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Home » $ z = \pm 1 $, both real, so imaginary part is 0. - Dyverse

$ z = \pm 1 $, both real, so imaginary part is 0. - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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a^2 = 1, b^2 = e^{i4\pi/3}, a^2 - b^2 = 1 - e^{i4\pi/3}, a^2 + b^2 = 1...
Numerator: $ 2(1 + e^{i4\pi/3}) $, denominator: $ 1 - e^{i4\pi/3} $, s...
S = 2i \cdot rac{e^{i2\pi/3} + e^{i2\pi/3}}{e^{i2\pi/3} - e^{-i2\pi/3...
But from earlier general form $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $, a...
eq b $, and $ |a| = |b| = 1 $, the only consistent value from examples...
oxed{0}
Question: Find the quadratic polynomial $ p(x) $ such that $ p(1) = 4...
Solution: Let $ p(x) = ax^2 + bx + c $. Using the given values:
a(1)^2 + b(1) + c = a + b + c = 4, \quad a(4) + 2b + c = 4a + 2b + c =...
Subtract the first equation from the second: $ 3a + b = 7 $. Subtract...

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