show that $f(z)$ is a polynomial in $z.$ 
Let $f(z); z = x + iy,$ be an analytic function and $u$ be its real part. If
  $u$ is a polynomial in the variables $x$ and $y,$ then show that $f(z)$ is a
  polynomial in $z.$

I need to confirm my attempt:
$u,v$ must satifies the C-R equation. So, $u_x=v_y\implies v=\int u_x dy,$ a polynomial in $x,y.$ Consequently $f=u+iv$ is a polynomial in $x+iy.$
 A: In general, that $v_y$ is a polynomial is not sufficient to show that $v$ is a polynomial.  Consider $v(x,y) = sin(x) + y^2$ and $v_y(x,y) = 2y$.  So $v = \int u_xdy$ does not imply that $v$ is a polynomial.
Food for thought: what's the relation between a polynomial's degree and its higher derivatives (in both variables)?
A: The partial derivatives of $u$ and $v$ in the expression $f = u+iv$ have to be zero up to some point because $u$ is a polynomial and $f$ satisfies the C-R equations. Therefore $u$ and $v$ are polynomials. Now write 
$$
f(x+iy) = u(x,y) + i v(x,y),
$$
which is a polynomial in $x$ and $y$ with coefficients in $\mathbb C$. Since this equation is valid for every $x,y$ real, it is also true if $y = 0$. So you have
$$
f(x) = u(x,0) + i v(x,0).
$$
The function $g(z) = u(z,0) + iv(z,0)$ is a complex function that is analytic (because it's a polynomial) and it coincides with $f$ when $x$ is a real number. By analytic continuation, $f$ and $g$ are equal all over the complex plane.
Hope it helps, 
