# Limits in complex numbers and functions

1. Oct 18, 2011

### MATHMAN89

1. The problem statement, all variables and given/known data

1. I'm trying to figure out how to take limits involving i and complex functions f(z)

The first problem is as follows:

lim(n $\rightarrow$ $\infty$ ) n*((1+i)/2))^n

The second is:

lim (z app. 0 ) of $$(sinz/z)(1/z^2)$$

The third is:

lim (z app. e^i*pi/3) of (z-e^i*pi/3) * (z/(z^3 + 1))

2. I have no idea how to get started on these. The first one seems to be divergent as I plugged in a few values for n and the expression kept getting larger.
For the second one, I know from real variabled calculus that lim (x app 0) of sinx/x is 1 but don't know how to use that in this problem.
The third one just seems crazy to me, but I do know that the denominator goes to 0 since we get cos pi + i sin pi = -1 and -1+1=0. We get 0/0 so need to use some tricks!

3. Any help?

Last edited: Oct 18, 2011
2. Oct 18, 2011

### Dick

They are all pretty different problems. For the first one, sure it's divergent. Can you prove it? For the second one take the log and use power series of sin(z) and log(1+z). For the third what's the limit of the denominator as z->exp(i*pi/3)? Give them another try.

3. Oct 18, 2011

### MATHMAN89

Dick,

I had a small typo in the way I wrote question 1. It's supposed to be (1+i)/2 not 1+i/2 inside the parens. That doesn't change the fact that its divergent though, does it?
I was just a bit confused because I plugged it in to wolfram alpha and it told me the answer was 0!

4. Oct 18, 2011

### MATHMAN89

Ooh for the third one I get the den. is 0 since It's cos pi + i*sin pi which is -1. the problem is that z - e ^ ipi/3 goes to 0 as well so it's a 0/0 problem.

From real variables I learned a few ways to deal with that : taking logs of both sides was one of them. I don't see how that would work here though.

5. Oct 18, 2011

### Dick

(1+i)/2 is different from 1+i/2 in an important way. What is that way? Think about real numbers, when is x^n divergent and when is it convergent?

6. Oct 18, 2011

### micromass

Staff Emeritus
The first one is convergent. Try to find the modulus of

$$n((1+i)/2)^n$$

7. Oct 18, 2011

### Dick

Let's try and talk about these one at a time. You just changed the third one as well. Can you factor (z^3-1) over the complex numbers? You probably can figure out the three roots. And take the log of the second one anyway without worrying about the details for now.

8. Oct 18, 2011

### MATHMAN89

Yes, Dick. Sorry for not writing the problems out correctly at first. They're all written correctly now.

I know that the three roots of z^3-1=0 are e^2i*pi/3, e^4i*pi/3, e^2i*pi

9. Oct 18, 2011

### MATHMAN89

Convergent when 0 ≤ x ≤ 1

And if we take abs ((1+i)/2) = sqrt (2)/2 < 1 so it is convergent.

10. Oct 18, 2011

### Dick

Convergent to 0, right? Do you see why that applies to complex numbers as well as reals?

11. Oct 18, 2011

### MATHMAN89

Yes, I see why, since the modulus is less than 1. but we then get a limit of the form infinity * 0.

12. Oct 18, 2011

### Dick

Sorry, I meant z^3+1 not z^3-1. That's what you have in the problem.

13. Oct 18, 2011

### MATHMAN89

I have an idea.

Could I rewrite the first problem as (n^1/n*(1+i)/2)^n and then take the log

14. Oct 18, 2011

### MATHMAN89

In that case it's e^i*pi/3, e^i*pi, e^i*5pi/3
so the first term cancels with one of the factors of the denominator! brilliant.

15. Oct 18, 2011

### MATHMAN89

Yes, Dick. Agreed on that point that ((1+i)/n) converges to 0. However, I still have the problem of the n on the outside.

16. Oct 18, 2011

### Dick

Yeah, I keep forgetting you changed it. Sorry. This is all getting kind of confusing. The usual way to deal with a infinity*0 problem is to use l'Hopital's theorem, right? Complex numbers aren't that different from reals. Pretend the problem is n*(1/2)^n.

17. Oct 18, 2011

### MATHMAN89

Aha, OK. I've got the first one now. Thank you so much.

I dont see how I get log(1+z) for the second one, though.

18. Oct 18, 2011

### Dick

Use power series. What's the power series expansion of sin(z)/z?

19. Oct 19, 2011

### MATHMAN89

sinz/z = 1 -z^2/2 +z^4/4! .... = cos z right?

20. Oct 19, 2011

### MATHMAN89

Still not sure how that gets me at my answer.