# complex analysis3

• May 4th 2011, 12:09 PM
shaheen7
complex analysis3
Suppose that f is an entire function, and there exists a nonnegative number \alpha such that
|f(z)| <= \alpha|z|^n
for all z in C. Show that f is a polynomial of degree at most n. (Hint: Use the Cauchy integral formula to show that f^(n+1)(w)=0 for all w in C.)

Thank you!
• May 4th 2011, 12:10 PM
shaheen7
\alpha is meant to be theta.
• May 5th 2011, 12:39 PM
Opalg
Quote:

Originally Posted by shaheen7
Suppose that f is an entire function, and there exists a nonnegative number \alpha such that
|f(z)| <= \alpha|z|^n
for all z in C. Show that f is a polynomial of degree at most n. (Hint: Use the Cauchy integral formula to show that f^(n+1)(w)=0 for all w in C.)

The Cauchy integral formula says that $f^{(n+1)}(w) = \frac{(n+1)!}{2\pi i}\oint_C\frac{f(z)}{(z-w)^{n+2}}dz$. The idea here is to take C to be a large circle, of radius R, centred at the origin. So make the substitution $z = Re^{i\theta}$, estimate the size of the integral and show that it can be made arbitrarily small if R is large enough.
• May 7th 2011, 12:38 PM
shaheen7
using circle of radius R and using modulus f(z) <= alpha modulus z ^ n
i got \int(f(z) / (z-w)^n+2) = (modulus f(z) * 2ipiR) / R^(n + 2) = alpha * R^n+1 *2ipi / R^n+2. Since R is in the denominator, do we get our result?
• May 7th 2011, 12:43 PM
Opalg
Yes, that looks about right. It shows that by choosing R large enough, you can make f^{n+1}(w) arbitrarily small. But since f^{n+1}(w) is independent of R, it follows that it must be 0. If that holds for all w, then f has to be a polynomial of degree at most n.
• May 7th 2011, 12:47 PM
shaheen7
by f^{n+1}(w) is independent of R, do you mean to say we can take R to infinity?