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Math Help - complex analysis2

  1. #1
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    complex analysis2

    By integrating a power series for 1/1+z^2 term-by-term, find a power series for arctan(z) centered at z0=0. Compute its radius of convergence.

    Any help would be great!!
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  2. #2
    Super Member girdav's Avatar
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    So, what is the power series of \frac 1{1+z^2} ?
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  3. #3
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    i know it equals (1/2i)[1/(z-i)-1/(z+i)]? where do I go from there?
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  4. #4
    Super Member girdav's Avatar
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    If you know the power series of \frac 1{1-Z} you can find the power series of \frac 1{1+z^2} by a substitution.
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  5. #5
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    would it be similar to 1/1-x which is 1 + x + x^2 + ... so 1/1-z = 1 + z + z^2 + z^3 onwards? and then how would one go about the substitution?
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  6. #6
    Super Member girdav's Avatar
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    Put Z:=-z^2.
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  7. #7
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    okay so when you put that in then it becomes 1 -z^2 + z^4 - z^6 ... but then what is the next step from that in order to find the power series for arctan(z). do you simply integrate the series we have found through substitution??
    and how would you get the radius of convergence??

    sorry to be a pain!!
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  8. #8
    Super Member girdav's Avatar
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    You have to integrate term by term the series \sum_{n=0}^{+\infty}(-1)^nz^{2n}.
    Use the fact that the radius of convergence is the same as the radius of convergence of its derivative.
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  9. #9
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    okay thank you i think ill be able to handle the integration part! im just confused about the radius of convergence. i know the interval of convergence but how do i go about getting the radius of convergence for this series?
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  10. #10
    Super Member girdav's Avatar
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    You can use the ratio test.
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