DeMoivre's theorem tells us that:
e^(niθ) = (cos(θ) + isin(θ))^n = cos(nθ) + isin(nθ)
In this case, θ = arctan(-√3) = -π/6 and r = 2
So therefore (-1 + i√3)^12 = (2e^(-πi/6))^12 = 2^12 * (1) = 4096
(-1 + sqrt(3) * i)^(12) =>
2^12 * (-1/2 + i * sqrt(3)/2)^(12) =>
4096 * (cos(2pi/3) + i * sin(2pi/3))^12 =>
4096 * (cos(2pi/3 * 12) + i * sin(2p/3 * 12)) =>
4096 * (cos(8pi) + i * sin(8pi)) =>
4096 * (1 + i * 0) =>
4096 * 1 =>
4096
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Verified answer
DeMoivre's theorem tells us that:
e^(niθ) = (cos(θ) + isin(θ))^n = cos(nθ) + isin(nθ)
In this case, θ = arctan(-√3) = -π/6 and r = 2
So therefore (-1 + i√3)^12 = (2e^(-πi/6))^12 = 2^12 * (1) = 4096
(-1 + sqrt(3) * i)^(12) =>
2^12 * (-1/2 + i * sqrt(3)/2)^(12) =>
4096 * (cos(2pi/3) + i * sin(2pi/3))^12 =>
4096 * (cos(2pi/3 * 12) + i * sin(2p/3 * 12)) =>
4096 * (cos(8pi) + i * sin(8pi)) =>
4096 * (1 + i * 0) =>
4096 * 1 =>
4096