Question 668545
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If a polynomial equation has a zero at *[tex \LARGE\ \alpha], then *[tex \LARGE x\ -\ \alpha] is a factor of the polynomial.  Complex zeros of polynomials with rational coefficients ALWAYS come in conjugate pairs.  That is to say, if *[tex \LARGE a\ +\ bi] is a zero of a polynomial with rational coefficients then *[tex \LARGE a\ -\ bi] is also a zero of that polynomial.


Given *[tex \LARGE 3i] and *[tex \LARGE 2\ -\ i] as zeros means that you have at least four zeros, namely *[tex \LARGE \pm{3i}] and *[tex \LARGE 2\ \pm\ i].  Also, since you want to derive the polynomial of least degree under the given circumstances, you cannot have any more than four zeros.  Hence, the given values are all of the zeros.


That means that *[tex \LARGE x\ -\ 3i\ ,\ x\ +\ 3i\ ,\ x\ -\ (2\ -\ i)] and *[tex \LARGE x\ -\ (2\ +\ i)] are the four factors of your polynomial.  Multiply them and collect like terms.  Hints for easier calculation: First, remember that the product of two conjugates is the difference of two squares.  Second, *[tex \LARGE i^2\ =\ -1].  Taken together these facts mean that the product of two complex conjugates is the SUM of two squares.


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