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Limit definition to compute some complex value function

  1. Nov 29, 2013 #1
    So there is something I don't understand in the definition of limit that is applied to some problem
    I have some intuition for like the rigorous limit definition but I don't have full understanding when applied to some problems.


    Use definition 2 to prove lim as z → i of z^2 = -1


    The book answer:
    We must show that for given E > 0 there is a positive number x such that

    |z^2 - (-1)| < E whenever 0 < |z - i| :

    so we express |z^2 - (-1) | in terms of |z - i|:

    z^2 - (-1) = z^2 + 1 = (z - i)(z + i) = (z - i)(z - i + 2i)

    It follows from the properties of absolute value defived in Sec 1.3
    that

    |z^2 - (-1)| = |z - i||z - i + 2i| <= |z - i| (|z - i| + 2)

    Now if |z - i| < x the right hand is less than x(x + 2) so to ensure that it is less than E, we can choose x to be smaller than either of the number E/3 and 1:

    |z - i||(|z - i| + 2) < E/3(1 + 2) = 2



    So Here there is alot of stuff that I don't understand like why did we need to express one value in terms of the other ? I still don't know follow what's going on here if someone could explain please because I want full understanding of these stuff. thank you.
     
  2. jcsd
  3. Nov 29, 2013 #2

    LCKurtz

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    I'm guessing you are OK to here, right? You are trying to make ##z^2-(-1)## small by getting ##z## near ##i##. Now, on the right side you have ##z-i## which you can make small. That will make the right side small if the rest of it ##|z-i+2i|## isn't too big. Well, how big can it be? ##|z+i|\le |z|+|i| = |z|+1##. That could presumably be very large, except for the fact we are making ##z## close to ##i##. Let's say we keep ##z## within ##1## unit of ##i## so ##|z-i|<1##. How big could ##z## be then? Well, ##|z|-|i|\le |z-i|<1## so ##|z|<1+|i|=2##. Putting this together gives ##|z+i|\le 3## so ##|z^2-(-1)|\le|z-i|\cdot 3##. That's where the overestimate of ##3## comes from. Now pick ##\delta=\min\{\epsilon/3,1\}##.
     
    Last edited: Nov 29, 2013
  4. Nov 29, 2013 #3
    Oke good I understand it now thank you it makes perfect sense.
     
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