How do you find the sixth roots of #64i#?

1 Answer
Sep 26, 2016

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 #theta = pi/12# we find:

#(cos (pi/12) + i sin (pi/12))^6 = cos (pi/2) + i sin (pi/2) = 0 + i (1) = i#

So #cos (pi/12) = i sin (pi/12)# is a sixth root of #i#.

We can form other sixth roots by adding any multiple of #(2pi)/6 = pi/3# to theta. So in general we find:

#(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 #64i# are:

#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)#