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Home » لنفترض $w = z^2$, إذن: - Blask

لنفترض $w = z^2$, إذن: - Blask

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 \boxed{\left( -\dfrac{4}{3}, \dfrac{3}{4} \right)}

📌 Question: The maximum imaginary part of a root of $ z^5 + z^3 + z + 1 = 0 $ can be expressed as $ \sin \theta $. Find $ \theta $ in degrees.

📌 Solution: Factor the polynomial:

📌 z^5 + z^3 + z + 1 = z^3(z^2 + 1) + 1(z + 1).

📌 Group terms:

📌 Roots: $ z^3 = -1 \implies z = e^{i\pi/3 + 2k\pi/3} $, and $ z^2 = -1 \implies z = e^{i\pi/2}, e^{i3\pi/2} $.

📌 The roots with maximum imaginary part are $ e^{i\pi/3} $ and $ e^{

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But we are given four values, so we reconstruct $ p(x) $.
Let $ p(x) = ax^3 + bx^2 + cx + d $
Use the values:
$ p(0) = d = 2 $
$ p(1) = a + b + c + d = 5 \Rightarrow a + b + c = 3 $
$ p(2) = 8a + 4b + 2c + d = 12 \Rightarrow 8a + 4b + 2c = 10 $
$ p(3) = 27a + 9b + 3c + d = 31 \Rightarrow 27a + 9b + 3c = 29 $
From (2): $ a + b + c = 3 $
Multiply by 2: $ 2a + 2b + 2c = 6 $
Subtract from (3):

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