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Homework Help: Complex number equations.

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data

    a) Find the modulus and argument of 6^(1/2) + 2^(1/2)i

    b) Solve the equation z^(3/4) = 6^(1/2) + 2^(1/2)i

    2. Relevant equations

    3. The attempt at a solution

    For part a) i used Pythagoras to find the modulus.

    ( (6^(1/2))^2 + (2^(1/2))^2 )^(1/2) = (6 + 2)^(1/2) = 8^(1/2)

    and to find the principle argument of z i said arctan(2^(1/2)/6^(1/2) = pi/6

    I assume this is correct.

    for part b)

    z^(3/4) = 6^(1/2) + 2^(1/2)i


    z^(3/4) = 8^(1/2)(cos(pi/6 + 2kpi) +isin(pi/6 + 2kpi))

    Then raising the R.H.S of the Equation to 4/3 to get z and Then using de Moivre's theorem.

    therefore z = 4(cos(4pi/18 + 8kpi/3) +isin(4pi/18 + 8kpi/3))

    This is where i am stuck. As k [itex]\in Z [/itex] i can get a value of theta which is contained within
    -pi < theta <= pi when k = 0.
    if i assign a value of 1 to K then i get a value of [itex]\vartheta[/itex] which is greater than pi.
    the answer i get when K = 0 is wrong and there are 3 solutions. So i must be doing something wrong elsewhere.

    I'm sorry in advance if i have made a silly error.
  2. jcsd
  3. Sep 10, 2014 #2


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    I want to focus first on your result for k=0. It looks right to me. What value are you supposed to get ? (this may be a silly comment but if you calculated a numerical value, were you in radians?)
  4. Sep 10, 2014 #3
    ImageUploadedByPhysics Forums1410375999.969665.jpg

    The answer 7b are the values I should be getting :/
  5. Sep 10, 2014 #4


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    First rise the equation to power 4. You get z3. Then take the third roots of the number with k=0.
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