  x^{m-1}y^{n}\Rightarrow$$
$$v(x,y)=\sum a_{mn}\cdot m\cdot x^{m-1}\cdot \frac{y^{n+1}}{n+1}+K(x)$$
Now use the $2$nd of the C-R equations: 
$$\dfrac{\partial v}{\partial x}=-\dfrac{\partial u}{\partial y}=-\sum a_{mn}\cdot n\cdot x^my^{n-1} \Rightarrow$$
$$\dfrac{\partial v}{\partial x}=-\sum a_{mn}\cdot n\cdot x^my^{n-1} $$
However, using the formula we found for $v(x,y)$ it follows that :
$$\dfrac{\partial v}{\partial x}=\sum a_{mn}\cdot m\cdot (m-1)\cdot x^{m-2}\frac{y^{n+1}}{n+1}+K'(x)$$
Thus, combining the above relationships: $$K'(x)=-\sum a_{mn}\cdot n\cdot x^my^{n-1} -\sum a_{mn}\cdot m\cdot (m-1)\cdot x^{m-2}\frac{y^{n+1}}{n+1}$$
where we must have a function of $x$ so the proper cancelations happen( Since $u$ is harmonic, the harmonic conjugate exists, therefore this proof works, which indicates that $K$ is a function of $x$ only). Can you find what are the restrictions on $a_{mn}$ such that $K'$ is a polynomial of $x$?
Now by integrating what is going to be left after cancelations, it is proved that $K(x)$ is a polynomial of $x$ (plus a constant). Therefore, $v(x,y) $ is a polynomial.
( I will add some details tomorrow, since I have little time now)
