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Math Help - Recurrence relation question

  1. #1
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    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..
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  2. #2
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    Quote Originally Posted by dhammikai View Post
    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.

    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 [itex]\theta[/itex]. 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.
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  3. #3
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    Thank you very much "HallsofIvy" now I have a clear answer. Thanks
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