- #1

AKG

Science Advisor

Homework Helper

- 2,565

- 4

## 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 m

^{th}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 f

^{2}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 f

^{2}= (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 n

^{th}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?