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Question 361896: Hi,
Just wondering if you could show me the steps of how to find all roots of the equation z^5 = i, i.e. find the 5 values of i^1/5
Thankyou,
franco
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! "Hi,
Just wondering if you could show me the steps of how to find all roots of the equation z^5 = i, i.e. find the 5 values of i^1/5
Thankyou,
franco"
z^5 = i
z = i^(1/5), find the 5 roots
360/5 = 72, so roots will be 72 degrees apart when plotted on complex plane
converting i to polar form
polar form is r(cos(t) + isin(t))
r = sqrt(0^2 + 1^2) = sqrt(1^2) = 1
t (angle theta) = 90 degrees
cos 90 = 0, sin 90 = 1
polar form of i is i
using de Moivre's formula
[r * (cost + isint)]^n = r^n * (cos(nt) + isin(nt))
r^(1/5) = 1^(1/5) = 1, n = 1/5, t = 90 + 360k
z^(1/5) = cos(1/5 * (90 + 360k)) + isin(1/5 * (90 + 360k)))
k from 0 to 4
k = 0
cos(1/5 * 90) + isin(1/5 * 90)
cos(18) + isin(18)
k = 1
cos(1/5 * 450) + isin(1/5 * 450)
cos(90) + isin(90)
i
k = 2
cos(1/5 * 810) + isin(1/5 * 810)
cos(162) + isin(162)
k = 3
cos(1/5 * 1170) + isin(1/5 * 1170)
cos(234) + isin(234)
k = 4
cos(1/5 * 1530) + isin(1530)
cos(306) + isin(306)
checking (cos(18) + isin(18))^5
use (a + b)^5 = a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5
from pascal's triangle
plug in cos(18) for a, and isin(18) for b
cos(18)^5 + 5*cos(18)^4*(isin(18)) + 10*cos(18)^3*(isin(18))^2 + 10*cos(18)^2*(isin(18))^3 + 5*cos(18)*(isin(18))^4 + (isin(18))^5
i = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i
cos(18)^5 + 5*i*cos(18)^4*sin(18) - 10*cos(18)^3*sin(18)^2 - 10*icos(18)^2*sin(18)^3 + 5*cos(18)*sin(18)^4 + isin(18)^5
cos(18)^5 - 10cos(18)^3*sin(18)^2 + 5cos(18)sin(18)^4 =
.7781 - .8215 + .0434 = 0 (rounded these to 4 digits)
5*i*cos(18)^4*sin(18) - 10*icos(18)^2*sin(18)^3 + isin(18)^5 =
1.2641i - .2669i + .0028i = i (rounded these to 4 digits)
so (cos(18) + isin(18))^5 = i
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