*** Find (√3+i)^12 ***

Note that, in polar form:
√3 + i = 2(cos π/6 + i sin π/6).

Therefore, by DeMovire's Theorem:
(√3 + i)^12 = [2(cos π/6 + i sin π/6)]^12
= 2^12 * [cos(12 * π/6) + i sin(12 * π/6)]
= 2^12 * (cos 2π + i sin 2π)
= 2^12, since cos(2π) = 1 and sin(2π) = 0.

I hope this helps!

I'll give you two answers, because I'm not sure if it's rad(3+i) or (rad(3) + i)...
If it's rad(3+i)^12, then that is the same as ((3+i)^(1/2))^12 = (3+i)^6 = ((3+i)^2)^3
To get the final answer for that, first square 3+i, then cube the resulting answer.

If it's (rad(3) + i)^12, then separate the exponents into (((rad(3)+i)^2)^2)^3
To get the answer, square (rad(3)+i), square your result, and then cube your next result.

Either way, it looks like you'll have a lot of multiplication to do, so don't forget that i^2 is negative.

I think,
(sqrt 3 +i ) ^12

=(1.732 +- 1)^12
= 1.732 +1)^12 = (2.732)^12 =172889.38...............1)
Next one,
=(1.732 -1)^12 = (.732)^12 =.0236.........................2)
1) & 2 ) expressions are the answer.

3^6 + 12×(√3)^11i + [(12×11)/(2)]×(√3)^10×i² + [(12×11×10)/(2×3)]×(√3)^9×i³ + ... + 12×√3×i^11 + i^12

=
729 + 2916i√3 - 16038 - 17820i√3 + 40095 + 21384i√3 - 24948 - 7128i√3 + 4455 + 660i√3 - 198 - 12i√3 + 1

= (729 - 16038 + 40095 - 24948 + 4455 - 198 + 1) + i√3×(2916 - 17820 + 21384 - 7128 + 660 - 12)

= 4096