How do you find the cube root of #-4sqrt2(1-i)#?

1 Answer
George C.
May 26, 2017

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 #n# is a positive integer, #r > 0# and #theta in (-pi, pi]# then (by convention) the principal #n#th root of:

#r(cos theta + i sin theta)#

is:

#root(n)(r)(cos (theta/n) + i sin (theta/n))#

There are #n-1# other non-principal #n#th roots.

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 #omega# or #omega^2 = bar(omega)# where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

#(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#

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