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Math Help - simple pole

  1. #1
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    simple pole

    Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:

     f(z) = \frac{cos(z)}{1+z^2} + 4z

    I dont get it sighs.. please help me
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  2. #2
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    Quote Originally Posted by Dgphru View Post
    Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:

     f(z) = \frac{cos(z)}{1+z^2} + 4z

    I dont get it sighs.. please help me
    Poles occur where the denominator is 0 + 0i.


    So if 1 + z^2 = 0

    z^2 = -1

    z = \pm i.


    So there are simple poles at z = i and z = -i.
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  3. #3
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    More generally, simple poles (poles of order 1) occur where the Laurent series for a function has a " (z- a)^{-1}" term but no other negative exponents. (The order of a pole is the "highest" negative exponent in the Laurent series expansion.)

    Here, the function is \frac{cos(z)}{z^2+ 1}+ 4z= \frac{cos(z)}{(z- i)(z+ i)}+ 4z.

    cos(z) is an analytic function for all z so its "Laurent" series is actually its Taylor series about that point and has no negative exponents. \frac{1}{z+ i} is analytic everywhere except at z= -i, so it has a Taylor series expansion about z= i and therefore so does the product \frac{cos(z)}{z+i}. Multiplying that Taylor series by \frac{1}{z- i} introduces a term with (z- i)^{-1}. Finally, just adding "4z" to the series does not change that. This function has a pole of order 1 at z= i.

    For z= -i, just swap "i" and "-i".
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