How do you find the cube root of #-4sqrt2(1-i)#?
1 Answer
Principal root:
#root(3)(-4sqrt(2)(1-i)) = sqrt(2)+sqrt(2)i#
There are two other roots.
Explanation:
By de Moivre's formula:
#(cos theta + i sin theta)^n = cos n theta + i sin n theta#
If
#r(cos theta + i sin theta)#
is:
#root(n)(r)(cos (theta/n) + i sin (theta/n))#
There are
Note that:
#-4sqrt(2)(1-i) = 8(-sqrt(2)/2+sqrt(2)/2i)#
#color(white)(-4sqrt(2)(1-i)) = 2^3(cos((3pi)/4)+i sin((3pi)/4))#
Hence:
#root(3)(-4sqrt(2)(1-i)) = 2(cos(pi/4)+isin(pi/4))#
#color(white)(root(3)(-4sqrt(2)(1-i))) = 2(sqrt(2)/2+sqrt(2)/2i)#
#color(white)(root(3)(-4sqrt(2)(1-i))) = sqrt(2)+sqrt(2)i#
The other roots can be found by multiplying by
#(sqrt(2)+sqrt(2)i)(-1/2+sqrt(3)/2i)#
#= -sqrt(2)/2+sqrt(6)/2i-sqrt(2)/2i-sqrt(6)/2#
#= (sqrt(6)-sqrt(2))/2+(sqrt(6)-sqrt(2))/2i#
#(sqrt(2)+sqrt(2)i)(-1/2-sqrt(3)/2i)#
#= -sqrt(2)/2-sqrt(6)/2i-sqrt(2)/2i-sqrt(6)/2i#
#= -(sqrt(2)+sqrt(6))/2-(sqrt(2)+sqrt(6))/2i#