Let z = -1+ i. Find the 3 cube roots of z.?

Question

Thank you!


Also, please explain! :)

Answer 1

z = √2 at angle (3π/4 + 2k π)
Roots are
z1 = 2^(1/6).(cos π/4 + i sin π/4) for k = 0
z2 = 2^(1/6).(cos 11π/12 + i sin 11π/12) for k = 1
z3 = 2^(1/6)(cos 19π/12 + i sin 19π/12) for k = 2

Answer 2

You need to use deMoivre's theorem (see reference below).

For a complex number z=r(cos x +i sin x), the roots are given by

z^1/n = r^1/n*(cos(x+2*k*pi/n)+i * sin(x +2*k*pi/n))

where r is the magnitude of the complex number
x is the angle of the complex number

where n = 0 ... n-1

For this particular case

0.794 + 0.794i
-1.084 +0.291i
0.291 - 1.084i

Answer 3

use trig form of complex number

z = sqrt2 (cos 135 + i sin 135)
z^1/3 = 2^1/6 (cos 45 + i sin 45)

change back to standard form
z^1/3 = 2^1/6 * 1/sqrt2 + i 2^1/6 * 1/sqrt2
z^1/3 = 2^(1/6-1/2) + i 2^(1/6-1/2)
z^(1/3) = 2^(-1/3) + i 2^(-1/3)

formated ==>
one divided by cuberoot2 plus one divided by cuberoot2 times i

.....1..........+........1...
-----------------------------i
cubert2.........cubert2

The other two roots are equally distributed over 360degrees
360 / 3 = 120 degrees

FIRST = 2^1/6 (cos 45 + i sin 45) = .794 + .794 i
SECOND = 2^1/6 (cos 165 + i sin 165) = -1.084 + .291 i
THIRD = 2^1/6 (cos 285 + i sin 285) = .291 - 1.084 i