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

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Answer 1 · 2016-08-13T23:55:06

#-64#

#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#