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Thread: Need help calculating convergence, series.

  1. #1
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    Need help calculating convergence, series.

    I have an exercise where I'm supposed to find convergence if any and if there are show how it's... constituted. The series:

    $\displaystyle \[
    \displaystyle \sum_{n=0}^{\infty} \left(\frac{z-1}{1-2z}\right)^n
    \]
    $

    I'd like help when it comes to ways of solving it. I solve it as a real problem, getting the zeros on the real axis. Then I use these to find the zeros in the img-plane.

    $\displaystyle \[
    \displaystyle \frac{\left|(\frac{z-1}{1-2z})^{n+1}\right|}{\left|(\frac{z-1}{1-2z})^n\right|}
    \]
    $ = $\displaystyle \[
    \left|(\frac{z-1}{1-2z})^{1}\right|
    \]
    $ = $\displaystyle \[
    \displaystyle \frac{\left|z-1\right|}{\left|1-2z\right|}
    $
    For what z is the fraction smaller than 1?
    $\displaystyle \[
    \displaystyle \frac{\left|z-1\right|}{\left|1-2z\right|} < 1
    \]$

    I get $\displaystyle 0 < z < \frac{2}{3}$ - treating z as a real variable, so centre on x-axis should be $\displaystyle \frac{1}{3}$

    Now to the imaginary part.
    $\displaystyle \[
    \displaystyle \frac{\left|(x+iy)-1\right|}{\left|1-2(x+iy)\right|} < 1
    \]$ which gives $\displaystyle \[
    \displaystyle \frac{\left|(\frac{1}{3}+iy)-1\right|}{\left|1-2(\frac{1}{3}+iy)\right|} < 1
    \]$

    I get $\displaystyle y = \pm \frac{i}{3}$ - which goes nicely with the real values.

    Centre = $\displaystyle \frac{1}{3}$
    Radius = $\displaystyle \frac{1}{3}$

    First on all, is this right? And is there a more solid way of doing this?
    Last edited by liquidFuzz; Sep 27th 2010 at 06:21 AM. Reason: missed a bbcode-tag.
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  2. #2
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    Opalg's Avatar
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    That is the right answer. I would do it this way, using the fact that $\displaystyle |z|^2 = z\overline{z}$, where $\displaystyle \overline{z}$ is the complex conjugate of z. Then $\displaystyle \frac{|z-1|}{|1-2z|} = 1$ is equivalent to $\displaystyle (z-1)(\overline{z}-1) = (1-2z)(1-2\overline{z})$. Multiply that out and rearrange it as $\displaystyle 3z\overline{z}-(z+\overline{z}) = 0$. Now "complete the square" (except that it's not a square, but the method is the same) to write this as $\displaystyle (z-\frac13)(\overline{z}-\frac13) = \frac19$. Finally, take the square root to get $\displaystyle |z-\frac13| = \frac13$, which is the equation of a circle centred at 1/3, with radius 1/3.
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  3. #3
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    Ahh... Conjugate. Your method is so simple that it's out right beautiful!

    Thank you very much!
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