You can put this solution on YOUR website!
Using Euler's:
y = -64*e^((i*(pi)/2))
y^(1/3) = (-64*e^((i*(pi)/2)) ) ^(1/3)
y^(1/3) = -4* ( e^(i*(pi)/2) ) ^(1/3)
Now, before multiplying through by 1/3 on the right notice that because e^(i*pi*a) = cos(pi*a) + i*sin(pi*a) the value repeats every 2k*pi, so insert a 2k*pi into the exponent:
y^(1/3) = -4[e^(i*(pi)/2 + 2k(pi)) ]^(1/3) [ eq 1 ]
Let the roots be r0, r1, r2 (corresponding to k=0,1,2):
k=0 —>
I will leave it to you to find the other two roots (set k=1 in [eq 1] and simplify , then set k=2 and simplify )
Let me know if you have any questions.
