Taking the limit of a complex function

  • Thread starter mancini0
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  • #1
mancini0
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Homework Statement



Hi guys. I was hoping you could help me find the limit of a complex function. So here goes:

The lim z --> i of [i(z)^3 - 1 ] / (z+i)

The Attempt at a Solution



If z approaches i, then (x,y) approaches (0,1)

Do I let z = x+iy, then expand out the cube and plug in 0's for x's and 1's for y's in one limit?

Or do I do two limits, one letting x go to 0, the other letting y go to one and compare these two limits?

Or do I just plug in i for z right off the bat and expand that out?

Basically, how is it possible to test the limit from every approach in the complex plane without doing some
kind of epsilon / delta proof.




Thank you very much if you respond.
 

Answers and Replies

  • #2
LeonhardEuler
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Did you write out that function correctly, because it seems that it is not an indeterminant form and just approaches 0. Maybe the bottom is supposed to be z-i?
 
  • #3
mancini0
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Well the question actually asks evaluate the limit or explain why it doesn't exist. So perhaps it doesn't exist.
 
  • #4
LeonhardEuler
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It does exist, but it's the quotient of two continuous functions, and the denominator does not approach 0, so that makes it very simple. Do you know anything about the limit of the quotient of continuous functions?
 
  • #5
mancini0
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Thank you very much. Yes, the limit of the quotient of two continuous functions is just the quotient of the limits.

But if I plug in i for z (since the function is continuous about i) I get:
i*i^3 - 1 / (i +i)

= i^4 -1 / 2i

= 0 / 2i = 0.

But if I take the limit as x,y approaches (0,1), where z = x+iy, I get:

lim x,y --> 0,1 of [ i(x+iy)^3 -1 ] / x+i(y+1)

expanding the cube leaves:
ix^3 -x^2y -iy^2x +y^3 -2x^2y -2ixy^2 / x+i(y+1)

From here, if I plug in x = 0 and y =1 simultaneously, i get 1 / 2i
which is not the result I expected, since plugging in i for z originally gave zero.
 
Last edited:
  • #6
LeonhardEuler
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You forgot the -1 at the end in the numerator
 
  • #7
mancini0
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Ahhh, I did. Thank you. But aren't there many ways for x,y to approach 0,1 on the complex plane? Haven't I only checked one of these directions?
 
  • #8
LeonhardEuler
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There are many ways, but I'm assuming you already know that polynomials are continuous on the complex numbers. Proving that polynomials are continuous would require an epsilon-delta proof. As an example of how this is done, look at z2 and prove that it is continuous:
[tex]|z^2-z_{0}^2| = |(z-z_0)^2 + 2zz_0 - 2z_{0}^2|\leq |z-z_0|^2 + 2z_0|z-z_0|[/tex]
In the complex plane what it means for 2 numbers to be close together is that |z-z0| is very small, so this is what we call delta in this context. There are two terms that we need to make sure are small, so let's make sure both contribute less than epsilon/2 by taking
[tex]\delta = min(\frac{1}{2},\frac{\epsilon}{2},\frac{\epsilon}{4z_0})[/tex]
(The half is in there to make sure [itex]\delta^2<\frac{\epsilon}{2}[/itex] when [itex]\epsilon>1[/itex])
so now
[tex]|z^2-z_{0}^2| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon[/tex]
 

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