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Finding the square root of a+bi (complex number)

  1. Mar 30, 2009 #1
    I was reading Roger Penrose' book "The Road to reality". He mentioned the square root of a+bi in terms of a and b. I am trying to figure his answer out for my self but am struggling. Here goes:


    [tex](x+yi)^2=a+bi[/tex]

    [tex]x^2+2xyi-y^2=a+bi[/tex]

    [tex]x^2-y^2=a[/tex]

    [tex]2xy=b[/tex]

    I can't rearrange these two equations to get x and y in terms of a and b. Even if I use a computer program to solve them for me, I get really complicated answers. Not like the solution in the book. Am I doing it wrong? Here is the solution he gives:

    I have checked it and it works quite cleverly.

    [tex]\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2}}[/tex]
     
  2. jcsd
  3. Mar 30, 2009 #2
    Solve for x in the second equation to get x = b/(2y), which you can then plug into the first equation to get a quartic in y:
    [tex]0 = y^4 + ay^2 + \left(\frac{b}{2}\right)^2[/tex]
    Complete the square (using y2 as the variable in order to separate it from a) and solve for y2.
    That's a great book. :)
     
  4. Mar 30, 2009 #3
    Thanks! I got to the quartic quite quickly in my attempt, but I got scared by it because it is a quartic :tongue:.
    Although you wrote that slightly wrong. It should have been - (b/2)² (not plus). Here is the working for anyone that is interested.

    [tex]0=y^4+ay^2-\frac{b^2}{4}[/tex]

    [tex]0=(y^2+\frac{a}{2})^2-\frac{a^2}{4}-\frac{b^2}{4}[/tex]

    [tex]y^2+\frac{a}{2}=\sqrt{\frac{1}{4}(a^2+b^2)}[/tex]

    [tex]y=\sqrt{\frac{1}{2}(-a+\sqrt{(a^2+b^2)})}[/tex]

    to get x

    [tex]x^2-\frac{1}{2}(-a+\sqrt{(a^2+b^2)}=a[/tex]

    [tex]x=\sqrt{\frac{1}{2}(a+\sqrt{(a^2+b^2)})}[/tex]
     
  5. Mar 30, 2009 #4
    Indeed. :) Good work.
     
  6. Jan 15, 2010 #5
    aren't you guys forgetting that the quadratic equation has a +/- option? That gives two solutions for x in the equation ax2 + bx + c = 0. Since 'x' itself is square, and we are given a quartic equation to solve, that's a total of 4 possible solutions for just the 'a' variable. The same goes for the 'b' variable, so, unless I'm missing some connection, 4 independent solutions of two variables each (a and b), gives a maximum total of 16 possible solutions just for the square root. If the +/- are connected for the two variables, that's a minimum of 4 possible solutions.

    I think a single solution is too simplistic, and that there's more than meets the eye. It's worth dissecting further to find the different possible solutions.
     
  7. Jan 15, 2010 #6
    There are indeed many solutions. Although this thread has been quite computational so it might not be apparent. You will get one solution for each analytic branch of the natural log that you choose.
     
  8. Jan 16, 2010 #7

    Gib Z

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    There are only two solutions.
     
  9. Jan 16, 2010 #8

    Mentallic

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    [tex]a-\sqrt{a^2+b^2}<0[/tex] for all real a,b so that eliminates the possibility of the negative solution from the [itex]\pm[/itex] in:

    [tex]x=\sqrt{\frac{1}{2}(a\pm\sqrt{(a^2+b^2)})}[/tex]

    and

    [tex]y=\sqrt{\frac{1}{2}(-a\pm\sqrt{(a^2+b^2)})}[/tex]

    since x,y must be real numbers.

    edit: except in the trivial case of b=0 but then we're dealing with real numbers so there's no point in all this :smile:
     
  10. Jan 17, 2010 #9
    I think you are trying to argue with the fundumental theorem of algebra here,
    Be careful
     
  11. Jan 17, 2010 #10
    True enough, my mistake.
     
  12. Mar 6, 2010 #11
    Could someone please explain how the above equation was derived?
     
  13. Mar 6, 2010 #12
    Even if a and b are not real the "i" will cancel out giving two solutions which are the same.
     
  14. Mar 6, 2010 #13

    Redbelly98

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    It's a lot easier to work this out in polar coordinates.

    z = r e

    √z = (r e)1/2

    = r1/2 eiθ/2

    (In other words, take the square root of the magnitude, and one half of the angle.)

    Replacing θ with θ+2π does not change the value of z, and it gives the other root:

    √z = r1/2 ei(π+θ/2)
     
  15. Mar 6, 2010 #14

    Galileo

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    It is easier, in my view, to work from polar coordinates.
    [itex]z=a+ib=|z|\exp(i\theta)=w^2[/itex].
    So [itex]w=\sqrt{|z|}(\cos(\theta/2)+i\sin(\theta/2))[/itex].

    Now use [itex]\cos(2\theta)=\sqrt{\frac{1}{2}(1+\cos \theta)}[/itex] and similar for the sine.
    Since [itex]\cos(\theta)=a/\sqrt{a^2+b^2}[/itex] you can write down the expression. The other solution is just minus this number.

    EDIT: Redbelly beat me to it :P
     
  16. Mar 6, 2010 #15

    Redbelly98

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    Great minds think alike :smile:

    I just realized this is an old thread, and was revived today because of the following question:
    See post #1, where the OP got to the following two equations:
    (1) x2 - y2 = a
    (2) 2xy = b​
    From (2), we know that x = b/2y
    Substitute for x in (1) to get
    b2/4y2 - y2 = a​
    Multiply through by y2, and get all terms on the right side of the equation, to get
    0 = y4 + ay2 - b2/4​
    Note the "-" sign which was missed in the post you quoted, but corrected in Post #3.
     
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