Question 350605: 3 roots 2i-2
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! 3 roots 2i-2
solving to find the 3rd roots of 2i - 2, if this is not what you need,
please repost
convert 2i - 2 to polar form
2i - 2 = -2 + 2i = x + yi = z
z = r * (cos(t) + i * sin(t))
r * (cos(t) + i * sin(t)) is polar form
t is the angle, theta
r = |z| = sqrt(x^2 + y^2) = sqrt((-2)^2 + 2^2)
r = sqrt((-2)^2 + 2^2), this is the magnitude of the complex number, or the distance it is from origin if plotted on a complex plane
r = sqrt(4 + 4) = 8^(1/2) = sqrt(8) = sqrt(4 * 2) = 2sqrt(2)
-2 < 0 and 2 > 0, so this is in quadrant 2 when plotted on a complex plane
(quadrants go counterclockwise)
tan t = |y/x| = |2/(-2)| = |-1| = 1
t = arctangent(1)
t = 45, since quadrant 2 --> t = 180 - 45 = 135 degrees
z = 2i - 2 = -2 + 2i = 2sqrt(2)(cos(135) + isin(135)) in polar form
we want third roots of 2sqrt(2)(cos(135) + isin(135)),
360 degrees / 3 = 120 degrees, these roots will be 120 degrees apart
using de Moivre's formula
[r * (cosx + isinx)]^n = r^n * (cos(nx) + isin(nx))
r = 8^(1/2) = sqrt(8) = sqrt(4 * 2) = 2sqrt(2)
r^(1/3) = (8^(1/2)^(1/3) = 8^(1/2 * 1/3) = 8^(1/6) = (8^(1/3))^(1/2) = 2^(1/2)
r^(1/3) = 2^(1/2) = sqrt(2)
z^(1/3) = sqrt(2)[cos(1/3 * (135 + 360k)) + isin(1/3 * (135 + 360k))]
k from 0 to 2, since n = 3
k = 0
sqrt(2)[cos(1/3 * 135) + isin(1/3 * 135)]
sqrt(2)[cos(45) + isin(45)]
sqrt(2)[sqrt(2)/2 + i(sqrt(2)/2)]
2/2 + i(2/2)
1 + i
k = 1
sqrt(2)[cos(1/3 * 495) + isin(1/3 * 495)]
sqrt(2)[cos(165) + isin(165)]
sqrt(2)cos(165) + isqrt(2)sin(165)
k = 2
sqrt(2)[cos(1/3 * 855) + isin(1/3 * 855)]
sqrt(2)[cos(285) + isin(285)]
sqrt(2)cos(285) + isqrt(2)sin(285)
check:
(1 + i)(1 + i)(1 + i)
(1 + 2i - 1)(1 + i) (since i^2 = -1)
2i(1 + i)
2i + 2i^2
2i - 2 = -2 + 2i, yes
[sqrt(2)cos(165) + isqrt(2)sin(165)]^3
inputted (sqrt(2)cos(165) + i*sqrt(2)sin(165))^3
into http://www.wolframalpha.com
you can multiply it out if you want
did come out to -2 + 2i, yes
[sqrt(2)cos(285) + isqrt(2)sin(165)]^3
inputted (sqrt(2)cos(285) + i*sqrt(2)sin(285))^3
into http://www.wolframalpha.com
you can multiply it out if you want
did come out to -2 + 2i, yes
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