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Prove: f(z) = log z cannot be analytic

  1. Jun 12, 2013 #1
    I'm having difficulty completing the last problem of an assignment due tomorrow evening. I feel as if I'm missing something; every time I attempt the problem I get stuck or confused.

    "22. Use the conclusion of Exercise 21 and Example 13 of Section 6, Chapter 1, to prove that f(z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve [itex]\gamma[/itex] that surrounds the origin. (Hint: What is the value of [itex]\int_{\gamma} f'(z) dz[/itex]?) (from Complex Variables, 2nd. edition by Stephen D. Fisher)


    As this problem refers to both Exercise 21 and Example 13 I will summarize them here as well:

    "21. Let [itex]\gamma[/itex] be a piecewise smooth simple closed curve, and suppose that F is analytic on some domain containing [itex]\gamma[/itex]. [Then] [itex]\int_{\gamma} F'(z) dz = 0[/itex]"

    "Example 13 Suppose that [itex]\gamma[/itex] is a piecewise smooth positively oriented simple closed curve. The value of the integral

    [itex]\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p}[/itex] , p not in [itex]\gamma[/itex] is

    [itex]\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p} = 0[/itex], p is outside [itex]\gamma[/itex], or

    [itex]\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p} = 1[/itex], p is inside [itex]\gamma[/itex]
    "



    I attempted to show that f(z) = log z is analytic by applying the Cauchy-Riemann equations.

    [itex]f(z) = log z = ln|z| + iarg(z) = ln|(x^2 + y^2)^\frac{1}{2}| + i arctan(y/x)[/itex]
    [itex]f(z) = u + iv[/itex] with
    [itex]u = ln|(x^2 + y^2)^\frac{1}{2}|[/itex] and
    [itex]v = arctan(y/x)[/itex]

    I then computed the partial derivatives of both u and v with respect to x and y and showed that u and v satisfy the Cauchy-Riemann equations. As a result, I expect f(z) = log z to be analytic.

    However, the question asks me to show that f(z) is not analytic...

    If I follow the hint given in the question:

    [itex]\int_{\gamma} f'(z) dz[/itex]
    [itex]\int_{\gamma} (log z)' dz[/itex]
    [itex]\int_{\gamma} \frac{1}{z} dz[/itex]

    I'm not too sure where I should go from here...if I simply integrate this I "just" get f(z) back (I'm not sure how to account for integrating over the path...should I write:

    [itex]\int_{\gamma} \frac{1}{z} dz = \int_{a}^{b} \frac{1}{\gamma(t)} \gamma'(t) dt[/itex]

    But then how should I proceed? I need to show that such an integral does not equal 0, because by exercise 21 if the integral equals 0 the function is analytic...


    Any help will be greatly appreciated! Thanks a lot in advance! :)
     
  2. jcsd
  3. Jun 12, 2013 #2
    The answers are there. What is the value of p for the function in your question?
     
  4. Jun 12, 2013 #3
    Oh! The value of p is 0. Then, since the origin (ie. 0) is interior to [itex]\gamma[/itex] by example 13 the value of the integral is non-zero and therefore f(z) = log z is not analytic by exercise 21!
     
  5. Jun 13, 2013 #4
    Yup, you've got it :)
     
  6. Jul 1, 2013 #5
    I kinda get what yall are sayin but...

    where did this example 13 business come from? I have to answer the same question but I don't think we ever learned that relation.
     
  7. Jul 1, 2013 #6

    Dick

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    Science Advisor
    Homework Helper

    It's the Cauchy Integral Theorem. You've got to have seen it.
     
  8. Jul 1, 2013 #7
    Aw dang well okay. Thanks for the help though
     
  9. Jul 2, 2013 #8

    Bacle2

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    Science Advisor

    Some other ideas, just to illustrate:

    1)Notice that the argument is not even continuous in the plane, let alone analytic.

    2)If you accept that Logz and e^z are" inverses " , then notice that e^z is not 1-1 in the plane ( it is actually oo->1), so that it cannot have a global inverse in the plane.
     
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