Describing Set of Complex Numbers for which This Converges

[Newtime]

Homework Statement

Describe the set of all $$z \in \mathbb{C}$$ such that the series $$\sum_{n=1}^{\infty} (1-z^2)^n$$ converges

Homework Equations

Basic analytic techniques.

The Attempt at a Solution

This is from a graduate complex analysis class, and I just have a feeling my answer is too obvious to be correct...

Basically, I used the necessary condition that the the sequence given by $$a_n = |(1-z^2)^n|$$ must tend to 0 as n tends to infinity. In other words, we know the modulus of the terms must go to zero. Then, viewing the series as a geometric series, $$|(1-z^2)| < 1$$. From there it was a few simple steps to conclude the set is $$\lbrace z\in \mathbb{C} \: | \: 0<|z|<2 \rbrace$$.

Even as I read over this, it makes sense. But like I said, it's much too simple to be correct...what am I forgetting here?

 

[Dick]

Your 'simple steps' are wrong. Whatever they are. |(1-z^2)|<1 looks ok. 0<|z|<2 doesn't. z=3/2 doesn't converge. z=i doesn't converge. I'd write z in polar form and try that again.

 

[Office_Shredder]

The reasoning described for determining the conditions necessary is missing a detail as well. Just because a_n goes to zero does not mean that the series will converge necessarily. It only works in this case specifically because we have a geometric series (and you can write out an explicit solution for the partial sums to see that it does converge)

 

[Newtime]

Ah of course! A silly mistake...thanks for pointing it out.

 

[Newtime]

Ah of course! A silly mistake...thanks for pointing it out.

Actually, it wasn't a silly mistake. Upon revisiting this problem, for some reason I'm not seeing the answer. Where should I go after $$\vert 1-z^2\vert<1 \leftrightarrow\vert 1- r^2e^{2i\theta} \vert < 1$$? I've tried using the triangle inequality and I've also tried expanding into sine's and cosine's to no avail. I know this must have some relatively simple trick, but I don't know what...a nudge in the right direction would be greatly appreciated.

 

[╔(σ_σ)╝]

Actually, it wasn't a silly mistake. Upon revisiting this problem, for some reason I'm not seeing the answer. Where should I go after $$\vert 1-z^2\vert<1 \equiv \vert 1- r^2e^{2i\theta} \vert < 1$$? I've tried using the triangle inequality and I've also tried expanding into sine's and cosine's to no avail. I know this must have some relatively simple trick, but I don't know what...a nudge in the right direction would be greatly appreciated.

$$-1 < 1- z^{2} < 1$$
$$-2 < -z^{2} < 0$$
$$0 <z^{2} < 2$$
$$0< \left|z^{2} \right| = \left|z \right|^{2} < 2$$

Can you complete it ?

 

[Newtime]

$$-1 < 1- z^{2} < 1$$
$$-2 < -z^{2} < 0$$
$$0 <z^{2} < 2$$
$$0< \left|z^{2} \right| = \left|z \right|^{2} < 2$$

Can you complete it ?

Well your last line implies $$\vert z \vert < \sqrt{2}$$ but how can we get from $$\vert 1-z^2 \vert <1$$ to $$-1 < 1-z^2 <1$$? The modulus of complex numbers and the absolute value of real numbers do not share this property.

 

[Dick]

Well your last line implies $$\vert z \vert < \sqrt{2}$$ but how can we get from $$\vert 1-z^2 \vert <1$$ to $$-1 < 1-z^2 <1$$? The modulus of complex numbers and the absolute value of real numbers do not share this property.

You are quite right about that. Write z=r*exp(i*theta) and try to write out what |1-z^2|<1 means in terms of r and theta.

 

[╔(σ_σ)╝]

Well your last line implies $$\vert z \vert < \sqrt{2}$$ but how can we get from $$\vert 1-z^2 \vert <1$$ to $$-1 < 1-z^2 <1$$? The modulus of complex numbers and the absolute value of real numbers do not share this property.

I need to stop browsing for the day. I am posting nonsense. Please, forgive my incorrect solution.

I do not know what possessed me to make such a claim.

I was even ordering complex numbers .

 

[Newtime]

You are quite right about that. Write z=r*exp(i*theta) and try to write out what |1-z^2|<1 means in terms of r and theta.

Right. So I have $$\vert 1-r^2e^{2i\theta} \vert < 1$$. This is where I get stuck. In real calculus one can easily get rid of those pesky vertical bars, but here, I don't know where to go. Like I said above, I've tried invoking the triangle inequality but I come to something which doesn't help since it is a lower bound. I've also tried expanding the inside of the modulus into sine's and cosine's but that gets messier, then a little nicer, then messier.

 

[Newtime]

I need to stop browsing for the day. I am posting nonsense. Please, forgive my incorrect solution.

I do not know what possessed me to make such a claim.

I was even ordering complex numbers .

Don't worry about it; I appreciate your help. And after all, it's almost the same error I made to begin with!

 

[Dick]

Right. So I have $$\vert 1-r^2e^{2i\theta} \vert < 1$$. This is where I get stuck. In real calculus one can easily get rid of those pesky vertical bars, but here, I don't know where to go. Like I said above, I've tried invoking the triangle inequality but I come to something which doesn't help since it is a lower bound. I've also tried expanding the inside of the modulus into sine's and cosine's but that gets messier, then a little nicer, then messier.

Yes, expand the inside in terms of sin's and cos's. You want to use that if a and b are real then |a+bi|=sqrt(a^2+b^2).

 

[Newtime]

Having solved the problem (finally), I see now my main (very, very silly) mistake: $$r^2e^{2i\theta} = r^2e^{i(2\theta)} = r^2(cos(2\theta) + isin(2\theta)) \neq r^2(cos^2(\theta) + isin^2(\theta))$$ which is the expansion I had been doing. So, it turns out the set is the interior of a lemniscate whose equation is immediate upon *properly* expanding into sine's and cosine's. Thanks for the help everyone.