- #1
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Homework Statement
1. Suppose that f(z) is holomorphic in C and that |f(z)| < M|z|n for |z| > R, where M, R > 0. Show that f(z) is a polynomial of degree at most n.
2. Let f(z) be a holomorphic function on a disk |z| < r and suppose that f(z)2 is a polynomial. Is f(z) a polynomial? Why or why not?
3. Suppose f(z) is holomorphic in |z| < R, with |f(z)| < M. Show that if |z| < r < R, then
|f(n)(z)| < Mn!(R-r)-n
Homework Equations
Schwarz's lemma
Maximum modulus principle
Definition: holomorphic = (one-time) complex differentiable
Fact: holomorphic = infinitely complex differentiable
Definition: analytic = has a power series expansion around each point
Fact: holomorphic = analytic
I know what the power series of f looks like in terms of the derivatives of f
Fact: f(n) = 0 for all n iff f is identically zero on an open set iff f is identically zero on its domain
The Attempt at a Solution
1. I thought of applying the maximum modulus principle to f(z)z-n, but this didn't yield the desired result in any obvious way. I also thought to say that if g(z) = f(z)z-n is bounded in absolute value outside of |z| < R, then the coefficients in its Laurent expansion (since it's meromorphic) must eventually be zero (i.e. exist N such that for all m > N, the mth coefficient is 0). My idea was that if they were not eventually zero, then no matter how small they are, eventually |z| will be so big that it would cause |g(z)| to blow up, contradicting that |g(z)| < M when |z| is big. But I couldn't make this argument rigorous, especially because Laurent series only need to be defined locally, I think.
2. My guess is yes. My guess is that f2 is a polynomial of degree n, and f(z) = [itex]\sum _{i=0}^{\infty}a_iz^i[/itex] then define g(z) = [itex]\sum _{i=0}^na_iz^i[/itex]. Then f2 = (f-g + g)2. I would then compare coefficients. Will this approach work?
3. I have little idea what to do. Like in 1, I would guess that somewhere I need to use the max. mod. principle or maybe Schwarz's lemma, but I can't clearly see how. I do see that
|f(n)(z)| < Mn!(R-r)-n
says:
|f(n)(z)|(R-r)n/n! < M
and that the thing on the left is similar to the nth term in the power series expansion of f about z. In particular, (R-r)n > (R-|z|)n. But what do I do with this?