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Solution to z^(2/3) when z=1+i

  1. Aug 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Find all solutions to the equation

    2. Relevant equations

    z^(2/3)

    z=1+i

    3. The attempt at a solution

    I put it in polar form and got √2e(∏/4)i

    Then took the power of that while adding 2∏ni and got

    (√2e(∏/4) + 2∏ni)^(2/3)

    so the argument here is ∏/6 + (4∏ni)/3

    The argument according to the mark sheet is ∏/6 + (2∏n/3)i

    where did i go wrong?

    thanks
     
  2. jcsd
  3. Aug 27, 2013 #2
    look like you are correct...
     
  4. Aug 27, 2013 #3

    Mark44

    Staff: Mentor

    What equation? The only equation you show is z = 1 + i. In case you're not clear on this, an equation has an equals sign (=) separating two expressions.

    Is the problem to evaluate z2/3 when z = 1 + i?
     
  5. Aug 27, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Either "find z^(2/3) where z= 1+ I" or "solve the equation "z^(3/2)= 1+ I". That's what Mark44 was complaining about.

    Yes, that is the correct polar form.

    You are correct that the "general" form of 1+ ix is [itex]\sqrt{2}e^{ix(\pi/4+ 2n\pi)}[/itex] so that the 2/3 power has argument [itex](2/3)(\pi/4+ 2n\pi)= \pi/6+ 4n\pi/3[/itex].

    If the problem is as stated, the "mark sheet" is incorrect.
     
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