How do you find the power #(sqrt3-i)^3# and express the result in rectangular form?

1 Answer
Mar 7, 2017

# (sqrt(3)-i)^3 = -8i#

Explanation:

Let #z=sqrt(3)-i#

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

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And we will put the complex number into polar form:

# |z| = sqrt(3+1) = 2 #
# arg(z) = tan^-1(-1/sqrt(3)) = -(pi)/6 #

So then in polar form we have:

# z = 2(cos(-(pi)/6) + isin(-(pi)/6)) #

By De Moivre's Theorem:

# z^3 = {2(cos(-(pi)/6) + isin(-(pi)/6))}^3 #
# \ \ \ = 2^3(cos(-(3pi)/6) + isin(-(3pi)/6)) #
# \ \ \ = 8(cos(-(pi)/2) + isin(-(pi)/2)) #
# \ \ \ = 8(0-i) #
# \ \ \ = -8i #