Complex Convergence with Usual Norm

[ChemEng1]

Homework Statement

Determine whether the following sequence {xn} converges in ℂunder the usual norm.
$x_{n}=n(e^{\frac{2i\pi}{n}}-1)$

Homework Equations

$e^{i\pi}=cos(x)+isin(x)$
ε, $\delta$ Definition of convergence

The Attempt at a Solution

I would like some verification that this response answers the problem statement. I have not worked with complex numbers since high school algebra. So this attempt may be very wrong.

The professor stated that for complex number to converge, both the real and imaginary parts have to converge.

$x_{n}=n(cos(\frac{2\pi}{n})+isin(\frac{2\pi}{n})-1)$, by substitution with Euler's formula.

$x_{n}=ncos(\frac{2\pi}{n})-n+isin(\frac{2\pi}{n})n$. This appears to converge to 0+0i as n→∞.

Real Part:$Re(x_{n})=ncos(\frac{2\pi}{n})-n=>n(cos(\frac{2\pi}{n})-1)=>n*0=0$

Imaginary Part:$Im(x_{n})=sin(\frac{2\pi}{n})n=>0*n=0$

Since both the real and imaginary parts of $x_{n}$ converge, then $x_{n}$ converges.

Is this close? Did I miss anything?

Thanks in advance,
Scott

 

[jbunniii]

The problem is that n is going to infinity at the same time as (cos(2*pi/n) - 1) and sin(2*pi/n) are going to zero. So you get something that looks like infinity * 0, an indeterminate form. This means that the sequence could either converge or diverge. You have to do a bit more work to find out which is the case.

Let's start with the imaginary part. You have sin(2*pi/n) * n. Multiplication by n is the same as division by 1/n. What can you say about

$$\frac{\sin(2\pi/n)}{1/n}$$

as $n \rightarrow \infty$?

 

[ChemEng1]

So you get something that looks like infinity * 0, an indeterminate form.

In a l'Hôpital's rule sense? We honestly did not cover that in my previous analysis course (so I don't remember any of the rigor behind the rule), but I remember from calculus that you take derivative of numerator and denominator until you no longer have an indeterminate form.

What can you say about

$$\frac{\sin(2\pi/n)}{1/n}$$

as $n \rightarrow \infty$?

$\frac{d[sin(\frac{2\pi}{n})]}{d[\frac{1}{n}]}\Rightarrow\frac{cos(\frac{2\pi}{n})(-\frac{2\pi}{n^{2}})}{-\frac{1}{n^{2}}}\Rightarrow2\pi cos(\frac{2\pi}{n})\Rightarrow 2\pi$

Then you repeat the same approach for the real portion which has a similar indeterminate form. And goes to 0.

So {xn} converges to $2i\pi$ then?

 

[Dick]

In a l'Hôpital's rule sense? We honestly did not cover that in my previous analysis course (so I don't remember any of the rigor behind the rule), but I remember from calculus that you take derivative of numerator and denominator until you no longer have an indeterminate form.



$\frac{d[sin(\frac{2\pi}{n})]}{d[\frac{1}{n}]}\Rightarrow\frac{cos(\frac{2\pi}{n})(-\frac{2\pi}{n^{2}})}{-\frac{1}{n^{2}}}\Rightarrow2\pi cos(\frac{2\pi}{n})\Rightarrow 2\pi$

Then you repeat the same approach for the real portion which has a similar indeterminate form. And goes to 0.

So {xn} converges to $2i\pi$ then?

Yes, it does. You guess this would be true using the taylor series of exp(2*pi*i/n).

 

[ChemEng1]

Thank you both for pointing me in the right direction!