How do you find the sixth roots of #64i#?
1 Answer
The sixth roots are:
#+-(1/2(sqrt(6)+sqrt(2)) + 1/2(sqrt(6)-sqrt(2))i)#
#+-(1/2(sqrt(6)-sqrt(2)) + 1/2(sqrt(6)+sqrt(2))i)#
#+-(-sqrt(2) + sqrt(2)i)#
Explanation:
De Moivre's formula tells us that:
#(cos theta + i sin theta)^n = cos n theta + i sin n theta#
Hence if
#(cos (pi/12) + i sin (pi/12))^6 = cos (pi/2) + i sin (pi/2) = 0 + i (1) = i#
So
We can form other sixth roots by adding any multiple of
#(cos ((1+4k)/12 pi) + i sin ((1+4k)/12 pi))^6 = cos (pi/2 + 2kpi) + i sin (pi/2 + 2kpi) = i#
Also we have:
#(r(cos theta + i sin theta))^n = r^n (cos n theta + i sin theta)#
#2^6 = 64#
Hence the sixth roots of
#2(cos ((1+4k)/12 pi) + i sin ((1+4k)/12 pi))" "# for#k = 0, 1, 2, 3, 4, 5#
You are probably aware that:
#sin (pi/4) = cos (pi/4) = sqrt(2)/2#
#sin (pi/6) = cos (pi/3) = 1/2#
#cos (pi/6) = sin (pi/3) = sqrt(3)/2#
#sin(alpha-beta) = sin alpha cos beta - sin beta cos alpha#
#cos(alpha - beta) = cos alpha cos beta + sin alpha sin beta#
So we find:
#sin (pi/12) = sin (pi/3 - pi/4)#
#color(white)(sin (pi/12)) = sin (pi/3) cos (pi/4) - sin (pi/4) cos (pi/3)#
#color(white)(sin (pi/12)) = (sqrt(3)/2) (sqrt(2)/2) - (sqrt(2)/2) (1/2)#
#color(white)(sin (pi/12)) = 1/4(sqrt(6)-sqrt(2))#
#cos (pi/12) = cos (pi/3 - pi/4)#
#color(white)(cos (pi/12)) = cos (pi/3) cos (pi/4) + sin (pi/3) sin (pi/4)#
#color(white)(cos (pi/12)) = (1/2) (sqrt(2)/2) + (sqrt(3)/2) (sqrt(2)/2)#
#color(white)(cos (pi/12)) = 1/4(sqrt(6)+sqrt(2))#
We also have:
#sin ((5pi)/12) = sin (pi/2 - pi/12) = cos (pi/12) = 1/4(sqrt(6)+sqrt(2))#
#cos ((5pi)/12) = cos (pi/2 - pi/12) = sin (pi/12) = 1/4(sqrt(6)-sqrt(2))#
So all the sixth roots are:
#+-2(cos (pi/12) + i sin (pi/12)) = +-(1/2(sqrt(6)+sqrt(2)) + 1/2(sqrt(6)-sqrt(2))i)#
#+-2(cos ((5pi)/12) + i sin ((5pi)/12)) = +-(1/2(sqrt(6)-sqrt(2)) + 1/2(sqrt(6)+sqrt(2))i)#
#+-2(cos ((9pi)/12) + i sin ((9pi)/12)) = +-(-sqrt(2) + sqrt(2)i)#
