Why f(x) = x^n - r e^{i \theta}? Seeking Analytic Explanation

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The discussion centers on the identity f(x) = x^n - r e^{i θ} and its relation to the roots of the function. Participants clarify that the right-hand side can be expressed as a product of the n-th roots of the complex number in polar form, r e^{i θ}. This leads to a rephrasing of the inquiry as a request for a proof of the Fundamental Theorem of Algebra, which states that every non-constant polynomial has at least one complex root. The n-th roots of unity are also introduced as essential components in understanding this identity.

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I'd like to figure out why

f(x) = x^n - r e^{i \theta} = \prod_{j = 0}^{n - 1} ( x - r^{1/n} e^{i \theta / n} e^{2 \pi i j / n} ),

as I've seen it used as an identity in a few courses but I cannot figure it out. Could somebody shed some light? I understand they are the roots of the function, but I'd like some kind of analytic description.

Thanks a lot!
 
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Hi,
I think you may have answered your own question: if you are willing to accept that the RHS above can be rewritten as
<br /> \prod_{j = 0}^{n - 1} ( x - a_j )<br />
where a_j (with j=0,1,...,n-1) are the n-th roots of the complex number in polar form r e^{i \theta}, then your question can be rephrased as: "can somebody point me to a proof of the Fundamental Theorem of Algebra"?

My 2 cents.
 
Last edited:
You have ##x^n = re^{i\theta}##, so taking the n'th root of each side you get
$$x = \sqrt[n]{re^{i\theta}}\,\omega_k, \quad k = 0,\dots,n-1$$
where the ##\omega_k## are the n'th roots of unity.
 

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