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Complex Limits

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Compute limit as n-> infinity of

    a)zn = exp(in^2)/(1+in^2)

    b)zn = 1/(n + i)

    2. Relevant equations


    3. The attempt at a solution

    These are 2 examples of a series of questions i have to complete. I can see that i need to calculate the limit for a complex function but i have not come accross this in lectures yet. Could you please point me in the right direction, maybe i need to use a certain rule/law/method?
  2. jcsd
  3. Jan 21, 2008 #2


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    Hint: look at the moduli first, that is, consider
    [tex]\lim_{n \to \infty} |z_n|[/tex]
  4. Jan 21, 2008 #3


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    Finding the limit in the complex numbers is pretty much the same as real numbers. As compuchip said, you need to look at [itex]|z_n|= \sqrt{(x_n)^2+ (y_n)^2}[/itex]. Although it won't give you the limit exactly, you should try calculating a few terms of the sequence to get an idea of what's happening. Have you done that?
  5. Jan 24, 2008 #4
    should i be using the subtitution of exp(ia) = cos a - isin a ? or is there a better method?
  6. Jan 24, 2008 #5
    yeah - i know that it tends to 0 - but im not too sure about what i use for the xn and yn parts and what that eventually gives me - if i can use exp(ia) = cos a - isin a i think i have crcked it?
  7. Jan 25, 2008 #6


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    You can always write it out in real and imaginary parts and notice that
    [tex]\lim_{z \to 0} f(z) = \left( \lim_{z \to 0} \operatorname{Re}(f(z)) \right) + \mathrm{i} \left( \lim_{z \to 0} \operatorname{Im}(f(z)) \right)[/tex].
    Then you can indeed use exp(i a) = cos(a) + i sin(a) for the first one, and use that the real and imaginary parts of a fraction can be determined by writing
    [tex]\frac{\alpha}{z} = \frac{\alpha}{z} \frac{\bar z}{\bar z}[/tex]
    with [itex]\bar z[/itex] the complex conjugate of z.

    So if you think you have a), can you show us your work?
    Last edited: Jan 25, 2008
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