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Math Help - Laurent Series - Complex Numbers

  1. #1
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    Laurent Series - Complex Numbers

    (a) Locate and classify the singularities (giving the order of any poles) of the function
    f(z) = z/(1-e^z)

    (b) Let f(z) = zsinh (1/(z+1))
    (i) Find the laurent series of f about -1, giving teh general term of the series for odd and even powers of (z+1)
    (ii) Write down a punctured open disk D, containing the circle C={z:|z+1|=1} on which f is represented by this series
    (iii) State the nature of the singularity of f at -1
    (iv) Evaluate

    integral zsinh(1/(z+1))dz where C={z:|z+1|=1}
    (c) Find the laurent series about 0 for the function

    f(z)=7z/((2z+1)(z-3)) on the set {z:|z|.3, giving teh general term of the series
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  2. #2
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    Quote Originally Posted by moolimanj View Post
    (a) Locate and classify the singularities (giving the order of any poles) of the function
    f(z) = z/(1-e^z)
    f(z) = - \frac{z}{e^z-1}
    We see that at z=0 we have a possible singulariy.
    Introduce some power series to this problem,
    - \frac{z}{1+z + \frac{z^2}{2!}+\frac{z^3}{3!}+.... - 1} = \frac{z}{z\left( 1 +\frac{z}{2!}+\frac{z^2}{3!}+... \right)} = \frac{1}{1+\frac{z}{2!}+\frac{z^2}{3!}+... }
    Now as z\to 0 we see that f(z) \to \frac{1}{1+0+0+...} = 1.
    Thus, z=0 is a "removable singularity".
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  3. #3
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    Thanks PerfectHacker

    Thanks for all your help with my questions to date PerfectHacker.

    You are the one!
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