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Homework Help: Convergence help

  1. Feb 1, 2012 #1
    [tex]\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n[/tex]

    z in C.

    This only converges for [itex]z>\frac{-1}{2}[/itex], correct?

    Thanks.
     
  2. jcsd
  3. Feb 1, 2012 #2

    vela

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    Re: Convergence

    No. For one thing, z is complex, so it doesn't make sense to say z > -1/2. In any case, show us how you reached this conclusion.
     
  4. Feb 1, 2012 #3
    Re: Convergence

    Ratio Test
     
  5. Feb 1, 2012 #4

    vela

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    Re: Convergence

    You need to explain what you did in more detail.
     
  6. Feb 1, 2012 #5
    Re: Convergence

    By the ratio test,

    [tex]\left|\frac{z}{z+1}\right|<1[/tex]

    Is this correct now?

    If so, how do I find the appropriate z for which it converges?
     
  7. Feb 1, 2012 #6

    jbunniii

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    Re: Convergence

    Your inequality is true if and only if

    [tex]\left|\frac{z}{z+1}\right|^2 < 1[/tex]

    You can rearrange this and simplify to find a simple inequality for z.

    Another way: your inequality is true if and only if

    |z| < |z + 1|

    which can be written more suggestively as

    |z - 0| < |z + 1|

    What does this mean geometrically?
     
  8. Feb 1, 2012 #7
    Re: Convergence

    Why did you square it?
     
  9. Feb 1, 2012 #8

    jbunniii

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    Re: Convergence

    Because you can then simplify [itex]|z + 1|^2[/itex]. If you're not sure how, try writing it as z + 1 times its complex conjugate.
     
  10. Feb 1, 2012 #9
    Re: Convergence

    I don't understand what you mean.

    z + 1 times it conjugate is

    [tex]z^2 - 1[/tex]
     
  11. Feb 1, 2012 #10

    jbunniii

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    Re: Convergence

    No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.
     
  12. Feb 1, 2012 #11
    Re: Convergence

    So am I supposed to get [itex]\bar{z}> -1[/itex]??
     
  13. Feb 1, 2012 #12

    jbunniii

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    Re: Convergence

    No. That doesn't even make any sense, as [itex]\bar{z}[/itex] is a complex number.

    [tex]|z+1|^2 = (z+1)(\overline{z + 1}) = (z + 1)(\bar{z} + 1) = z\bar{z} + z + \bar{z} + 1 = |z|^2 + z + \bar{z} + 1[/tex]

    So what does your inequality become?

    Hint: you can simplify [itex]z + \bar{z}[/itex].
     
  14. Feb 1, 2012 #13
    Re: Convergence

    I defined [itex]z = x + yi[/itex] and then did what you said. But we had [itex]|z| < |z+1|[/itex] That is how I obtained my solution.
     
  15. Feb 1, 2012 #14

    jbunniii

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    Re: Convergence

    But you didn't obtain a correct solution, or at least you haven't posted it here.

    OK, forget the squaring for now and let's focus on this inequality:

    |z| < |z + 1|

    What does this mean geometrically?

    I will write it more suggestively as follows:

    |z - 0| < |z - (-1)|

    What values of z satisfy this inequality? It's a certain region of the complex plane. What region?

    This kind of thing is often easier to answer if you first determine what region is specified by the EQUALITY:

    |z - 0| = |z - (-1)|
     
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