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Finding derivative of Z^(7/3)

  1. Sep 18, 2016 #1
    Consider the principal branch of the function

    f(z)= z7/3

    Find f'(-i) and write it in the form a+bi

    My attemp is :

    I know zc = exp(c logz)
    and the derivative of that is : (c/z) * exp(c Logz)
    That is in this case (7/3)*(i) *exp((7/3)*Log-i) = f'(-i)
    I know that Log(-i) = Log(1) + i(-pi/2)= -i pi/2
    and exp((7/3)(-i pi/2)) = (cos(7pi/6)-i*sin(7pi/6))
    and I get (7/3)*(i)(cos(7pi/6)-i*sin(7pi/6)) as final answer

    But I see the answer is : -(7/6)(1+isqrt(3))!
    What is wrong with my algorithm ?
     
    Last edited: Sep 18, 2016
  2. jcsd
  3. Sep 18, 2016 #2

    ehild

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    Include (i) into the exponent.
     
  4. Sep 18, 2016 #3
    -i is in the exponent
     
  5. Sep 18, 2016 #4

    Mark44

    Staff: Mentor

    Is there some reason you can't use the derivative formula ##\frac d {dz} z^k = k z^{k - 1}##?
    Then ##f'(-i) = \frac 7 3 (-i)^{4/3} = \frac 7 3 (e^{-i \pi/2})^{4/3}##
    Simplify this last expression to get to the form in your textbook.
     
  6. Sep 18, 2016 #5

    Ray Vickson

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    Easier: if ##f(z) = z^n## then ##f'(z) = n z^{n-1}##. Apply that to ##n = 7/3## and ##z = -i##.

    Anyway,
    $$\exp((7/3)(-i \pi/2)) = \cos(7\pi/6)-i \sin(7\pi/6) =- \frac{\sqrt{3}}{2} + \frac{1}{2} i $$.
     
  7. Sep 18, 2016 #6

    Mark44

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    That's what I said.
     
  8. Sep 18, 2016 #7

    ehild

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    i is also a factor:
    Write the multiplicative i in exponential form and collect the exponents.
     
  9. Sep 18, 2016 #8
    Thank you so much for your help. I actually mixed up this problem with the formula

    Zm/n = exp((m/n)Log|z|)exp(i(m/n)(Arg z + 2kpi))
    but now I see I could solve it easier:smile:
     
  10. Sep 18, 2016 #9

    Ray Vickson

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    For some reason, your message did not appear on my screen until after I pressed the "enter" key. That sort of thing happens to me quite often, and I have no idea why.

    In fact, I did not see your message until I had logged off and later, logged on again!
     
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