How do you find all the real and complex roots of #(z-1)^3 =8i#?

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Answer 1 · 2016-12-29T01:43:12

The roots are all complex:

#z_1 = 1+sqrt(3)+i#

#z_2 = 1-sqrt(3)+i#

#z_3 = 1-2i#

Using de Moivre's formula, note that:

#(cos(pi/6) + i sin(pi/6))^3 = cos(pi/2)+i sin(pi/2) = i#

So one cube root of #i# is:

#cos(pi/6)+ i sin(pi/6) = sqrt(3)/2 + 1/2i#

The other cube roots can be formed by adding #(2pi)/3 = (4pi)/6# to the angle a couple of times:

#cos((5pi)/6) + i sin((5pi)/6) = -sqrt(3)/2 + 1/2i#

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

The cube roots of #8i# will be double these values since #8 = 2^3#.

So the cube roots of #8i# are:

#sqrt(3)+i#

#-sqrt(3)+i#

#-2i#

We find the required values of #z# by adding #1# to each of these to give solutions:

#z_1 = 1+sqrt(3)+i#

#z_2 = 1-sqrt(3)+i#

#z_3 = 1-2i#