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Question: Given that $e^{i \theta}$ is a root to the equation $$z^3 - (\sqrt{3} +1)z^2+\lambda z - 1 =0,$$ where $0< \theta < \pi/2$ and $\lambda \in \mathbb{R}$, show that this equation can be factorized as $$(z^2 - 3z \cos \theta +1) (z-1) = 0.$$

My problem: I know that since all coefficients are real numbers, the complex roots are in conjugate pairs. Multiplying out the conjugate pairs I am able to get the quadratic factor. However, I'm not sure how to show that $z-1$ is a factor without first finding the value of $\lambda$? I know that there are three roots, and the last root has to be a real root since otherwise a complex conjugate root also exists, making it four roots in total. I'm sure I'm not supposed to solve for $\lambda$ first because the next part of this question asks us to find the value of $\lambda$ given the above factorization.

This is a homework assignment, so please give me some hints! Thank you.

Micah
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troggz
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2 Answers2

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Suppose the third factor is $z-a$. What is the constant term of the cubic?

Empy2
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Since $e^{i\theta}$ is a root, so is its conjugate. Their product is 1.

From the constant term, the product of all the roots is 1.

Therefore 1 is a root.

You now know all the roots. Take it from there.

marty cohen
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