How do you use demoivre's theorem to simplify (1-i)^12?

1 Answer
Aug 13, 2016

-64

Explanation:

z = 1 - i will be in 4th quadrant of argand diagram. Important to note for when we find the argument.

r = sqrt(1^2 + (-1)^2) = sqrt(2)

theta = 2pi - tan^(-1)(1) = (7pi)/4 = -pi/4

z = r(costheta + isintheta)

z^n = r^n(cosntheta + isinntheta)

z^12 = (sqrt(2))^12(cos(-12pi/4) + isin(-12pi/4))

z^12 = 2^(1/2*12)(cos(-3pi) + isin(-3pi))

z^12 = 2^6(cos(3pi) - isin(3pi))

cos(3pi) = cos(pi) = -1

sin(3pi) = sin(pi) = 0

z^12 = -2^6 = -64