Find all complex roots of #z^3= -8i#?

1 Answer
Apr 16, 2017

Answer:

The roots are: #2i#, #-sqrt(3)-i# and #sqrt(3)-i#

Explanation:

See https://socratic.org/s/aDTc9kmp for one method.

In trigonometric form:

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

By de Moivre we have:

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

Hence one of the cube roots of #-8i# is:

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

The others can be found by adding multiples of #(2pi)/3#...

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

#2(cos (-pi/6+(4pi)/3) + i sin (-pi/6+(4pi)/3)) = 2(cos ((7pi)/6) + i sin ((7pi)/6)) = -sqrt(3)-i#

Here are the three roots in the complex plane...

graph{(x^2+(y-2)^2-0.005)((x-sqrt(3))^2+(y+1)^2-0.005)((x+sqrt(3))^2+(y+1)^2-0.005) = 0 [-5, 5, -2.5, 2.5]}