Let \(f = u + iv\) be an entire function such that \(\Re(f(x+iy))\) is polynomial in \(x\) and \(y\). Show that \(f(z)\) is polynomial in \(z\).
Solution
To clear up notation: write \(f(z) = u(x, y) + iv(x, y)\), here we’re assuming that \(u\) is polynomial in \(x\) and \(y\).
If \(u\) is polynomial in \(x,y\), then so is \(v\).
This follows from the fact that \(u\) is a harmonic conjugate of \(v\), and the explicit process computing the conjugate will result in a polynomial. Gamelin describes this process in detail, see Ch.2 Section 5 on Harmonic functions where he proves the formula \[ v(x, y) = \int_{y_{0}}^{y} \frac{\partial u}{\partial x}(x, t) \dt -\int_{x_{0}}^{x} \frac{\partial u}{\partial y}\left(s, y_{0}\right) \,ds + C .\]
Since \(f(x, y)\) is a polynomial in \(x, y\), \(f(z)\) must be a polynomial in \(z\).
Since \(f\) is entire, it’s equal to its Laurent series everywhere, so \[ f(z) = \sum_{k\geq 0} c_k z^k, \qquad c_k = {f^{(k) }(0) \over k!} = {1\over 2\pi i} \int_{S^1} {f(\xi) \over \xi^{k+1} } \dxi .\] Thus \(f\) will be a polynomial if \(c_{N} = 0\) for all \(N\) large enough, which will be true if \(f^{(N)}(z) = 0\) for large enough \(N\). But we can write \[ \dd{}{z} f(z) = \dd{}{x} f(x, y) \implies 0 = \qty{\dd{^N}{x^N}} f(x, y) = \qty{\dd{^N}{z^N}} f(z) \da f^{(N)}(z) ,\]