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Find the square roots of a = root3 + root3*i

  1. Jan 20, 2015 #1
    • Member warned about not using the homework template
    I don't recall ever doing this but maybe I have.

    z2 = a = p [cos Ψ + i sin Ψ] = √3 + i*√3

    p = √6
    Ψ = π/4

    Using the formula in the notes, z = 61/4 * exp[i*(π/4 + 2π*k)/2], k = 0, 1.
     
    Last edited: Jan 20, 2015
  2. jcsd
  3. Jan 20, 2015 #2

    Mark44

    Staff: Mentor

    Do you have a question?

    Also, your problem would be much more readable if you used LaTeX or the symbols available from the Advanced Menu, under the ##\Sigma## icon; e.g., √, ψ, and π.
     
  4. Jan 20, 2015 #3
    When I multiply it out, I get -a.
     
  5. Jan 21, 2015 #4

    Mark44

    Staff: Mentor

    I don't. What do you get for the two roots of ##\sqrt{6}(cos(\pi/4) + i sin(\pi/4))##?

    Note that in polar form, what you're calling a is (##\sqrt{6}, \pi/4##).
     
  6. Jan 21, 2015 #5
    That's odd. I used my TI-89 to check my work, and it gave me -a. Maybe I had a typo.
     
  7. Jan 21, 2015 #6

    Mark44

    Staff: Mentor

    Possibly, or maybe it's in degree mode when it should be in radian mode.

    Either way, this is an easy enough problem that a calculator shouldn't be needed. ##cos(\pi/4) = sin(\pi/4) = \frac{\sqrt{2}}{2} \approx .707##

    Again, what did you get for the two roots?
     
  8. Jan 21, 2015 #7
    From the original post, z2 = {61/4*ei*9π/8, 61/4*ei*π/8}
     
  9. Jan 21, 2015 #8

    Mark44

    Staff: Mentor

    There's your mistake. Those are the correct roots of the equation z2 = ##\sqrt{6}(cos (\pi/4) + i sin (\pi/4))##, but they are values of z, not z2.

    If you square each of them, you get ##\sqrt{6}(cos (\pi/4) + i sin (\pi/4))##.
     
  10. Jan 21, 2015 #9
    Ah. For some reason I thought that those were the square roots, not that each of those was a square root. Thanks for the help. My brain is still on vacation. Haha.
     
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