[SIDE NOTE: De Moivre's Formula:
if #z# is a complex number in the form #r(cos(x)+i*sin(x))#, then #z^n=r^n(cos(n*x)+i*sin(n*x))#]
for this problem, use the extension of De Moivre's Formula (Roots of Complex Numbers):
if #z# is a complex number in the form #r(cos(x)+i*sin(x))#, then the n'th roots of z (#z^(1/n)#) are:
#r^(1/n)(cos((x+2pik)/n)+i*sin((x+2pik)/n))#, where k is any integer
note that #1296=1296+0i=1296(cos(0)+i*sin(0))=6^4(cos(0)+i*sin(0))#
if you let the complex number #z# equal 1296, then #r=6^4# and #x=0#
to find the 4th roots, let #n=4#
so the roots are: #(6^4)^(1/4)(cos((0+2pik)/4)+i*sin((0+2pik)/4))#
#=6(cos((pik)/2)+i*sin((pik)/2))#, where k is any integer
as long as you choose 4 consecutive values for k, you'll get all the solutions. i would choose: #k=0,1,2,3#
#k=0rarr6(cos((pi(0))/2)+i*sin((pi(0))/2))=6(1)=6#
#k=1rarr6(cos((pi(1))/2)+i*sin((pi(1))/2))=6(i)=6i#
#k=2rarr6(cos((pi(2))/2)+i*sin((pi(2))/2))=6(-1)=-6#
#k=3rarr6(cos((pi(3))/2)+i*sin((pi(3))/2))=6(-i)=-6i#
you can check by raising the roots to the 4th power and using #i^2=-1# to simplify
