Can Complex Limits Involve Infinity or Non-Existence?

[e(ho0n3]

[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 - I am 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?

 

[Dick]

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.

 

[e(ho0n3]

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 z_n converges then so does |z_n|, but that the converse is generally false. I would imagine that this also holds for functions so your last statement is necessarily true.

 

[Dick]

It's certainly false that |f(z)|->L implies f(z)->L. It is not false if L is zero. Think about it.

 

[e(ho0n3]

I got it know. Thank you for the insight.