Problem: Let ω=cos7272π+isin7272π. The imaginary part of the complex number k=8∏13(1+ω3k−1+ω2⋅3k−1) is equal to sinα for some angle α between −2π and 2π, inclusive. Find α.
Solution
Solution: 727=36−2. Our product telescopes to 1−ω371−ω313=1−ω61−ω12=1+ω6, which has imaginary part sin72712π, giving α=72712π.
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