Does $f^{(n)} = 0$ imply that complex $f$ is a polynomial?

complex-analysis polynomials

Hint: Can you do the case $n=1$? Induct on $n$.


Yes.

Consider that there is one unique solution to the differential equation $y^{(n)}(x) = 0$ where $y^{(j)}(0) = c_j\cdot (j-1)!$ for $0< j<n$ and $y(0) = c_0$.

This solution is the polynomial $y = c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + \cdots + c_1x+ c_0$.

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