How do you use demoivre's theorem to simplify #4(1-sqrt3i)^3#?

1 Answer
May 24, 2017

Answer:

#-32#

Explanation:

DeMoivre's theorem for exponents says that any complex number #z# can be written as #r(costheta+isintheta)#, or #rcolor(white)"." "cis"(theta)# for short.

It continues to say that when raising an imaginary number to a certain power, the result is:

#z^n = r^n color(white)".""cis"(ntheta)#

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To simplify this, let's first calculate #(1-isqrt3)^3#.

We find #r# by using Pythagoras' theorem:

#r = sqrt(1^2+sqrt(3)^2) = sqrt4 = 2#

We find #theta# by taking the inverse tangent of #(Im(z))/(Re(z)#.

#theta = tan^-1((-sqrt3)/(1))=-pi/3#

So #(1-isqrt3) = 2"cis"(-pi/3)#. Therefore:

#(1-isqrt3)^3 = 2^3"cis"(3(-pi/3)) = 8"cis"(pi)#

Finally, we expand #"cis"(pi)#.

#8"cis"(pi)=8cospi+8isinpi = 8(-1)+8i(0) = -8#

And don't forget to multiply by 4!

#-8*4 = -32#

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So #4(1-isqrt3)^3 = -32#.

Final Answer