Results 1 to 9 of 9

Math Help - Complex square roots

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    34

    Complex square roots

    Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by Rine198 View Post
    Find the complex square roots of 5-12i by solving (x+iy)^2=5-12i for x,y real.

    (x+iy)^2=x^2-y^2+2xyi=5-12i


    Now, identify real and imaginary parts and solve the system.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    34
    so i've identified that

    5=x^2-y^2 (real part)

    -12=2xy (imaginary part)

    and i tried to do simultaneous equation with it, but i still don't get how i can get the answer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by Rine198 View Post
    5=x^2-y^2 (real part)

    -12=2xy (imaginary part)

    and i tried to do simultaneous equation with it, but i still don't get how i can get the answer?

    Right, now y=-6/x, substitute in the first equation and you'll obtain a biquadratic equation on x .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2010
    From
    Mumbai
    Posts
    91
    Thanks
    2
    Quote Originally Posted by Rine198 View Post
    so i've identified that

    5=x^2-y^2 (real part)

    -12=2xy (imaginary part)

    and i tried to do simultaneous equation with it, but i still don't get how i can get the answer?
    Show what you have done.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2011
    Posts
    34
    but how do i factorise x^4-36-5x^2=0, which is what i got after substituting y=-6/x
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Let u = x^2 such that x^4-5x^2-36 = u^2-5u-36 = 0

    This one does factorise - what two number multiply to -36 and sum to -5?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Mar 2011
    Posts
    34
    So i've found that x=+/-3 and y=+/-2, but i still don't get how to find the complex square roots from this?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    The square roots are 3-2i,-3+2i .

    Equivalently : \pm(3-2i) .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum of Square Roots...
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: September 25th 2011, 06:19 PM
  2. Square roots of complex numbers.
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: November 7th 2010, 11:34 PM
  3. Replies: 4
    Last Post: October 24th 2009, 07:22 AM
  4. Square roots of a complex number.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 16th 2009, 04:40 AM
  5. Square Roots and Complex Numbers!!!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 27th 2007, 11:59 AM

Search Tags


/mathhelpforum @mathhelpforum