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Posers and Puzzles

Posers and Puzzles

  1. Standard member Mexico
    Quis custodiet
    24 Feb '08 03:33 / 1 edit
    Whats the square root of -i ??
  2. 24 Feb '08 03:50
    -sqrt(2)/2+sqrt(2)i/2
    and
    sqrt(2)/2-sqrt(2)i/2
    I think.
  3. Standard member Mexico
    Quis custodiet
    24 Feb '08 04:01
    Originally posted by GregM
    -sqrt(2)/2+sqrt(2)i/2
    and
    sqrt(2)/2-sqrt(2)i/2
    I think.
    Don't suppose you want to show me how to get that?
  4. 24 Feb '08 04:41 / 2 edits
    Originally posted by Mexico
    Don't suppose you want to show me how to get that?
    we seek a complex number a+bi, whose square: (a^2 - b^2) + (2ab)i = -i

    so we seek a simultaneous solution to

    (1) a^2 - b^2 = 0 and

    (2) 2ab = -1.

    clearly by equation (1), a=b will give us one solution. then by (2), a=b= [sqrt(2)/2]*i, and our complex number a+bi = [sqrt(2)/2]i - [sqrt(2)/2]i^2 or finally:
    a+bi = [sqrt(2)/2]i - sqrt(2)/2

    the other solution is given by a=-b. so then a = [sqrt(2)/2] and b = -[sqrt(2)/2].
    so a + bi = [sqrt(2)/2] - [sqrt(2)/2]i.
  5. Standard member Mexico
    Quis custodiet
    24 Feb '08 04:42
    Thank you....
  6. 24 Feb '08 04:44
    you're welcome it's easier than it looks with all the sqrt(---) notation. i wish equation editor worked outside of microsoft word haha
  7. 24 Feb '08 15:47 / 1 edit
    Another way is to use the identity:

    r(cos x + i sin x) = e^(ix)

    So -i = e^(-i PI/2)

    => sqrt(i) = +/- e^(-i PI/4)
    = +/- [cos PI/4 - i sin PI/4]
    = +/- [sqrt(2)/2 - i sqrt(2)/2
  8. 25 Feb '08 05:23
    Originally posted by Mexico
    Don't suppose you want to show me how to get that?
    -i = e^(3pi/2 * i)
    sqrt(-i) = (-i)^(1/2) = (e^(3pi/2 * i)) ^ (1/2) = e^(3pi/2 * i * 1/2) = e^(3pi/4 * i)

    r*e^(theta * i) = r*(cos(theta) + i*sin(theta)), so that
    e^(3pi/4 * i) = cos(3pi/4) + i*sin(3pi/4) = -sqrt(2)/2 + sqrt(2)i/2

    But all numbers have two square roots (and, in general, n nth roots). The second square root is always the negative of the first, because if x = sqrt(y) then (-x)^2 = (-1 * x)^2 = (-1)^2 * (x)^2 = 1*x^2 = y, so that -x is another square root of y. So the other root is

    -1*(-sqrt(2)/2 + sqrt(2)i/2) = sqrt(2)/2 - sqrt(2)i/2.
  9. Subscriber AThousandYoung
    It's about respect
    29 Feb '08 15:25
    Originally posted by Mexico
    Whats the square root of -i ??
    (-i)^(1/2)
    (-1)^(1/2)*i^(1/2)
    i*-i = 1
  10. Standard member PBE6
    Bananarama
    29 Feb '08 17:24
    Further to what GregM showed, use can use de Moivre's theorem to find the nth root of a complex number. Given a complex number c = a + bi = r*cis(x), where cis(x) = sin(x) + i*cos(x), the roots are:

    c^(1/n) = r^(1/n) * (sin((x+2k*pi)/n) + i*cos((x+2k*pi)/n)) for k = 0,1,...(n-1)
  11. Standard member preachingforjesus
    Iron Pillar
    29 Feb '08 17:57
    simplest way:

    transform < a , b > = a + bi

    to < r , y > where r = radius and y = angle from positive real numbers.

    then < r , y > ^n = < r^n , y*n >

    to get all roots of a number

    < r , y > ^(1/n) = < r^(1/n) , [y/n + (2k pi /n)] > k= (1, 2, ... , n)
  12. Standard member PBE6
    Bananarama
    29 Feb '08 21:12
    The cool thing about the roots of an imaginary number is that when you plot them on the imaginary plane, they form a regular n-gon. If you can find one of the roots by inspection, this makes finding the other roots a snap.

    For example, one of the cube roots of 1 is 1. To find the others, just rotate the vector (1,0) on the imaginary plane 120 degrees (360/3) until you're back where you started. The other two roots are then (-SQRT(2), SQRT(2)) and (-SQRT(2), -SQRT(2)).
  13. 01 Mar '08 09:29
    this material is too complex for me.

    and i've always loved that fact about the n-th roots of a number forming a regular n-gon. it's unreal.
  14. 01 Mar '08 12:34
    Originally posted by Mexico
    Whats the square root of -i ??
    Do negative numbers have square roots? I don't think so, but my maths teacher says they do in a complicated way...
  15. 01 Mar '08 16:51
    Originally posted by Mexico
    Whats the square root of -i ??
    You can check quaternions math. It's cool.