How do you use DeMoivre's theorem to simplify (1+isqrt3)^3?
1 Answer
Mar 28, 2017
(1+isqrt(3))^3 = -8
Explanation:
Let
First let us plot the point
And we will put the complex number into polar form:
|omega| = sqrt(1+3) = 2
arg(omega) = tan^(-1)sqrt(3) = pi/3
So then in polar form we have:
omega = 2(cos((pi)/3) + isin((pi)/3))
By De Moivre's Theorem we have:
omega^3 = {2(cos (pi/3) + isin(pi/3))}^3
\ \ \ = {2}^3{(cos(pi/3) + isin(pi/3))}^3
\ \ \ = 8(cos(3pi/3) + isin(3pi/3))
\ \ \ = 8(cos pi + isin pi)
\ \ \ = 8(-1 + 0)
\ \ \ = -8
