How do you cube #1-sqrt3i#?

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Answer 1 · 2015-10-15T16:42:49

-8

You will have to multiply it one at a time.

#(1-isqrt3)^3#

#=(1-isqrt3)(1-isqrt3)(1-isqrt3)#

#=(1-isqrt3)(1-isqrt3-isqrt3-3)#

#=(1-isqrt3)(1-2isqrt3-3)#

#=1-2isqrt3-3-isqrt3-6+3isqrt3#

#=color(red)(-8)#

Answer 2 · 2015-10-15T20:36:13

#1 - sqrt(3)i = 2(cos (-pi/3) + i sin (-pi/3))#

So #(1 - sqrt(3)i)^3 = 2^3(cos (-pi) + i sin (-pi)) = 2^3 * -1 = -8#

De Moivre's formula tells us that:

#(cos theta + i sin theta)^n = cos (n theta) + i sin (n theta)#

Now

#1 - sqrt(3)i = 2(1/2 - sqrt(3)/2i) = 2(cos (-pi/3) + i sin (-pi/3))#

So

#(1 - sqrt(3)i)^3 = (2(cos (-pi/3) + i sin (-pi/3)))^3#

#= 2^3(cos (-pi/3) + i sin (-pi/3))^3 = 8(cos (-pi) + i sin (-pi))#

#= 8*-1 = -8#