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Math Help - radius of convergence question..(complex function)

  1. #1
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    radius of convergence question..(complex function)

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

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

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

    where is my mistake?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i was by my teacher that the radius of convergence
    is what smaller then the number which makes the denominator 0.
    if
    f(x)=\frac{1}{1-z}
    then
    the radius is 1 and
    because 1-1=0
    so
    it is analitical on
    |z|<1

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

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

    where is my mistake?
    Your logic is a bit messy... Let's clean this up a bit.

    Remember that for a geometric series

    S_n = a + ar + ar^2 + ar^3 + \dots = \sum_{n = 0}^{\infty}ar^n, we can write this in closed form as

    S_n = \frac{a}{1 - r} when |r| < 1.


    If you have f(z) = \frac{1}{1 - z}, then we can say a = 1, r = z.

    So f(z) = 1 + z + z^2 + z^3 + \dots = \sum_{n = 0}^{\infty}z^n.

    This is convergent for |r| < 1, so |z| < 1.

    The radius of convergence is 1.


    Can you follow some similar logic for Questions 2 and 3?
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    when i apply your word there i get the same answer as before
    could you do the others too?
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    No, you have a try.

    The second is almost as easy as the first, simply apply a transformation so that it is of the form \frac{a}{1 - r}. At the moment the denominator is r - 1... What can you do to turn it into 1 - r?

    On second viewing, I get the same answer as you. The answer given is wrong, since

    f(z) = \frac{-2}{z - 1} = \frac{-2}{-(1 - z)} = \frac{2}{1 - z}.

    So a = 2 and r = z.


    Therefore f(z) = 2 + 2z + 2z^2 + 2z^3 + \dots = \sum_{n = 0}^{\infty}2z^n, which is convergent for |r| < 1. So |z| < 1. The radius of convergence is 1.
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  5. #5
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    f(x)=\frac{-2}{z-1}
    f(x)=\frac{1}{\frac{-z}{2}+\frac{1}{2}}
    i cant getto tht form that is the best i got
    ??
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  6. #6
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    f(x)=\frac{-2}{z-1}
    f(x)=\frac{1}{\frac{-z}{2}+\frac{1}{2}}
    i cant got to that form, that is the best i got
    ??
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  7. #7
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    Quote Originally Posted by transgalactic View Post
    f(x)=\frac{-2}{z-1}
    f(x)=\frac{1}{\frac{-z}{2}+\frac{1}{2}}
    i cant got to that form, that is the best i got
    ??
    See the edit on my previous post for Question 2.

    For Question 3 you have

    f(z) = \frac{3}{z + 2} = \frac{3}{1 + z + 1} = \frac{3}{1 - (-z - 1)}.

    What do you think a and r are?
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  8. #8
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    but for question two the answer is|z|>1 not
    |z|<1

    your answer contradicts the answer of the book
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  9. #9
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    the original says the develop exaple2+example3 by 1<|z|<2
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  10. #10
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    Like I just told you, either your book's answer is wrong or you have copied the question down wrongly.

    For Question 3, I have already put the expression into \frac{a}{1 - r} form for you. Surely you can read off a and r, and be able to solve |r| < 1 for z...
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