Use the Definition of a Limit to Find a Complex Limit (z->i)

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Homework Help Overview

The discussion revolves around finding the limit of the expression [z^2+(1+i)z+2] as z approaches i, utilizing the epsilon-delta definition of a limit in the context of complex analysis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the transformation of the expression into a form suitable for applying the epsilon-delta definition. There are attempts to factor the expression and questions about the validity of long division with complex numbers. Some suggest substituting variables to simplify the limit evaluation.

Discussion Status

There is ongoing exploration of different methods to approach the limit, including variable substitution and factoring. Participants are questioning their understanding of bounds in the context of complex numbers and discussing the implications of their findings. No explicit consensus has been reached, but various productive directions have been suggested.

Contextual Notes

Participants express concerns about the lack of examples and resources for understanding complex limits, indicating a need for further clarification on applying the epsilon-delta definition in this context.

MelissaJL
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Homework Statement


The lim(z->i) of [z^2+(1+i)z+2] using the epsilon-delta proof.

Homework Equations


z=x+iy
Triangle Inequality: |z+w|<or=|z|+|w|

The Attempt at a Solution


For every epsilon>0, there exists a delta>0 such that
|(z^2+(1+i)z+2)-(i)|<epsilon whenever 0<|z-i|<delta

I'm not sure how to change z^2+(1+i)z+2 into z-i. Or am I approaching the situation wrong?
 
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MelissaJL said:

Homework Statement


The lim(z->i) of [z^2+(1+i)z+2] using the epsilon-delta proof.

Homework Equations


z=x+iy
Triangle Inequality: |z+w|<or=|z|+|w|

The Attempt at a Solution


For every epsilon>0, there exists a delta>0 such that
|(z^2+(1+i)z+2)-(i)|<epsilon whenever 0<|z-i|<delta

I'm not sure how to change z^2+(1+i)z+2 into z-i. Or am I approaching the situation wrong?
I think you're on the right track. z2 + (1 + i)z + 2 - i factors into (z - i)(z + 1 + 2i). I came up with this by long division.
 
When I try the long division I get a remainder of i. Is that okay? I haven't ever done long division with complex numbers.
 
MelissaJL said:
When I try the long division I get a remainder of i. Is that okay? I haven't ever done long division with complex numbers.
Are you dividing ##z^2+(1+i)z+2## by ##z-i##? Mark divided ##z^2+(1+i)z+2-i## by ##z-i##.
 
MelissaJL said:

Homework Statement


The lim(z->i) of [z^2+(1+i)z+2] using the epsilon-delta proof.

Homework Equations


z=x+iy
Triangle Inequality: |z+w|<or=|z|+|w|

The Attempt at a Solution


For every epsilon>0, there exists a delta>0 such that
|(z^2+(1+i)z+2)-(i)|<epsilon whenever 0<|z-i|<delta

I'm not sure how to change z^2+(1+i)z+2 into z-i. Or am I approaching the situation wrong?

Substitute ##z = i + t## into the original expression ##f(z) = z^2 + (1+i) z + 2## and expand it all out to get a new, equivalent, expression ##q(t)##. Now take ##t \to 0##. Believe me when I tell you it is much easier that way.
 
Ahh I see what I did wrong with the long division. I definitely forgot that -i in there. So now I have:
|(z+i)(z+1+2i)|<epsilon for 0<|z-i|<delta
|z+i||z+1+2i|<epsilon but what can I do with z+1+2i to isolate z+1?
 
MelissaJL said:
Ahh I see what I did wrong with the long division. I definitely forgot that -i in there. So now I have:
|(z+i)(z+1+2i)|<epsilon for 0<|z-i|<delta
|z+i||z+1+2i|<epsilon but what can I do with z+1+2i to isolate z+1?
No. You should have (z - i) as one of the factors, not (z + i). For the other factor, you'll need to establish some bounds on it.
 
Oh sorry that was a typo. I'm very dyslexic. On my paper I have it written down as |z-i||z+1+2i|<epsilon
 
What do you mean by bounds on it?
 
  • #10
trying the modulus wouldn't do anything for me right now since I still have that pesky z there in z+1+2i, can I expand z out as z=a+ib?
 
  • #11
You might want to review how you did this kind of problem when you were working with real functions. How did you find a bound when doing those limits?
 
  • #12
MelissaJL said:
What do you mean by bounds on it?
You want to establish the largest and smallest values it (|z + 1 + 2i|) can have.
 
  • #13
MelissaJL said:
trying the modulus wouldn't do anything for me right now since I still have that pesky z there in z+1+2i, can I expand z out as z=a+ib?
That's no help because you would be trading one unknown, z, for two unknowns, a and b.
 
  • #14
Oh like making an upper or lower bound on it by taking one of the factors to be less than 1 then plugging it back into the other inequality??
 
  • #15
MelissaJL said:
Oh like making an upper or lower bound on it by taking one of the factors to be less than 1 then plugging it back into the other inequality??
Something like that. vela's recommendation to review some similar limits with real-valued function is good advice.

BTW, this thread is probably a better fit in the Calc section, so I moved it.
 
  • #16
Oh okay, thank you. So if I want to make an upper bound I first want to take 0<|z+1+2i|<1 but by definition of absolute value it turns into -1<z+1+2i<1 and if I want to take this and make it z-i I have to add -i-1-2i to each side which gives -i-2-2i<z-i<-i-2i. So |z-i|<-i-2i
 
  • #17
MelissaJL said:
Oh okay, thank you. So if I want to make an upper bound I first want to take 0<|z+1+2i|<1 but by definition of absolute value it turns into -1<z+1+2i<1
No, it doesn't work this way with complex numbers, since you can't compare complex numbers using < or >. For instance, if |z| < 1, all that means is that z is a complex number inside the unit circle with center at the origin.
MelissaJL said:
and if I want to take this and make it z-i I have to add -i-1-2i to each side which gives -i-2-2i<z-i<-i-2i. So |z-i|<-i-2i
 
  • #18
MelissaJL said:
When I try the long division I get a remainder of i. Is that okay? I haven't ever done long division with complex numbers.

Why don't you try out the suggestion I made in Post #5?
 
  • #19
Ray Vickson said:
Why don't you try out the suggestion I made in Post #5?
Ray, I don't see how this is helpful. Melissa already has the limit (which is i), and now she needs to prove that this is the limit using a delta-epsilon proof.
 
  • #20
My professor wants us to follow the definition the way he defined it in class but never gives us any examples. How do you put bounds on complex numbers then? Does anyone have any good resources for Complex Analysis? All I have for a textbook is that old Ian Stewart and David Tall's Complex Analysis: Hitchhikers guide to the plane. There's no examples in it either and the practice questions don't have answers. I never know if I'm doing anything right and clearly I'm not.
 
  • #21
Mark44 said:
Ray, I don't see how this is helpful. Melissa already has the limit (which is i), and now she needs to prove that this is the limit using a delta-epsilon proof.

It's a lot neater and cleaner in the new representation---or at least, I think it is. What I really want to get across to Melissa is that sometimes it is simpler and more "insightful" to change variables in some way. I personally always find it easier to do epsilon-delta on polynomials about 0 than about some non-zero point, but maybe Melissa won't.
 
  • #22
At this point I could just really use some good resources for me to be an autodidact about this course. Any online books, Youtube users, websites with good solid examples and explanations.
 
  • #23
If I were to make a new thread later on finding resources, what category should I put it under?
 
  • #24
MelissaJL said:
If I were to make a new thread later on finding resources, what category should I put it under?
I think here (https://www.physicsforums.com/forums/math-learning-materials.178/ ) would be a good choice.
 
Last edited by a moderator:
  • #25
MelissaJL said:
How do you put bounds on complex numbers then?
You can put a bound on the distance between complex numbers. You can't say z>w, but it's perfectly fine to say |z|>|w| because the modulus of a complex number is a real number.

Write |z+1+2i| = |(z-i) + (1+3i)| and then use the triangle inequality.
 

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