find analytic function whose real part is given 
Let $u(x,y)$ be a harmonic polynomial of real coefficients and $$f(z)=2u(\frac{z}{2}, \frac{z}{2i})-u(0,0)$$
  prove that for any $z\in \mathbb{C}$, $\text{Re} f(z)=u(x,y)$.

I tried to use the condition $u_{xx}+u_{yy}=0$ to construct harmonic conjugate $v$ then deduce $f$, but it comes out a lot of integrals, and I don't know how to simplify. A similar question goes here where $u$ is degree one wrt $y$ but here it's more general. Thanks for any help.
 A: We can consider $u(x,y)$ as a sum of polynomials with all its terms of the same degree, so is, homogeneus. Consider the one of degree $m>0$, call it $u_m(x,y)$.Next, every harmonic homogeneus polynomial of two variables of degree $m$ is a linear combination of at most two linearly independent homogeneus harmonic polynomials. Such linearly independent homogeneus polynomials can be chosen to be the imaginary part and the real part of the complex polynomial $(x+iy)^m$ (Here) E.g. For degree 3 these are, $P_1(x,y)=x^3-3xy^2$ and $P_2(x,y)=3x^2y-y^3$
$$P_{m_1}(x,y)=\Re (x+iy)^m=\binom{m}{0}x^m-\binom{m}{2}x^{m-2}y^2+\binom{m}{4}x^{m-4}y^4+\,...$$
$$P_{m_2}(x,y)=\Im (x+iy)^m=\binom{m}{1}x^{m-1}y-\binom{m}{3}x^{m-3}y^3+\binom{m}{5}x^{m-5}y^5+\,...$$
Now, calculate these polynomials for $x=z/2$ and $y=z/(2i)$, with $z=\bar x+i\bar y$
$$P_{m_1}(\frac{z}{2},\frac{z}{2i})=\binom{m}{0}\frac{z^{m}}{2^{m}}-\binom{m}{2}\frac{z^{m-2}}{2^{m-2}}\frac{z^2}{(2i)^2}+\binom{m}{4}\frac{z^{m-4}}{2^{m-4}}\frac{z^4}{(2i)^4}+\,...$$
$$=\binom{m}{0}\frac{z^{m}}{2^{m}}+\binom{m}{2}\frac{z^m}{2^m}+\binom{m}{4}\frac{z^m}{2^m}+\,...=\frac{z^{m}}{2^{m}}\left(\binom{m}{0}+\binom{m}{2}+\binom{m}{4}+\,...\right)=$$
$$=\frac{z^{m}}{2^{m}}2^{m-1}=\frac{z^{m}}{2}$$
$$P_{m_2}(\frac{z}{2},\frac{z}{2i})=\binom{m}{1}\frac{z^{m-1}}{2^{m-1}}\frac{z}{(2i)}-\binom{m}{3}\frac{z^{m-3}}{2^{m-3}}\frac{z^2}{(2i)^3}+\binom{m}{5}\frac{z^{m-5}}{2^{m-5}}\frac{z^5}{(2i)^5}+\,...$$
$$=-i\binom{m}{1}\frac{z^{m}}{2^{m}}-i\binom{m}{3}\frac{z^m}{2^m}-i\binom{m}{5}\frac{z^m}{2^m}+\,...=\frac{z^{m}}{2^{m}}\left(\binom{m}{1}+\binom{m}{3}+\binom{m}{5}+\,...\right)=$$
$$=-i\frac{z^{m}}{2^{m}}2^{m-1}=-i\frac{z^{m}}{2}$$
Now,
$$u_m(x,y)=k_{m_1}P_{m_1}(x,y)+k_{m_2}P_{m_2}(x,y)$$
$$2u_m(\frac{z}{2},\frac{z}{2i})=2k_{m_1}P_{m_1}(\frac{z}{2},\frac{z}{2i})+2k_{m_2}P_{m_2}(\frac{z}{2},\frac{z}{2i})=k_{m_1}z^{m}-k_{m_2}iz^{m}$$
$$\Re\left(2u_m(\frac{z}{2},\frac{z}{2i})\right)=k_{m_1}\Re z^m+k_{m_2}\Im z^m=u_m(\bar x,\bar y)$$
Finally, $u$ is sum of homogeneus polynomials of several degrees: $u(x,y)=\sum_{m=1}^n u_m(x,y)+C$ (with $u(0,0)=C)$
$$\Re\left(f(z)\right)=\Re\left(2u(\frac{z}{2}, \frac{z}{2i})-u(0,0)\right)=\Re\left(\sum_{m=1}^n2u_m(\frac{z}{2},\frac{z}{2i})+C\right)=$$
$$=\sum_{m=1}^n\Re\left(2u_m(\frac{z}{2},\frac{z}{2i})\right)+C=\sum_{m=1}^nu_m(\bar x,\bar y)+C=u(\bar x,\bar y)$$
A: Since $u(x, y) = u\left(\frac{z+\bar z}{2}, \frac{z - \bar z}{2}\right) = P(z, \bar z)$, we can define $P(z, \bar z)$ to be a harmonic polynomial, such that
$\frac{\partial^2 P}{\partial z\partial \bar z}$ is zero. Therefore every monomial in $P$ can't be the form $z^i\bar z^j$ otherwize the Laplace is not zero.
Write in the form of $P(z, \bar z) = c_0 + \sum_{k = 1}^n c_kz^k + d_k \bar z^k$ and thus $$u(x, y) = c_0 +\sum_{k=1}^n c_k(x+iy)^k + d_k(x-iy)^k.$$
From that the coefficients of $u$ are real numbers, we have $c_k = d_k$, and $$u(x, y) = c_0 +\sum_{k=1}^n c_k(x+iy)^k + c_k(x-iy)^k, c_i\in \mathbb R.$$
From definition of $f$, we have $f(z) = 2u(z/2, z/2i)-c_0=c_0+2 \sum_{k=1}^nc_kz^k$.
Notice that $\text{Re}(2z^k)=z^k+\bar z^k$ we can easily get $\text{Re}(f(z)) = u(x,y)$.
