Question

Complex question about entire functions!

Answer 1

If f is entire and satisfies \[|f(z)| \le A|z|^n \] for some positive A and n how can I show that f is a polynomial of degree at most n?

Answer 2

If f were of degree greater than n, then at values where |z| times the constant associated with the highest power of z is greater than A, f(z) will be greater than A. There is no A which will never be overtaken by sufficiently high powers of z, no matter how small the constant multiplying those powers is in comparison to A. Basically, if z*(the constant next to the highest power of z)>>>A, then the other powers of z will become insignificant at even higher values of z and |f(z)|>>A|z^n|.

Answer 3

How would I prove this using the Maclaurin expansion of f?

Answer 4

The MacLaurin expansion will end at a certain power of f. If that power is higher than n, then the nth derivative of |z|^n is a constant but the nth derivative of f is still a function in z, which means that at infinity f must necessarily be greater than z^n.

Answer 5

http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem
This theorem says that if \( |f(z) \le M|z^n|\) then \( f(z) + Mz^n\) will have same number of zeros inside \( |z| \le 1/M\). Of course if \( f(z) \) is an entire function then it will be a polynomial of degree \( n\) which follows from Fundamental theorem of algebra.

Answer 6

if \( f(z) \) were of degree > n, say \(m \) then choose \( z = \) very large so that \( 1/z \to 0\) ... divide both sides \( z^m\) and take \( |z| \to \infty \) the condition is no longer valid.

Answer 7

Or even in Rouche's theorem choose the polynomial \( M z^n = 1\), choose the region to be entire complex plane. \( M z^n = 1\) will have n roots (in entire plane) so does \(f(z) + Mz^n = 0 \) in entire complex plane. rest follows from fundamental theorem of algebra.