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Square root of a complex number

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data
    find the square root of: 4+4(sqrt3)i

    put the answer in a+bi form

    2. Relevant equations
    finding the nth roots:

    √(r(cos(theta)+isin(theta)))=√r(cos(theta+2(pi)(k)/(n))+isin(theta+2(pi)(k)/(n)

    where k=0,1


    3. The attempt at a solution

    first i converted my equation to trigonometric form: 8(cos(pi/3)+isin(pi/3))

    then using "finding the nth root" equation above i got: 2√2(cos(pi/6)+isin(pi/6) and 2√2(cos(7pi/6)+isin(7pi/6)

    i then converted the two answers to a+bi form and got 2√6+√(2)i and -2√6-√(2)i

    im pretty sure i did something wrong. If u need me to go more in depth on anything just post a comment
     
  2. jcsd
  3. Apr 11, 2010 #2

    Mark44

    Staff: Mentor

    Your answers in polar form look fine, but you made a mistake when you converted to Cartesian form.

    2√2(cos(pi/6)+isin(pi/6)) = 2√2(√3 /2 + i/2) = √2(√3 + i) = √6 + i√2. The other root has a similar mistake.

    You can check your work by taking either of your square roots and squaring it. You should get 4 + 4√3i.
     
  4. Apr 13, 2010 #3
    Woo hoo! Thanks for the help
     
  5. Apr 14, 2010 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since it only asked for the square root you could also try
    [itex](a+ bi)^2= a^2- b^2+ (2ab)i= 4+ 4i\sqrt{3}[/itex]
    so that you must have [itex]a^2- b^2= 4[/itex] and [itex]2ab= 4\sqrt{3}[/itex]

    From [itex]2ab= 4\sqrt{3}[/itex], [itex]b= 2\sqrt{3}/a[/itex] and the first equation becomes [itex]a^2- 12/a^2= 4[/itex].

    Multiplying both sides by [itex]a^2[/itex], [itex]a^4-12= 4a^2[/itex] which is the same as [itex](a^2)^2- 4a^2- 12= 0[/tex]. Solve that quadratic equation for [itex]a^2[/itex] and then solve for a and b.
     
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