• Support PF! Buy your school textbooks, materials and every day products Here!

Checking Some Complex Limits

  • Thread starter e(ho0n3
  • Start date
  • #1
1,357
0
[SOLVED] Checking Some Complex Limits

Homework Statement
Find the limit of each function at the given point or explain why it does not exist.

(a) f(z) = (1 - Im z)-1 at 8 + i
(b) f(z) = (z - 2) log |z - 2| at 2

The attempt at a solution
(a) f(z) is a real valued function and it seems to me that it approaches infinity as z approaches 8 + i. The book states the limit doesn't exist. I don't get it.

(b) This one is also real valued. Can I safely apply l'Hospital's rule? I'm worried because of log |z - 2|. I know |z - 2| is not differentiable at 2 but since I'm taking a limit, I need not worry right? I get that the limit is 0. Is there another way to evaluate the limit without l'Hospital's rule or using power series?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Saying that the limit is infinity is just saying that it doesn't exist in a certain way. b) is not real valued. But if you consider the absolute value of f(z) you should be able to show that it converges to zero. That |z-2| is not differentiable at zero is not a problem, yes, because you are taking a limit. If |f(z)|->0 then f(z)->0.
 
  • #3
1,357
0
Saying that the limit is infinity is just saying that it doesn't exist in a certain way.
Hmm...I was thinking that a limit doesn't exist in the sense of say cos x as x -> infinity.

b) is not real valued.
Right. I ignored the (z - 2). Sorry

But if you consider the absolute value of f(z) you should be able to show that it converges to zero. That |z-2| is not differentiable at zero is not a problem, yes, because you are taking a limit. If |f(z)|->0 then f(z)->0.
According to my book, it says that if a sequence zn converges then so does |zn|, but that the converse is generally false. I would imagine that this also holds for functions so your last statement is necessarily true.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
It's certainly false that |f(z)|->L implies f(z)->L. It is not false if L is zero. Think about it.
 
  • #5
1,357
0
I got it know. Thank you for the insight.
 

Related Threads for: Checking Some Complex Limits

  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
817
  • Last Post
Replies
6
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
2
Views
348
Replies
1
Views
1K
Replies
3
Views
685
Top