Write [2(cos 5π/16 + i sin 5π/16)]^4 in the standard a+bi?
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By De Moivre's Theorem
[2(cos 5π/16 + i sin 5π/16)]^4 = 2^4 (cos(20π/16) + i sin(20π/16)) =
16(-1/√(2) - i/√(2)) = -8√(2)(1 + i).
[2(cos 5π/16 + i sin 5π/16)]^4 = 2^4 (cos(20π/16) + i sin(20π/16)) =
16(-1/√(2) - i/√(2)) = -8√(2)(1 + i).