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Radius of convergence question(complex function)

  1. Jul 10, 2010 #1
    i was by my teacher that the radius of convergence
    is what smaller then the number which makes the denominator 0.
    if
    [tex]f(x)=\frac{1}{1-z}[/tex]
    then
    the radius is 1 and
    because 1-1=0
    so
    it is analitical on
    |z|<1

    so if i apply the same logic
    [tex]f(x)=\frac{-2}{z-1}[/tex]
    1 still makes denominator 0
    and
    it is analitical on
    |z|<1
    but the correct answer is
    it is analitical on
    |z|>1

    for
    [tex]f(x)=\frac{3}{z+2}[/tex]
    -2 makes denominator 0
    so |z|<-2 (but its illogical because |z| is a positive numbe)

    where is my mistake?
     
  2. jcsd
  3. Jul 10, 2010 #2

    Gib Z

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    More precisely, your teacher was using the theorem that if it is known a power series diverges for some number P, then it diverges for all complex numbers greater in magnitude than P, |z|>|P|. So for the first example in your post, since the function is not analytic at z=1, the radius of convergence is certainly 1 or smaller, definitely not bigger.
     
  4. Jul 10, 2010 #3
    so why the other example dont work by that logic
    ?
     
  5. Jul 10, 2010 #4

    Gib Z

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    Because you are wrong and the radius of convergence for that is indeed |z|< 1.

    It could never possibly be |z|>1, the radius of convergence is always in the shape of a ball from some center, not an annulus. By the comparison test, if the power series converges for some complex value P, it converges for all |z|<|P|
     
  6. Jul 10, 2010 #5

    vela

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    It depends on how you expand the function actually. For example, you can write

    [tex]\frac{1}{1-z} = 1+z+z^2+\cdots[/tex]

    which converges for |z|<1, but you could also write it as

    [tex]\frac{1}{1-z} = \frac{1}{z(1/z-1)} = -\frac{1}{z}\left(1+\frac{1}{z}+\frac{1}{z^2}+\cdots\right) = -\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3}-\cdots[/tex]

    This series converges for |z|>1.

    You didn't tell us what the actual series were in each example, so talking about a radius of convergence doesn't really make sense. All of the functions are analytic everywhere except at the poles.
     
  7. Jul 10, 2010 #6

    Gib Z

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    That's a Laurent Series, not a Power Series though. The concept of Radius of Convergence is specific to Power series.
     
  8. Jul 10, 2010 #7
    the original says the develop exaple2+example3 by 1<|z|<2
     
  9. Jul 10, 2010 #8

    Gib Z

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    It sound's like then you are being asked to find Laurent Series for those functions valid in that annulus. Radius of Convergence refers only to Power Series.
     
  10. Jul 10, 2010 #9
  11. Jul 10, 2010 #10

    Gib Z

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    Okay then...you should start a new thread with all your working, and this time, with the complete entire original problem. You didn't say anything Laurent series in the original post.
     
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