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Thread: Power series (Complex)

  1. #1
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    Power series (Complex)

    Find the power series about the origin for the given function.

    (1) $\displaystyle \frac{z^3}{1-z^3}$ , |z| < 1

    (2) $\displaystyle z^2cosz$

    I will try to attempt the first one:

    (1) $\displaystyle \frac{1}{(1-z)^3} = (\frac{1}{1-z})'$

    $\displaystyle \frac{1}{1-z} = 1 + z + z^2 + z^3 + ....$

    $\displaystyle \frac{z}{1-z} = z + z^2 + z^3 + z^4 + ...$

    $\displaystyle \frac{z^3}{1-z} = z^3 + z^6 + z^9 + z^{12} + ...$

    $\displaystyle = \sum_{n = 1}^{\infty} z^{3n}$

    Somehow, I think my reasoning is wrong. The answer is correct, but I am not sure I took the correct steps.

    I have no idea how to approach (2). Thanks for looking and any help provided!
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  2. #2
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    Quote Originally Posted by shadow_2145 View Post
    (1) $\displaystyle \frac{z^3}{1-z^3}$ , |z| < 1
    $\displaystyle \frac{1}{1-z^3} = 1 + z^3+z^6+... $ now multiply by $\displaystyle z^3$

    (2) $\displaystyle z^2cosz$
    $\displaystyle z^2 \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$
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