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Math Help - Use CR Equations to show f'(z^n)=nz^n-1

  1. #1
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    Use CR Equations to show f'(z^n)=nz^n-1

    So the first part to this question was,

    Use the CR equations to show that z^n is entire for any n using polar coordinates.

    hence,

    du/dr=n(r^n-1)(cosnx) dv/dr=n(r^n-1)(sinnx)

    du/dx=(-r^n)(nsinnx) , dv/dx=(r^n)(ncosnx).

    Which via the CR equalities, are differentiable for all C, for any n => Entire

    The second part to the question is, using your answer above show that

    (z^n)'=nz^n-1

    Okay I have two problems with this. Firstly,

    if f is entire then the derivative may be writen as

    f'(z)=du/dr +idv/dr = dv/dx-idu/dx

    => f'(r,x)=nr^(n-1)(cosnx+isinnx)= nr^n(cosnx+isinnx)

    First off, they are not the same! so why does this not hold true.

    Secondly if , f'(r,x)=nr^(n-1)(cosnx+isinnx), is correct.

    i cannot see how it equates to (z^n)'=nz^n-1 ?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Use CR Equations to show f'(z^n)=nz^n-1

    Quote Originally Posted by olski1 View Post
    if f is entire then the derivative may be writen as f'(z)=du/dr +idv/dr = dv/dx-idu/dx
    It should be f'(z)=(u_r+iv_r)(\cos \theta -i\sin \theta) .
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