# Recurrence relation question

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

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

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

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

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

(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:

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

Is that above answer is true?
But I saw in other way like $\displaystyle Z_{n+1}=Z_{n}^{2} + \lambda$ and then $\displaystyle Z^2=r^2e^{i2\theta}$ 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, $\displaystyle x= r cos(\theta)$ and $\displaystyle y= r sin(\theta)$ so $\displaystyle a+ bi= r cos(\theta)+ i r sin(\theta))$ But it is also true that [tex]e^{i\theta}= cos(\theta)+ i sin(\theta) so that can be written as $\displaystyle a+ bi= re^{i\theta}$. Then $\displaystyle (a+ bi)^2= r^2e^{i(2\theta)}f$, $\displaystyle (a+ bi)^3= r^3e^{i3\theta}$, 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 $\displaystyle z_{10}$ 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