Question 465666
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The actual complex number for the given zero is *[tex \Large 0\ +\ 2i], but we know that complex roots ALWAYS come in conjugate pairs, which is to say if *[tex \Large a\ +\ bi] is a root, then *[tex \Large a\ -\ bi] is also a root -- guaranteed.


So our required polynomial has at least three factors, some constant *[tex \Large a\ \in\ \mathbb{R},\ \ ] and then the two complex factors: *[tex \Large x\ +\ 2i] and *[tex \Large x\ -\ 2i]


A little FOIL keeping in mind that *[tex \Large i^2\ =\ -1] and the distributive property gets us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax^2\ +\ 4a]


which is the family of polynomial functions of <i>least degree</i> that have *[tex \Large 2i] as a zero.


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