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Math Help - Poles or Removable Singularities

  1. #1
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    Poles or Removable Singularities

    hi,

    Given the equation below, it a pole of order 5 when z = 0

    f(z) = sin(z)/z^6


    Question--> I can't seem to get how to achieve the pole. The answer was given by the prof so i assume it's right.

    Why is it not RS when we apply limit z => 0 ? isn't the whole equation set out to be 0 , then using L'hopital rule to find the answer?


    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Chris0724 View Post
    hi,

    Given the equation below, it a pole of order 5 when z = 0

    f(z) = sin(z)/z^6


    Question--> I can't seem to get how to achieve the pole. The answer was given by the prof so i assume it's right.

    Why is it not RS when we apply limit z => 0 ? isn't the whole equation set out to be 0 , then using L'hopital rule to find the answer?


    Thanks
    Because \frac{\sin(z)}{z^6} \approx z^{-5} near z=0

    CB
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  3. #3
    MHF Contributor

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    And that is because \lim_{x\to 0}\frac{sin(x)}{x}= 1.
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  4. #4
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    Quote Originally Posted by Chris0724 View Post
    hi,

    Given the equation below, it a pole of order 5 when z = 0

    f(z) = sin(z)/z^6


    Question--> I can't seem to get how to achieve the pole. The answer was given by the prof so i assume it's right.

    Why is it not RS when we apply limit z => 0 ? isn't the whole equation set out to be 0 , then using L'hopital rule to find the answer?


    Thanks
    Also, just do long division to calculate the Laurent series:

    \frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots}{z^6}=\frac{1  }{z^5}-\frac{1}{6z^3}+\cdots

    The answer tells you immediately it's a pole of order five.
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