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Complex analysis proof

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that f is an entire function. Define g(z)=f*(z*), where * indicates conjugates. I know from another problem that g(z) is also entire. Suppose also that f(z) maps the real axis into the real axis, so that f(x+0i)is in R for at x in R. Show that f(z)=g(z) for all z in C.

    2. Relevant equations
    There is a hint that says, consider f(z)-g(z).
    I know that if the derivative of something is 0, then it is constant, and if we know that value at one point we know it is that value everywhere.

    3. The attempt at a solution
    I know that on the real axis, f(z)=f(z*)=f*(z*), so that f(z)-g(z)=0 for all z such that z=x+0i. I was hoping to either show that the derivative is constant, or perhaps use something about isolated zeros? Is it true that a non-constant entire function has isolated zeros?
  2. jcsd
  3. May 22, 2010 #2
    Yes, you can only have isolated zeroes. You can also study the smewhat stronger result: Principle of Analytic Continuation. Whenever you have that two analytic functions are equal to each other on a set of points z_n such that all the z_n are different and there is limit point, then the two functions must be equal to each other everywhere on the domain.
  4. May 22, 2010 #3
    Okay, so the proof would go like this:
    1. show that f(z)=f(z*)=f*(z*)=g(z) when z in on the real axis
    2. This the function f(z)-g(z) has zeros whenever z in on the real axis
    3. However, since f and g are analytic, f-g is analytic, and must have isolated zeros if it is not constant
    4. Thus f-g is constant b/c the real axis does not consist of isolated points. Thus, f-g=o everywhere, and f=g
    Does that sound right?
  5. May 22, 2010 #4
    Yes, that's correct. But I think it is better exercise for you to give a proof from first principles. Because even if you have isolated zeroes it is possible that the two functions would have to be equal (this will be the case if the isolated zeroes have an accumulation point).

    You can simply prove that all the derivatives of the function
    h(z)= f(z) - g(z) are equal to zero at some arbitrary point in the real axis. This is then a good exercise to understand the slightly more general proof of the Principle of Analytic Continuation.
  6. May 22, 2010 #5
    So even if I proved that all the derivatives of f-g are equal to zero at some arbitrary point on the real axis, how would that help me extend out? We haven't yet learned that Principle. If I know all the derivatives at a single point for f and g, I know that I can find a Taylor series that is valid everywhere because both f and g are entire. Then their Taylor series would be equal, and they would be equal. So I guess I need to prove that either all the derivatives of f-g are zero at a single point. How would I do that?
  7. May 22, 2010 #6
    Yes, that's a correct way to attack the problem. Let's consider some particular point on the real axis, say, z = 0. If we put h(z) = f(z) - g(z), then we have that h(z) is analytic and that

    h(0) = 0

    The derivative of h at z = 0 is the limit of [h(z) - h(0)]/(z-0) = limit of z to zero of h(z)/z. Can you compute this limit?
  8. May 22, 2010 #7
    Okay, so we know that h(z) is entire, so it must be differentiable everywhere, particularly at z=0. Thus, we can take the limit of any sequence of points we want and it will be the derivative. Thus, let z_n =1/n, so that the limit of z_n is 0. Then, h(z)=0 at every z_n, so that h'(z)=0 at z=0. Then, how would we prove it for h"(0)?
  9. May 22, 2010 #8
    To be more precise, h'(0) = 0 because

    h(z_n)/z_n = 0

    for all n and thus the limit of this sequence is zero.

    You can write the second derivative also as a differential coefficient involving only h. But in this particular problem you can use a shortcut. The point z = 0 was an arbitrary point in the real axis, we could have aken any other point x on the real axis instead. So, we know that h'(z) = 0 everywhere on the real axis. This means that you can replace h by h' in the proof yielding tha the second derivative is also zero. This means that you can easily set up a proof of the general result that all derivatives are zero using induction.

    However, it is still a good exercise to not use that h' is zero everywhere on the real axis and use only that you have a set of points z_n with limit point zero where f and g agree.
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