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Complex Analysis: Identity Theorem

  1. Mar 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Let f be a function with a power series representation on a disk, say [itex]D(0,1)[/itex]. In each case, use the given information to identify the function. Is it unique?
    (a) [itex]f(1/n)=4[/itex] for n=1,2,[itex]\dots[/itex]
    (b) [itex]f(i/n)=-\frac{1}{n^2}[/itex] for n=1,2,[itex]\dots[/itex]

    A side question:
    Is corollary 1 from my textbook given below just the Uniqueness Theorem?
    2. Relevant equations
    Corollary 1
    Let [itex]g_1[/itex] and [itex]g_2[/itex] be holomorphic on the open connected set [itex]O[/itex], such that [itex]g_1=g_2[/itex] on a set that has an accumulation point within [itex]O[/itex]. Then [itex]g_1=g_2[/itex] throughout [itex]O[/itex].

    3. The attempt at a solution
    (a)
    Consider [itex]g(z)=f(z)-4[/itex]. [itex]g[/itex] has a zero set [itex]\{1/n\}[/itex] that converges to [itex]0[/itex]. Therefore, by, the identity theorem
    $$
    g(z)=0\implies f(z)=4
    $$
    Finally, we note that [itex]f(z)=4[/itex] satisfies the conditions given in the problem statement for part (a). Also, in solving the first part of this problem we've demonstrated the hypothesis for Corollary 1, thereby showing that [itex]f(z)[/itex] is unique.
    (b)
    Consider [itex]g(z)=f(z)-z^2[/itex]. [itex]g[/itex] has a zero set [itex]\{i/n\}[/itex] that converges to [itex]0[/itex]. Therefore, by the Identity Theorem,
    $$
    g(z)=0\implies f(z)=z^2
    $$
    Finally, we note that [itex]z^2[/itex] satisfies the conditions given in the problem statement for part (b). Also, in solving the first part of this problem we've demonstrated the hypothesis for Corollary 1, thereby showing that [itex]f(z)[/itex] is unique.
     
  2. jcsd
  3. Mar 4, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    I'm not sure what you mean with "zero set [...] that converges to 0" - a set does not converge, and what is a zero set?
    You can probably skip some steps you make twice, but the proof is fine.
     
  4. Mar 4, 2015 #3
    A zero sequence would be a more apt descriptor then. So, a zero sequence is a sequence [itex]\{z_k\}[/itex] such that [itex]f(z_k)=0[/itex] for all k and where [itex]\lim_{k\to\infty} z_k\to a[/itex]. In my case a=0.
     
  5. Mar 4, 2015 #4

    mfb

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    2016 Award

    Staff: Mentor

    Ah, okay.
     
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