# Recurrence relation question

• Dec 31st 2009, 06:22 PM
dhammikai
Recurrence relation question
Let $Z_{n+1}=Z_{n}^{2}+2i$ and $Z_0=0.5+02.i$. Find $Z^{10}$ value and hence deduce the $Z_n$ when $n \longrightarrow \alpha$

here my calculations are
$Z_1=(0.5+0.2i)^2+2i=(0.2+2.2i)$
$Z_2=(0.2+2.2i)^2+2i=(-4.8+2.88i)$
$Z_3=(-4.88+2.88i)^2+2i=(14.7456-25.648i)$
.......

Is that above answer is true?
But I saw in other way like $Z_{n+1}=Z_{n}^{2} + \lambda$ and then $Z^2=r^2e^{i2\theta}$ bt I couldn't understand this..
• Jan 1st 2010, 06:25 AM
HallsofIvy
Quote:

Originally Posted by dhammikai
Let $Z_{n+1}=Z_{n}^{2}+2i$ and $Z_0=0.5+02.i$. Find $Z^{10}$ value and hence deduce the $Z_n$ when $n \longrightarrow \alpha$

here my calculations are
$Z_1=(0.5+0.2i)^2+2i=(0.2+2.2i)$

(0.5+ 0.2i)^2= 0.5^2- 0.2^2+ (2(0.5)(0.2))i= 0.21+ .2i so (0.5+ 0.2i)+ 2i= 0.21+ 2.2i. You seem to have dropped the hundredths place and that might be important.

Quote:

$Z_2=(0.2+2.2i)^2+2i=(-4.8+2.88i)$
$Z_3=(-4.88+2.88i)^2+2i=(14.7456-25.648i)$
.......

Is that above answer is true?
But I saw in other way like $Z_{n+1}=Z_{n}^{2} + \lambda$ and then $Z^2=r^2e^{i2\theta}$ bt I couldn't understand this..
If you think of a+ bi as represnting a point, (a, b) in the "complex plane", then it can also be written in polar coordinates with distance from the origin r and angle $\theta$. Of course, $x= r cos(\theta)$ and $y= r sin(\theta)$ so $a+ bi= r cos(\theta)+ i r sin(\theta))$ But it is also true that [tex]e^{i\theta}= cos(\theta)+ i sin(\theta) so that can be written as $a+ bi= re^{i\theta}$. Then $(a+ bi)^2= r^2e^{i(2\theta)}f$, $(a+ bi)^3= r^3e^{i3\theta}$, etc.

Unfortunately, that "polar form" doesn't work well with addition so you would have to change back each time to add the "2i".

In fact, that recursion formula is the one used in calculating Julia sets and the Mandelbrot set so I don't see any method other than 'brute strength" calculation nor do I see that knowing $z_{10}$ will tell you very much about the limit as n goes to infinity.
• Jan 3rd 2010, 07:17 PM
dhammikai
Thank you very much "HallsofIvy" now I have a clear answer. Thanks