How do you use DeMoivre's theorem to simplify #(1+isqrt3)^3#?

1 Answer
Steve M
Mar 28, 2017

# (1+isqrt(3))^3 = -8#

Explanation:

Let # omega=1+isqrt(3) #

First let us plot the point #omega# on the Argand diagram:

Wolfram Alpha

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#

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