Find all complex roots of #z^3= -8i#?
1 Answer
Apr 16, 2017
The roots are:
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
#2(cos (-pi/6) + i sin (-pi/6)) = sqrt(3)-i#
The others can be found by adding multiples of
#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]}