Problem

Expand the expression (1+i3)6 using De Moivre's theorem.

Answer

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Answer

Finally, we convert back to rectangular form. Since cos(2π)=1 and sin(2π)=0, our final answer is 64(1+i⋅0)=64.

Steps

Step 1 ：De Moivre's theorem states that (cos(θ)+isin(θ))n=cos(nθ)+isin(nθ). We'll use this to simplify the expression.

Step 2 ：First, we convert the complex number 1+i3 into polar form. The magnitude r is 12+(3)2=2. The angle θ is arctan⁡(31)=π3. So 1+i3=2(cos(π3)+isin(π3)).

Step 3 ：Now we can apply De Moivre's theorem. (1+i3)6=(2(cos(π3)+isin(π3)))6=26(cos(6⋅π3)+isin(6⋅π3))=64(cos(2π)+isin(2π)).

Step 4 ：Finally, we convert back to rectangular form. Since cos(2π)=1 and sin(2π)=0, our final answer is 64(1+i⋅0)=64.

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