Question 428149
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Your answer is exactly correct.  For a polynomial equation with real coefficients, complex roots always occur in conjugate pairs.  The irrational conjugate roots theorem says: Let *[tex \Large p(x)] be any polynomial with rational coefficients. If *[tex \Large  a\ +\ b\sqrt{c}] is a root of *[tex \Large  p(x)], where *[tex \Large \sqrt{c}] is irrational and *[tex \Large a] and *[tex \Large b] are rational, then another root is *[tex \Large a\ -\ b\sqrt{c}].


So, if you have a complex root *[tex \Large a\ +\ bi] then you are guaranteed to have another complex root *[tex \Large a\ -\ bi].  Likewise, if you have an irrational root *[tex \Large p\ +\ r] where *[tex \Large p\ \in\ \mathbb{Q}] and *[tex \Large r\ \in\ \overline{\mathbb{Q}}], then you are guaranteed to have another irrational root *[tex \Large p\ -\ r]


Therefore *[tex \Large 2\ -\ 3i] is a root guarantees that *[tex \Large 2\ +\ 3i] is a root and *[tex \Large 0\ +\ \sqrt{7}] guarantees *[tex \Large 0\ -\ \sqrt{7}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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