a) The polar form of 1+i3 is
2(cos(3π)+isin(3π))We have that r=2,θ=3π,n=3.
According to the De Moivre's Formula
k=0:
32(cos(3π/3+2π(0))+isin(3π/3+2π(0)))
=32(cos(9π)+isin(9π))
k=1:
32(cos(3π/3+2π(1))+isin(3π/3+2π(1)))
=32(cos(97π)+isin(97π))
=32(−cos(9π)+isin(9π))
k=2:
32(cos(3π/3+2π(2))+isin(3π/3+2π(2)))
=32(cos(913π)+isin(913π))
=32(−cos(94π)−isin(94π))
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