Prove that a function is a polynomial of degree at most one given $v(x,y) \geq x$

Question

The question is Let $f(z) = u(x,y) + iv(x,y)$ be an entire function satisfying $v(x, y) ≥ x$ for all $z = x + iy$. Then show that f(z) is a polynomial of degree at most one.
I know that I have to somehow show $f''(z) = 0$
I tried showing $|e^{if(z)}|=e^{-v}<e^{-x}$ which is of no use. Also, I haven't been able to determine any condition on $v_x(x,y)$. please suggest how should I approach this question.

Answer 1

Here's a sketch:

1. Show $\left|e^{z+if(z)}\right| \leq 1$. Use a theorem to conclude that $e^{z+if(z)}$ is constant.
2. Use the fact that the derivative of $z\mapsto e^z$ is never zero to conclude* that $z+if(z)$ is a constant $C$.
3. Solve for $f(z) = i(z-C)$.

*Note: If you cannot see how the derivative of $z\mapsto e^z$ relates to showing this, then try some other method. This is not the only way to reach that conclusion.