Posted by
**Anonymous** on
.

by using the substitution w = z^3, find all the solutions to z^6 - 8z^3 +25 = 0 in complex numbers, and describe them in polar form, using @(theta) to denote the angle satisfying tan@ = 3/4 ( note simply leave @ as it is, don't calculate it).

i got up to z^3 = 4+3i and 4-3i then got stuck !

4 + 3i = 5 Exp[i theta]

The equation z^3 = Q for real positive Q has three solutions:

z = cuberoot[Q] Exp[2 pi n i/3]

for n = 0, 1 and 2, because

Exp[2 pi n i/3]^3 = Exp[2 pi n i] = 1

So, in this case you find:

z = 5^(1/3) Exp[i theta/3 + 2 pi n i/3 ]