How do you set it up? It is like this. For instance The FIRST root of unity, is simply 1.

(A + i*sqrt(1 – A^{2) )}1 = 1 this means A + i*sqrt(1 – A^{2) = 1 + 0*i
So A = 1 and i*sqrt(1 — A}2) = 0 This leads to 1 — A^{2 = 0 —> A}2 = 1 —> A = 1 or -1

The answer: A = 1 AND (A = 1 or A = -1) since the AND is involved, it follows that A =1 ONLY

(1 + i*sqrt( 1 — 1) ) = 1, and 1 is the 1st root of unity.

The SQUARE ROOT of UNITY
[A + i*sqrt(1 — A^{2) ]}2 = 1
BINOMIAL EXPANSION
A^{2 + 2*i*sqrt(1 — A}2 ) — (1 - A^{2 ) = 1
Set the Real Side to be A}2 - 1 + A^{2 = 1 and IMAGINARY side = 0 = 2*i*sqrt(1 — A}2 ) = 0
(A^{2 + A}2 ) — 1 + 1 = 1 + 1 → 2*(A^{2) = 2; A}2 = 1 so A = 1 OR —1 on Real Side
On the Imaginary side 2*i*sqrt(1 - A^{2) = 0 thus sqrt( 1 — A}2) = 0 —> 1 - A^{2 = 0 —> A}2 = 1 thus A = 1 OR -1
So the TRUE solution is the one that is the solution on BOTH sides.
(A = 1 OR A = —1 ) AND (A = 1 OR A = -1) thus A = 1 OR A = -1
Then substituting A into the equation you get..
(1 + i*sqrt(0) ) OR ( -1 + i*sqrt( 1 - 1) ) yielding + 1 or —1

For the cube root of unity…

[A + i*sqrt(1 — A^{2) ]}3 = 1 → A^{3 + 3*(A}2)*i*sqrt(1 – A^2 ) - 3*A*(1 - A^2) - i*sqrt(1 — A^{2)*(1 — A}2) = 1
The REAL SIDE is set = 1 and IMAGINARY side = 0
REAL SIDE: A^{3 — 3*A*(1 — A}2) = 1 and i*sqrt(1 — A^2)*[3*(A2) — (1 — A^{2) ] = 0
4*(A}2) - 3*A - 1 = 0 AND [ i*qrt(1 — A^{2) = 0 OR [ 4*(A}2) — 1] = 0 The TRUE solution will be the one that is a SOLUTION for BOTH the REAL side and the IMAGINARY side. Since the IMAGINARY side is alot easier to handle we get…
i*sqrt(1 — A^{2) = 0 yielding A = 1 OR A = —1 but ALSO it has: 4*(A}2) — 1 = 0 → 4*(A^2) = 1 thus 2*A = 1 OR 2*A = —1
So, for the IMAGINARY side, the POSSIBLE solutions for A is A = 1 OR A = —1 OR A = 1/2 OR A = - 1/2
SUBSTITUTING, we get
[ -1 +i*sqrt( 0 ) ]^3 = (-1)^{3 = -1 , so A = —1 is NOT a solution.
A = 1
[1 + i*sqrt(1 - 1)]}3 = 1, so A =1
A = 1/2 we get [1/2 +i*sqrt( 1 – 1/4 ) = 1/2 + i*sqrt(3) / 2
At A = —1/2 we get -1/2 +-i*sqrt(3) / 2 Substituting these in, we can determine if ALL or SOME of them is the solution.