Question

given z^6=1, find all complex solutions for z. Can anyone guide me through the steps required to solve this?

Answer 1

Use polar coordinates for z, \[z = re^{iφ}\] then set radius and arcs equal, but observe periodicity.

Answer 2

This can be written as z^6 - 1 = 0
then z^6 - 1^6 = 0

Continue with sum and difference of cubes formula:

(z^3 + 1)(z^3 - 1) = 0
(z + 1)(z^2 - z + 1)(z - 1)(z^2 + z + 1) = 0

This yields z = ±1 and z = (±1 ± i√3)/2. That's all 6.

Answer 3

Hope this helps, I have time constraint or would hv gone into details

Answer 4

thanks