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Find limit of complex function

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Let z = x + iy and let f(z) = 3xy + i(x - y2). Find limz→3 + 2i f(z).

    2. Relevant equations

    The definition of a limit.

    3. The attempt at a solution

    I did f(3 + 2i) = 18 - i

    It seems pretty clear that it is a continuous function, but I can't prove it.

    So I tried using the definition of a limit of a function:

    Let ε be a strictly positive real number.

    Take |x + iy - 3 -2i| ≤ |x - 3| + |y - 2| < δ

    Then, |3xy + i(x - y2) - 18 + i| ≤ |3xy - 18| + |1 + x - y2|

    Clearly I need to get to some multiple of |x - 3| + |y - 2| which would give me a relation between δ and ε.
    The problem is I can get there. I tried adding and subtracting some terms in the absolute values, but I have trouble getting rid of the y2 and the 3xy.

    Any help would be awesome!
     
  2. jcsd
  3. Sep 28, 2012 #2
    |x - 3| + |y - 2| < δ implies |x - 3| < δ and |y - 2| < δ, which mean x in (3 - δ, 3 + δ) and y in (2 - δ, 2 + δ). With these you should be able to estimate the upper bound of |3xy - 18| + |1 + x - y2|.
     
  4. Sep 28, 2012 #3
    Perhaps a more intuitive way of looking at it is to let [itex]z = 3+2i + \delta z [/itex] and [itex]\delta z = \epsilon e^{i \phi} [/itex], then take limit [itex]\epsilon \rightarrow 0 [/itex] and check that end result does not depend on [itex]\phi [/itex].
     
  5. Sep 28, 2012 #4
    Thanks, those both make sense, but I'm going to try and finish it off with the original ε,δ proof.

    So here's what I get:

    We have |x - 3| < δ and |y - 2| < δ, therefore x in (3 - δ, 3 + δ) and y in (2 - δ,2 + δ)
    so |x| < 3 + δ and |y| < 2 + δ

    hence 3|x||y| + 18 < 36 + 15δ + 3δ2
    so |3xy - 18| ≤ 3|x||y| + 18 < 36 + 15δ +3δ2 (1)

    Also |y|2 < 4 + 4δ + δ2
    Thus 1 + |x| + |y|2 < 8 + 5δ + δ2
    So |1 + x - y2| ≤ 1 + |x| + |y|2 < 8 + 5δ + δ2 (2)

    If we sum (1) and (2) we get:

    |3xy - 18| + |1 + x - y2| < 44 + 20δ + 4δ2 which we want to be less or equal to ε.

    ie, 44 + 20δ + 4δ2 ≤ ε.

    But then how do I get a specific δ? Since it's a quadratic equation it will have two different values for δ.

    Tell me if you see anything wrong with my logic
     
  6. Sep 28, 2012 #5
    You should not expand |3xy - 18| into |x||y| + 18. If do expand, then you have a positive constant (18), which, of course, can never be less than any arbitrary ε no matter what positive δ you choose. Instead, prove that |3xy - 18| < |3(3 + δ)(2 + δ) - 18| = 15δ + 3δ2.

    Ditto for the other term.
     
  7. Sep 30, 2012 #6
    Ok. But if I do that won't I still get a quadratic equation relating δ and ε? So how do I get a specific δ?
     
  8. Sep 30, 2012 #7
    You should get δ(a + bδ). Then you can always choose δ = min(1, ε/(a + b)).
     
  9. Sep 30, 2012 #8

    Mute

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    Homework Helper

    Are you sure? This function doesn't satisfy the Cauchy-Riemann equations. How are you sure there is not some path you can take such that the limit is not 18-i? (It may turn out that for this particular point the limit is path independent, but the fact this function doesn't satisfy the Cauchy-Riemann equations implies that there is some point for which the limit ##z \rightarrow z_0## is path-dependent)
     
  10. Sep 30, 2012 #9
    You are confusing continuity with holomorphicity.
     
  11. Sep 30, 2012 #10

    Mute

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    Homework Helper

    I am pointing out the failure of the CR equations as a warning that it signals the possibility that the function might not be continuous at the point. If the function cleared the CR equations, then the OP would know right away that the function is continuous an epsilon-delta limit will work. In the present case, the OP can only try and and see if it works, so I would hesitate to say it is "clear" that the function is complex-continuous.

    However, perhaps you are right that I misspoke saying the failure of CR implies there was a point for which the limits weren't path independent; I should have merely said it was possible.
     
  12. Sep 30, 2012 #11
    The CR conditions have no implications w.r.t. continuity. A function may not be holomorphic anywhere, yet be continuous everywhere. Every complex function whose real and imaginary parts are random continuous functions of two arguments is continuous. Why is it so? Because the sum of two continuous complex functions is continuous.

    Holomorphicity is a much, much stronger condition than continuity.
     
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