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Math Help - Proofs about an analytic function

  1. #1
    MHF Contributor arbolis's Avatar
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    Proofs about an analytic function

    Let f(z) be an analytic function in the open disk D(0,R) which satisfies |f(z)| \leq |z|^n for some n \in \mathbb{N}^* and for all z \in D(0,R).
    1)Prove that f has a zero in z=0 or order \geq n.
    2)Prove that \frac{f(z)}{z^n} has a removable singularity at z=0.
    3)Assume that f is entire and that |f(z)| \leq |z|^n for some n \in \mathbb{N}^* and for all z \in \mathbb{C}. Show that f(z)=c z^n with c \in \mathbb{C}.
    Attempt:1) Let g(z)=|z|^n. Then we have g(0)=0. Since |f(z)| \leq g(z), it follows that f(0)=0. Does this also mean that |f^{(n)}(z)|\leq n! |z|? If so, then I think I've proven part 1).
    Part 2): I think they want me to prove that \lim _{z \to 0} z\frac{f(z)}{z^n}=l where l \in \mathbb{C}.
    Using the fact that \frac{|f(z)|}{|z|^n} \leq 1, I get that -z \leq \frac{f(z)}{z^{n-1}} \leq z which tend to 0 as z tends to 0 because of the sandwich theorem. Proved. (I don't think my proof is valid).
    3)I'm not sure.
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  2. #2
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    For the first one argue with the series expansion at 0. Could g(z)=\frac{f(z)}{z^n} be bounded around 0 if there was a coefficient a_k\neq 0 with 0\leq k\leq n in the Taylor series for f?

    For the second one, if g is bounded near 0, could it have a pole or essential singularity there?

    For the third apply Liouville to g.
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  3. #3
    MHF Contributor arbolis's Avatar
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    I'm still clueless with 1) and 3).
    With 2), -1 \leq g(z) \leq 1. I just saw a theorem in my class notes which states that if g is analytic in D(0,R)-z_0, then z_0 is a pole of g if and only if \lim _{z \to z_0} f(z)=\infty. So clearly there is no pole at z=0 for g. I must still show that there isn't an essential singularity. Writing f(z)=\sum _{m=0}^{+\infty} a_m z^m since it's analytic, it's obvious that dividing it by |z|^n won't make appear an infinity of negative powers of z. Therefore since g(0) isn't defined, it can only be a removable singularity.
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  4. #4
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    Quote Originally Posted by arbolis View Post
    With 2), -1 \leq g(z) \leq 1.
    Be careful, you made the same mistake in your first post: We are dealing with complex numbers, which aren't ordered, so the inequality above is meaningless.

    For the first one note that |g(z)| \leq 1 and apply the same reasoning as in (2) and use the fact that f(z) has a zero of order n at 0 iff f(z)=z^nh(z) where h is analytic and h(0)\neq 0.

    For the third one, using the second one you get that g(z) has an entire extension whose modulus is bounded by 1
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