Calculate #root(6)(-64)# means you have to find a real number #x# such that #x^6=-64#. Such number doesnt exist because if it were positive, then never will get a negative number as product, if it were negative, then
#(-x)·(-x)·(-x)·(-x)·(-x)·(-x)=# positive number (there are an even number of factors (6) and never will get #-64#)
In summary that #root(6)(-64)# has no real solutions. There is no number #x# such that #x^6=-64#
But in complex set of numbers there are 6 solutions
First put #-64# in polar form which is #64_180#
Then the six solutions #r_i# from i=0 to i=5 are
#r_0=root(6)64_(180/6)=2_30#
#r_1=root(6)64_((180+360)/6)=2_90#
#r_2=2_((180+720)/6)=2_150#
#r_3=2_((180+1080)/6)=2_210#
#r_4=2_270#
#r_5=2_330#
Who are these numbers?
#r_0=2(cos30+isin30)=sqrt3+i#
#r_1=2i#
#r_2=-sqrt3+i#
#r_3=-sqrt3-i#
#r_4=-2i#
#r_5=sqrt3-i#
