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Homework Help: Complex Numbers

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Express z=-1+4i in polar for then find z^4 converting to Cartesian form

    2. Relevant equations
    r = sqrt(x^2+y^2)
    theta = y/x
    z= r cos (theta) + i r sin (theta)

    3. The attempt at a solution
    r= sqrt(-1^2+4^2)
    = sqrt(17)

    theta = tan a = 4/1
    a = tan^-1 (4/1)
    = 1.3258
    pi-a = 1.8158 (to find the argument from the real axis)

    in polar form = sqrt(17) cis 1.8158

    Then to cartesian form

    z^4 = r^4 cos 4(theta) + i r^4 sin 4(theta)
    = (sqrt (17))^4 cos 4(1.8158) + i (sqrt(17))^4 sin 4(1.8158)
    = 289 cos 7.2632 + i 289 sin 7.2632
    z = 4sqrt(160.9760) + 4sqrt(240.0161) i
    z = 3.5620 + 3.9360 i

    Im not sure if this is correct but hopefully i am on the right track. I was not sure if I was converting to Cartesian form correctly or if i should introduce z^4 at the polar form so that z^4 = r^4 cis 4(theta)
  2. jcsd
  3. Apr 24, 2012 #2


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    Homework Helper

    This is where you went wrong. You had found z4 correctly, but then in order to convert back to z
    (which you shouldn't be doing anyway because the question asked you to convert z4 back into Cartesian form, not z)
    you should have simply calculated [itex]289\cdot \cos(7.2632)[/itex] and [itex]289\cdot \sin(7.2632)[/itex]

    Now something else you should work on is avoiding the decimal approximations. If you calculate what you have, you won't get the exact answer. For example, [itex]289\cdot \cos(7.2632)+i\cdot289\cdot\sin(7.2632)\approx 160.976+240.016i[/itex] but the exact answer to the problem is [tex]\left(-1+4i\right)^4=161+240i[/tex]

    So let [itex]\alpha = \tan^{-1}(4)[/itex] therefore the argument is [itex]\pi-\alpha[/itex] and so we have


    And we can simplify these cos and sin expressions, and after that, we can then find z4. Now, can you simplify the expressions [itex]\cos(\tan^{-1}\theta)[/itex] and [itex]\sin(\tan^{-1}\theta)[/itex] ? And also, what about expressing [itex]\cos(4x)[/itex] and [itex]\sin(4x)[/itex] in terms of [itex]\sin(x)[/itex] and [itex]\cos(x)[/itex]?
  4. Apr 24, 2012 #3

    I think that the argument required for z (which according to convention is measured from the real axis in an anticlockwise manner) is just a = tan^-1(4/1) and not (∏ - a).
  5. Apr 24, 2012 #4
    I am so sorry!!

    I did not notice the -1 in z!
  6. Apr 24, 2012 #5


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    Homework Helper

    grzz, you can go back and edit your first post if you made a mistake somewhere. Better than making a second post :wink:
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