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Complex Analysis

  1. Feb 3, 2007 #1


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    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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

    3. 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


    |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?
  2. jcsd
  3. Feb 3, 2007 #2

    Gib Z

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    2. I haven't done complex analysis yet, but ignoring the disk part, f(z) is not nessicarily a polynomial. eg if f^2 (z)=z.
  4. Feb 3, 2007 #3


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    1. This is a generalisation of Liouville's theorem, which says that if a function is analytic on a disc, and bounded as z tends to infinity, then z is bounded.

    Now, I'm not sure if your proof will work, since I've not come across Laurent series in my course, but I remember seeing a nice couple of line proof of this. It relies on knowledge of Cauchy's integral theorem for a disc; have you come across this yet? It states that if gamma is a cycle in [itex]U={z \in \mathbb{C}:|z-a|<R}, 0<R<+\infty[/itex], and [itex]f:U\rightarrow\mathbb{C}[/itex] analytic, then for all w in U not on gamma, and for all integers [itex]k\geq 0[/itex]:[tex]n(\gamma,w)f^{(k)}(w)=\frac{k!}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-w)^{k+1}}dz[/tex]

    Try applying this to your question:take u in C, and T[itex]\geq[/itex]R; take the integral over [itex]\gamma:=|z|=T[/itex] and let k=n+1, then n(gamma, u)=1 (the winding number around a circle one anticlockwise is equal to 1). From here, the proof is straighfoward.

    (I am assuming that you've come across CIF before; apologies if you've not!)
  5. Feb 3, 2007 #4


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    not sure which "Maximum Principle" you are referring to, but as far as I know Maximum Principle that we often use in complex analysis requires your function to be harmonic on some domain.

    Ok, I had a quick go at Q1. and it appears to me that your attempt seems reasonable. To make it rigorous, write out f(z) in a power series, then examine the effect on |f(z)|/|z|^n as |z| -> infinity, the conditions should then ensure that coefficients going to zero.

    Q2: Gib Z, I think f^2 (z) = z contradicts the fact that f is holomorphic. Again writing f(z) as a power series and than square it... the fact that f^2 (z) is polynomial means that the sum terminates at some deg n.
    To make it rigorous you may have to identify the rule that translate the coefficients from f^2(z) to f(z).

    Q3: indeed your suggestion seems reasonable. again write out f(z) in power series form and look at term by term differentiation and see if you can invoke |f(z)| < M and |z| < R to put a bound on it.

    NB: I must confess that I have not put in a huge amount of time in thinking about this.. . but those are the things I would try....
    may come back to it when I have time
    Last edited: Feb 3, 2007
  6. Feb 3, 2007 #5
    Many people who know complex analysis complain about things like the maximum principle being taught in a complex analysis class as being true holomorphic functions without mention to it being true for holomorphic function because holomorphic functions are harmonic. I guess its also important to point out that harmonic functions are automatically harmonic.

    Your approach is the correct one (kind of). Noting that f(z)/z^n must be bounded outside a ball. Now what you need to use is precisely Q3. The result in Q3 is called the cauchy estimates. Anyways use the cauchy estimates on
    [itex]f^{(n+1)} [/itex] to show that [itex]f^{(n+m)}(0)=0[/itex] for all m>0. Therefore the power series expansion about zero has only finitely many terms. i.e., f is a polynomial.

    Q2. I'm a little confused about making the domain not be all of [itex]\mathbb{C}[/itex]. It seems that if f^2 is a polynomial then it holomorphic on all of [itex]\mathbb{C}[/itex]. We also have [itex]\frac{\partial f^{2}}{\partial \overline{z}}=2f\frac{\partial f}{\partial \overline{z}}=0[/itex]. This must hold when evaluated at all values of [itex]z\in\mathbb{C}[/itex]. f can have at most n isolated zeros for whatever degree of f^2=n is. So we must have [itex]\frac{\partial f}{\partial \overline{z}}=0[/itex] for all z except possibly n points isolated points. By continuity its 0 on all of [itex]\mathbb{C}[/itex]. This means that f is holomorphic on ALL of C as well. Now I think you can use Q1 an put a bound (depending on |z|^(2n)) and then use that to put a similar bound on f (depending on |z|^(n)). Then by Q1 you are done.

    Sorry is you don't know about [itex]\frac{\partial f}{\partial \overline{z}}=0[/itex] iff holomorphic, I could not think of an easier way to approach it right away.
    But again I'm a little suspicious about having a restricted domain to begin with. Try this and see if you can complete it.

    For this you have to know the Cauchy integral formula. You have to get the generalized cauchy integral formula. This is the cauchy integral formula but for the deriatives of f. Then use what is sometime called the ML estimate. It is basically that the modulus of an integral is less than the modulus of the supremum of a function. Times the length of the curve about which you integrate. This should be in any book that you are using. (ML estimate I mean and the generalized cauchy integral formula).

    EDIT: Q2 is still bothering me. If you think about [itex]f^{2}(z)=z+1[/itex]. Then then we can choose a branch of sqrt(z) that is holomorphic to the right of x=-1. So holomorphic for |z|<1.f^2 is a polynomial but f isn't. I guess in my previous approach I assumed the function was well defined, but possibly not holomorphic outside the restricted domain.
    Last edited: Feb 3, 2007
  7. Feb 3, 2007 #6


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    cristo, that works for 1, thanks! I will use that formula (which is in my book and was covered last year, but not covered this year in the review of last year). I let w = 0, and let k be an arbitrary integer greater than n. I get that |f(k)(0)| < Mk!Tn-k for all T > R. As T tends to infinity, this expression tends to 0, hence f(k)(0) = 0 for all k > n. Thus f(n+1) is a function whose power series expansion about 0 has all 0 coefficients, and from facts that I already know, this makes it identically 0 on all of C, as desired.

    hrc969, thanks for the help on 2. I can choose f(z)2 = z + 2r. Then I can choose:

    [tex]f(z) = \sqrt{z + 2r}[/tex]

    with the appropriate determination of [itex]\sqrt{\ }[/itex] such that the "branch cut" goes along the interval of the real axis I = {x in R : x < -2r}. This isn't holomorphic everywhere, but it is on C - I, hence in particular on |z| < r. And I see what to do with Q3 now as well, using the Cauchy integral formula for higher derivatives (the one cristo gave). Thanks.
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