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Let f(z) be analytic with a zero of order k

  1. Jan 8, 2012 #1
    Let f(z) be analytic with a zero of order k ....

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

    Let f(z) be analytic with a zero of order k at z0. Show that f '(z) has a zero of order k-1 at z0.

    2. Relevant equations

    f(z) has a zero of order k at z0 if Ʃcn(z-z0)n (here n goes from k to ∞, and k ≠ 0)

    3. The attempt at a solution

    Well, we of course f '(z) = kck(z-z0)k-1 + (k+1)ck+1(z-z0)k + .... ,

    which looks to me like the definition of a function with a zero of order k at z0. ??????
     
  2. jcsd
  3. Jan 8, 2012 #2

    Dick

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    Re: Let f(z) be analytic with a zero of order k ....

    It looks to me like f'(z) has a zero of order k-1 at z0. Why do you think it's order k?
     
  4. Jan 8, 2012 #3
    Re: Let f(z) be analytic with a zero of order k ....

    If it had order k-1, then ck-1 would not equal zero.

    At least according to http://chanarchive.org/content/63_x/3693517/1267659070786.jpg [Broken]

    A function f(z) analytic in Dr(z0) has a zero of order k at the point z=z0 iff its Taylor series given by f(z) = ∑cn(z-z0)n has c0 = c1 = ... = ck-1 = 0 and ck ≠ 0
     
    Last edited by a moderator: May 5, 2017
  5. Jan 8, 2012 #4

    Dick

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    Re: Let f(z) be analytic with a zero of order k ....

    Perhaps this definition is faulty. It should have read

    f(z) has a zero of order k at z0 if Ʃcn(z-z0)n (here n goes from k to ∞, and [itex]c_k[/itex] ≠ 0)

    The order of a zero is the index of the smallest nonzero coefficient in the power series expansion.
     
  6. Jan 8, 2012 #5

    Dick

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    Re: Let f(z) be analytic with a zero of order k ....

    Ah. In f '(z) = kck(z-z0)k-1 + (k+1)ck+1(z-z0)k + ...., "c_(k-1)" is the coefficient of (x-x0)^(k-1). That's k*c_k. When they say c_k, it means different things in f(z) and f'(z). That's an imaginative way of being confused!
     
  7. Jan 8, 2012 #6
    Re: Let f(z) be analytic with a zero of order k ....

    I see what the problem was. lol
     
  8. Jan 10, 2012 #7
    Re: Let f(z) be analytic with a zero of order k ....

    Another question:

    Locate the poles of the following functions and determine their orders.

    (z6 + 1)-1



    z-5sin(z)



    (z2sin(z))-1




    I'm not sure where to start with these. I haven't bought my textbook yet, and Google isn't giving me any good results.
     
  9. Jan 10, 2012 #8

    Dick

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    Re: Let f(z) be analytic with a zero of order k ....

    That's pretty weak. Look up the definition of what a pole is. Now start with the first one 1/(z^6+1). If it's going to have a pole at a point z then z^6+1=0, right? Where can that happen?
     
  10. Jan 10, 2012 #9
    Re: Let f(z) be analytic with a zero of order k ....

    z = i

    for one
     
  11. Jan 10, 2012 #10

    Dick

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    Re: Let f(z) be analytic with a zero of order k ....

    Ok. Yes, that's one. You are supposed to be doing complex analysis here. The question of where x^6+1=0 isn't complex analysis. It's complex arithmetic. You should have covered this before. It's de Moivre. What are all six of the solutions?
     
    Last edited: Jan 10, 2012
  12. Jan 10, 2012 #11
    Re: Let f(z) be analytic with a zero of order k ....

    We never learned that, but according to Wikipedia .....

    (cos(x) + i sin(x))n = cos(nx) + i sin(nx).

    Not sure exactly how to invoke it.

    z6 = r6ei(6ø) and e = -1, so ........

    Is that on the right track?
     
  13. Jan 10, 2012 #12

    Dick

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    Re: Let f(z) be analytic with a zero of order k ....

    I would complain to your teachers they never taught you that, if you dare. I wouldn't. I think they probably did and you forgot. You already know i works. That's because i=exp(i*pi/2) and i^6=exp(i*pi/2)^2=exp(6*i*pi/2)=exp(3*i*pi)=(-1). Doesn't this ring kind of a bell that angles might have something to do with this? Hint: exp(i*pi/6) also works.
     
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