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How to deal with f(z) that is only analytic outside the unit disk?

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    f(z) is analytic for |z|≥1. Let C be the unit circle. Show that the integral [itex]\frac{1}{2i\pi}\int_C\frac{f(w)}{wz-z^2}dw[/itex] is 0 if [itex]|z|<1[/itex], is [itex]\frac{f(z)}{z}[/itex] if [itex]|z|>1[/itex]



    2. Relevant equations



    3. The attempt at a solution
    For [itex]|z|<1[/itex] case, I tried to write the integral as
    [itex]\frac{1}{z2\pi i}\int_C\frac{f(w)}{w-z}dw[/itex] and write the integrand as a series
    [itex] \frac{f(w)}{w}\sum_{n=0}^\infty(\frac{z}{w})^n [/itex] which converges uniformly, then interchange integral with summation ... I tried to show that every term under the summation is 0 but didn't make it...
     
  2. jcsd
  3. Oct 15, 2012 #2
    I think you have that worded wrong. How about I write it like I think it should be:

    Let [itex]f(z)[/itex] be entire. Then:

    [tex]\frac{1}{2\pi i}\mathop\oint\limits_{|z|=1} \frac{f(w)}{wz-z^2}dw=
    \begin{cases} 0 & |z|>1 \\ \frac{f(z)}{z} & 0<|z|<1
    \end{cases}
    [/tex]
     
  4. Oct 15, 2012 #3
    I guess the main difference between our wording is that we have exactly the opposite range of [itex]|z|[/itex] ? I think the problem I wrote was correct since [itex]f(z)[/itex] is only analytic outside the unit circle. So it seems to have exactly the opposite answer as what we originally expected. The normal case we encountered that can apply Cauchy's Integral Formula is that [itex]f(z)[/itex] is analytic inside and on a circle of certain radius
     
  5. Oct 15, 2012 #4
    Ok, I think I understand now.

    How about we do an example and then see if you can generalize it. Let:

    [tex]f(w)=\frac{1}{w-1/4}[/tex]

    That's certainly analytic outside the unit disc and not analytic inside the unit disc. So now, let's take [itex]z=1/5[/itex] so that [itex]|z|<1[/itex]

    Then we have:

    [tex]\mathop\oint\limits_{|z|=1}\frac{1}{1/5(w-1/4)(w-1/5)}dw[/tex]

    Now, that's zero right? In fact, any integral:

    [tex]\mathop\oint\limits_{|z|=1}\frac{1}{(w-a)(w-b)}dw[/tex]

    with a and b inside the unit circle is zero. You can prove that right? Just compute the residues and add them. But that's just for a simple pole and you did not state what the order of the pole for f(w) inside the unit disc was.
     
    Last edited: Oct 15, 2012
  6. Oct 15, 2012 #5

    Thanks for your example~ Things are getting complicated if no explicit expression of [itex]f(z)[/itex] is given. Is there a way of showing
    [itex]\int_{|w|=1}\frac{f(w)dw}{w^{n+1}}=0[/itex] for [itex]n=0,1,2,...[/itex] ?
    Or it is a false statement ?

    Also for the [itex]|z|>1[/itex] part, can I just specify a circle with radius [itex]R>|z|[/itex], and claim that since [itex]f(z)[/itex] is analytic within the annulus [itex]1<|z|<R[/itex], Cauchy's integral formula could be applied ? Or this theorem could only be applied for cases where the the function is analytic within the whole circle rather than an annulus ?
     
  7. Oct 15, 2012 #6
    How about this example:

    [tex]f(w)=w+\frac{1}{w-1/2}[/tex]

    Then again it's not analytic in the unit circle but analytic outside it. Letting z=1/5, the integral is:

    [tex]\mathop\oint\limits_{|w|=1} \frac{f(w)}{w-1/5}dw\neq 0[/tex]

    So at this point, either the question is phrased wrong or I'm stuck.

    Also. Cauchy's Integral Theorem applies to simply-connected regions like a disc. The annulus you talked about is not simply-connected.
     
  8. Oct 15, 2012 #7
    Maybe try looking at [itex]f(\frac{1}{z})[/itex]?
     
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