roots in standrard form?

Question

a) cube roots of -27i
b)fourth roots of 1
c) fourth roots of -8-8radical 3i

only know how to get only one of the roots.thanks

Answer 1

Start by writing each number in exponential form. For example, -27i = 27e^(3i*pi/2). Then take the requested root of the modulus. Here, the modulus is 27 and the requested root is the third root, so you get 27^(1/3) = 3. Then divide the angle by the number of the root. The angle is 3*pi/2, but we need to write it as (3*pi/2) + 2k*pi, where k can be any integer. Then we divide by three, the number of the root. We will get a number of answers equal to the number of the root, so k needs to vary from 0 to one less than the number of the root. That gives us (3*pi/2 + 0)/3 = pi/2, (3*pi/2 + 2*pi)/3 = (7*pi/2)/3 = 7*pi/6, and (3*pi/2 + 4*pi)/3 = (11*pi/2)/3 = 11*pi/6. That means that the exponential forms of the three third roots of -27i are 3e^(i*pi/2), 3e^(7i*pi/6), and 3e^(11i*pi/6). You can then rewrite them in rectangular form, using Re^(t*i) = R*cos(t) + iR*sin(t).

Offhand, though, the fourth roots of 1 are 1, -1, i, and -i.

Answer 2

use de moivre formula
a)z=0-27i
b)z=1+0i
c) z=fourth roots of -8-8radical 3i

first find the place of z in unit circle.
for choice a
angle=270 degree

for choice b
angle zero or 2 pie

for choice c
angle= 240


remember z^1/n=(rcisx)^1/n

=(r^1/n cis x/n)
do not forget to add the period to the angle k2pi
k=0,1,2,3...

Answer 3

a) 3i
b) 1 & i