Limits in complex numbers and functions

[MATHMAN89]

Homework Statement

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 [tex](sinz/z)^{(1/z^2)}[/tex]

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?

 

[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.

 

[MATHMAN89]

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.

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 into wolfram alpha and it told me the answer was 0!

 

[MATHMAN89]

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.

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.

 

[Dick]

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 into wolfram alpha and it told me the answer was 0!

(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?

 

[micromass]

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 into wolfram alpha and it told me the answer was 0!

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

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

 

[Dick]

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.

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.

 

[MATHMAN89]

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.

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

 

[MATHMAN89]

(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?

Convergent when 0 ≤ x ≤ 1

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

 

[Dick]

Convergent when 0 ≤ x ≤ 1

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

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

 

[MATHMAN89]

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

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

 

[Dick]

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

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

 

[MATHMAN89]

I have an idea.

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

 

[MATHMAN89]

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

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.

 

[MATHMAN89]

Why? |z^n|=|z|^n. If |z|<1 then |z|^n converges to 0. So so does z^n, yes?

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.

 

[Dick]

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.

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.

 

[MATHMAN89]

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.

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

I don't see how I get log(1+z) for the second one, though.

 

[Dick]

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

I don't see how I get log(1+z) for the second one, though.

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

 

[MATHMAN89]

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

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

 

[MATHMAN89]

Still not sure how that gets me at my answer.

 

[Dick]

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

Noo. Take the series for sin(z) and divide by z. Your factorials are wrong and it isn't cos(z).

 

[Dick]

Still not sure how that gets me at my answer.

Take the log of (sin(z)/z)^(1/z^2). Expand everything in powers. Take the limit as z->0. You'll see how it works if you do it.

 

[MATHMAN89]

Take the log of (sin(z)/z)^(1/z^2). Expand everything in powers. Take the limit as z->0. You'll see how it works if you do it.

I did this but I still get something like e^ infinity*0

so do I use L'hopital's rule here?

 

[Dick]

I did this but I still get something like e^ infinity*0

so do I use L'hopital's rule here?

It's hard for me to tell what you did. Start from log((sin(z)/z)^(1/z^2)).

 

[MATHMAN89]

It's hard for me to tell what you did. Start from log((sin(z)/z)^(1/z^2)).

Hi Dick,

I know how to do this problem in the real numbers, I use L' Hopital's rule twice and eventually get that the limit is e^-1/6.

Do I need to do the problem differently, though, since z = x + iy ?
If so, do I start with assuming x is 0 and then taking the limit as y approaches zero and then assume y is 0 and take the limit as x approaches 0 to find the real and complex parts of my answer?

 

[Dick]

Hi Dick,

I know how to do this problem in the real numbers, I use L' Hopital's rule twice and eventually get that the limit is e^-1/6.

Do I need to do the problem differently, though, since z = x + iy ?
If so, do I start with assuming x is 0 and then taking the limit as y approaches zero and then assume y is 0 and take the limit as x approaches 0 to find the real and complex parts of my answer?

The more 'complex' way to do this is expanding the function in terms of power series in z. Then letting z->0. It would probably be instructive to do it that way too. Of course, you'll get the same answer.

 

[MATHMAN89]

The more 'complex' way to do this is expanding the function in terms of power series in z. Then letting z->0. It would probably be instructive to do it that way too. Of course, you'll get the same answer.

AHA! got it! Thanks

 

[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.

So I am left with
z/((z-e^i*pi)(z-e^5i*pi/3)
How can I evaluate this when I plug in? Just use Euler?

 

[Dick]

So I am left with
z/((z-e^i*pi)(z-e^5i*pi/3)
How can I evaluate this when I plug in? Just use Euler?

Sure. Just use Euler. If it were up to me, I wouldn't really care much if you simplified it or not. But I'm not your teacher.

 

[MATHMAN89]

So I am left with
z/((z-e^i*pi)(z-e^5i*pi/3)
How can I evaluate this when I plug in? Just use Euler?

I think I got it.
I ended up with
(1-i√3)/6
Sound right?? Should be if I did all the calculations right.

 

[Dick]

I think I got it.
I ended up with
(1-i√3)/6
Sound right?? Should be if I did all the calculations right.

That looks like what I got.

 

[MATHMAN89]

Just checked on WOLFRAM. Correct! VICTORY!

 

[MATHMAN89]

I only have one problem left I'm trying to do now, and it's not one that I posted because I thought it would be the easiest.

Can you see https://www.physicsforums.com/showthread.php?p=3567888&posted=1#post3567888
?
I appreciate your help, Dick. You're the best!