Answer:
- 1, cos(2π/3)+i·sin(2π/3), cos(4π/3) +i·sin(4π/3)
- 1, i, -1, -i
Explanation:
Euler's formula comes into play here. That tells you ...

Then 1 can be written as ...

Then the n-th root of 1 will be ...

Since the trig functions are periodic with period 2π, useful values of k are from 0 to n-1.
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1) The cube roots of 1 are ...
cos(2πk/3) +i·sin(2πk/3) . . . . for k = 0 to 2
= {1, cos(2π/3) +i·sin(2π/3), cos(4π/3) +i·sin(4π/3)}
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2) The fourth roots of 1 are ...
cos(2πk/4) +i·sin(2πk/4) = cos(kπ/2) +i·sin(kπ/2)
= {1, i, -1, -i}