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Another calculus proof

  1. Mar 14, 2006 #1
    Prove that [tex] \lim_{z \rightarrow -i} 1/z = i [/tex] using the definition of a limit

    ok from teh defition of limit we know that
    [tex] |z+i| < \delta [/tex]

    also we need to show that [tex] |\frac{1}{z} - i| < \epsilon [/tex]
    [tex] |\frac{1}{z} - i| = |\frac{1}{z} + z - z + i| \leq |\frac{1}{z} - z| + |z+i| < |\frac{1}{z} - z| + \delta [/tex]

    stuck here... do i just say the above is less than delta 1 and then pick an epsilon which si the min of delta 1 and delta?

    i have (clarly) forgotten what to do about these kinds of proofs, please help*!
  2. jcsd
  3. Mar 14, 2006 #2


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    Well, this is a pretty silly problem. You obviously don't need to use limits to get the answer. But if you must, your approach won't work, since |z-1/z| isn't approaching zero.

    You want to prove that, given any [itex]\epsilon[/itex]>0, there exists a [itex]\delta[/itex]>0 such that if:




    I'm sure you know this, but what you've written doesn't make it very clear that this is what you're trying to prove. So [itex]\epsilon[/itex] is some number, and you want to find the corresponding [itex]\delta[/itex] that makes the above true. Use:


    Do you see where to go from here?
  4. Mar 14, 2006 #3
    [tex] \frac{|i+z|}{|iz|} < \frac{\delta}{iz} = -\frac{i\delta}{z} [/tex]

    now im not sure ...
  5. Mar 14, 2006 #4


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    The magnitude of a complex number is always real. Once you fix that, you need to pick delta so that this expression alway evaluates to something less than epsilon. (by the way, you used what you're trying to prove in your second step there. Not that there's anything wrong with that, it just shows what a silly problem this is)
    Last edited: Mar 14, 2006
  6. Mar 14, 2006 #5
    so that denominator turns into x^2 + y^2??
  7. Mar 14, 2006 #6


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    OK. Now, given z is in a circle of radius delta around -i, what is the maximum value of the expression?
  8. Mar 14, 2006 #7
    so the magnitude of the numerator is
    [tex] |i+z| = \sqrt{x^2 + (y+1)^2} [/tex] ??

    the denominator is simply [tex] \sqrt{x^2 + y^2} [/tex]
  9. Mar 14, 2006 #8
    so then the max vcalue would be the distance that z is awawy from teh edge of the circumference?

    z + delta - |-i-z| ??
  10. Mar 14, 2006 #9


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    Like I've been saying, you want to calculate the maximum value that |1/z-i| could acheive given that |z+i|<delta. Once you know this, you know how to choose delta, given an epsilon, that guarantees |1/z-i|<epsilon. This is the essence of the epsilon-delta definition of a limit. The maximum value of the expression you found:

    [tex]\frac{\delta}{|z|^2} [/tex]

    can be found geometrically when you consider that z must lie inside a circle of radius delta around the point i. To find the max of this expression, you want to minimize the denominator.
    Last edited: Mar 14, 2006
  11. Mar 14, 2006 #10
    minimize the denominator...
    not sure what that means here.
    to minimize te z we need to make z small by finding z's smallest value in that cirle u describe
    but the smallest value would be when z = -i - delta?j The center loess the radius on the left?
  12. Mar 15, 2006 #11


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    Well, you might have a sign error there, but what's important is the minimum value of |z|2. And do you understand why you need this?
  13. Mar 15, 2006 #12
    always try to simplify the expression:

    1/z = (1*z')/(z*z') = z'/|z^2| = ...
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