How do you find all solutions to #x^6-64i=0#?
1 Answer
Use de Moivre's theorem to find all
Explanation:
Given:
#x^6-64i = 0#
Add
#x^6 = 64i#
#color(white)(x^6) = 2^6(0 + 1i)#
#color(white)(x^6) = 2^6(cos (pi/2) + i sin (pi/2))#
From de Moivre's theorem, we have:
#(cos theta + i sin theta)^n = cos n theta + i sin n theta#
Hence principal root:
#x_1 = root(6)(64i) = 2(cos (pi/12) + i sin (pi/12))#
The other
#x_2 = 2(cos ((5pi)/12) + i sin ((5pi)/12))#
#x_3 = 2(cos ((9pi)/12) + i sin ((9pi)/12))#
#x_4 = 2(cos ((13pi)/12) + i sin ((13pi)/12))#
#x_5 = 2(cos ((17pi)/12) + i sin ((17pi)/12))#
#x_6 = 2(cos ((21pi)/12) + i sin ((21pi)/12))#
These trigonometric values can be expressed in terms of square roots:
#cos (pi/12) = 1/4(sqrt(6)+sqrt(2)) " " sin (pi/12) = 1/4(sqrt(6)-sqrt(2))#
#cos ((5pi)/12) = 1/4(sqrt(6)-sqrt(2)) " " sin ((5pi)/12) = 1/4(sqrt(6)+sqrt(2))#
#cos ((9pi)/12) = -sqrt(2)/2 " " sin ((9pi)/12) = sqrt(2)/2#
#cos ((13pi)/12) = -1/4(sqrt(6)+sqrt(2)) " " sin ((13pi)/12) = -1/4(sqrt(6)-sqrt(2))#
#cos ((17pi)/12) = -1/4(sqrt(6)-sqrt(2)) " " sin ((17pi)/12) = -1/4(sqrt(6)+sqrt(2))#
#cos ((21pi)/12) = sqrt(2)/2 " " sin ((21pi)/12) = -sqrt(2)/2#
So here are the
graph{((x-1/2(sqrt(6)+sqrt(2)))^2 + (y-1/2(sqrt(6)-sqrt(2)))^2 - 0.01)((x-1/2(sqrt(6)-sqrt(2)))^2 + (y-1/2(sqrt(6)+sqrt(2)))^2 - 0.01)((x+sqrt(2))^2+(y-sqrt(2))^2-0.01)((x+1/2(sqrt(6)+sqrt(2)))^2 + (y+1/2(sqrt(6)-sqrt(2)))^2 - 0.01)((x+1/2(sqrt(6)-sqrt(2)))^2 + (y+1/2(sqrt(6)+sqrt(2)))^2 - 0.01)((x-sqrt(2))^2+(y+sqrt(2))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}