# limit of sequences in complex plane

• Feb 22nd 2010, 09:09 PM
jzellt
limit of sequences in complex plane
Zn = ((1+i)/squareRoot(3))^n

Zn = n(1/i)^n

• Feb 22nd 2010, 09:14 PM
Prove It
Quote:

Originally Posted by jzellt
Zn = ((1+i)/squareRoot(3))^n

Zn = n(1/i)^n

What are you trying to do? I assume from the title, that you are taking limits, but the limit as n approaches what? And are we finding limits of the sequences or are we summing them?
• Feb 22nd 2010, 09:28 PM
jzellt
Sorry...

Find the limit as n approaches infinity of each sequence that converges. If the sequence diverges, explain why.
• Feb 22nd 2010, 11:08 PM
Prove It
Quote:

Originally Posted by jzellt
Zn = ((1+i)/squareRoot(3))^n

Zn = n(1/i)^n

$\lim_{n \to \infty}\left(\frac{1 + i}{\sqrt{3}}\right)^n = \lim_{n \to \infty}\left[\frac{\sqrt{2}}{\sqrt{3}}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)\right]^n$

$= \lim_{n \to \infty}\left[\left(\frac{\sqrt{6}}{3}\right)^n\left(\cos{\frac{ n\pi}{4}} + i\sin{\frac{n\pi}{4}}\right)\right]$

$= \lim_{n \to \infty}\frac{(\sqrt{6})^n}{3^n} \cdot \lim_{n \to \infty}\left(\cos{\frac{n\pi}{4}} + i\sin{\frac{n\pi}{4}}\right)$

Can you go from here?
• Feb 22nd 2010, 11:30 PM
Prove It
Quote:

Originally Posted by jzellt
Zn = ((1+i)/squareRoot(3))^n

Zn = n(1/i)^n

$z_n = n\left(\frac{1}{i}\right)^n$

$= n\left(\frac{i}{i^2}\right)^n$

$= n\left(-i\right)^n$

$= n\left[\cos{\left(-\frac{\pi}{2}\right)} + i\sin{\left(-\frac{\pi}{2}\right)}\right]^n$

$= n\left[\cos{\left(-\frac{n\pi}{2}\right)} + i\sin{\left(-\frac{n\pi}{2}\right)}\right]$

So $\lim_{n \to \infty}z_n = \lim_{n \to \infty}n\left[\cos{\left(-\frac{n\pi}{2}\right)} + i\sin{\left(-\frac{n\pi}{2}\right)}\right]$

$= \lim_{n \to \infty}n \cdot \lim_{n \to \infty}\left[\cos{\left(-\frac{n\pi}{2}\right)} + i\sin{\left(-\frac{n\pi}{2}\right)}\right]$.

Can you go from here?
• Feb 22nd 2010, 11:38 PM
CaptainBlack
Quote:

Originally Posted by Prove It
$\lim_{n \to \infty}\left(\frac{1 + i}{\sqrt{3}}\right)^n = \lim_{n \to \infty}\left[\frac{1}{\sqrt{3}}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)\right]^n$

$= \lim_{n \to \infty}\left[\left(\frac{1}{\sqrt{3}}\right)^n\left(\cos{\frac{ n\pi}{4}} + i\sin{\frac{n\pi}{4}}\right)\right]$

$= \lim_{n \to \infty}\frac{1}{(\sqrt{3})^n} \cdot \lim_{n \to \infty}\left(\cos{\frac{n\pi}{4}} + i\sin{\frac{n\pi}{4}}\right)$

Can you go from here?

$1+i=\sqrt{2}\; \left[\cos(\pi/4)+i \sin(\pi/4)\right]$

CB
• Feb 22nd 2010, 11:47 PM
Prove It
Quote:

Originally Posted by CaptainBlack
$1+i=\sqrt{2}\; \left[\cos(\pi/4)+i \sin(\pi/4)\right]$

CB

Yes it is, thanks.

Will edit.