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Math Help - IS F(Z) = Z6^ analytic

  1. #1
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    IS F(Z) = Z6^ analytic

    Hi guys, would be very grateful if someone could post a solution to this question


    "Check if F(Z) = Z^6 is analytic"
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    One way:


    f'(z)=\displaystyle\lim_{h\to 0}\dfrac{(z+h)^6-z^6}{h}=\ldots=6z^5

    That is, for every z\in\mathbb{C} there exists f'(z) (finite) .
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  3. #3
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    Thank You, just wondering if thats the full solution or do i continue getting the limit, i am unsure how to continue getting the limit.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by davy337 View Post
    Thank You, just wondering if thats the full solution or do i continue getting the limit, i am unsure how to continue getting the limit.

    Use the Newton's Bynomial Theorem, (z+h)^6=z^6+6z^5h+\ldots+h^6 , cancel z^6 and divide numarator and denominator by h .
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  5. #5
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    I feel terrible asking but would it be possibe to get the full solution from you ? apoligies for taking up your time
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by davy337 View Post
    I feel terrible asking but would it be possibe to get the full solution from you ? apoligies for taking up your time

    I wouldn't mind, but is not good for you. Better study the corresponding theory and ask your doubts.
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  7. #7
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    Quote Originally Posted by FernandoRevilla View Post
    I wouldn't mind, but is not good for you. Better study the corresponding theory and ask your doubts.
    I would like to support the above advice and remind members of MHF policy (clearly stated in the forum rules).
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