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Math Help - limit of sequences in complex plane

  1. #1
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    limit of sequences in complex plane

    Zn = ((1+i)/squareRoot(3))^n


    Zn = n(1/i)^n


    Please Help...Not sure how to do this...
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  2. #2
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    Quote Originally Posted by jzellt View Post
    Zn = ((1+i)/squareRoot(3))^n


    Zn = n(1/i)^n


    Please Help...Not sure how to do this...
    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?
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  3. #3
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    Sorry...

    Find the limit as n approaches infinity of each sequence that converges. If the sequence diverges, explain why.
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  4. #4
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    Quote Originally Posted by jzellt View Post
    Zn = ((1+i)/squareRoot(3))^n


    Zn = n(1/i)^n


    Please Help...Not sure how to do this...
    \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?
    Last edited by Prove It; February 22nd 2010 at 10:48 PM.
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  5. #5
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    Quote Originally Posted by jzellt View Post
    Zn = ((1+i)/squareRoot(3))^n


    Zn = n(1/i)^n


    Please Help...Not sure how to do this...
    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?
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  6. #6
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    Quote Originally Posted by Prove It View Post
    \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
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    1+i=\sqrt{2}\; \left[\cos(\pi/4)+i \sin(\pi/4)\right]

    CB
    Yes it is, thanks.

    Will edit.
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