Question 391373
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Complex zeros of polynomial functions with real coefficients come in conjugate pairs.  That means that if *[tex \Large \alpha\ +\ \beta i] is a zero of a given polynomial function, then *[tex \Large \alpha\ -\ \beta i] must also be a zero of that same polynomial function.


The conjugate of *[tex \Large 2\ +\ i] is *[tex \Large 2\ -\ i] and the conjugate of *[tex \Large 0\ +\ 6i] is *[tex \Large 0\ -\ 6i].  Now we know that we have at least 4 zeroes, so the minimum degree polynomial function is 4th degree.


If *[tex \Large z] is a zero of the polynomial function *[tex \Large \rho(x)], then *[tex \Large x\ -\ z] is a factor of *[tex \Large \rho(x)].


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ \left(x\ -\ (2\ +\ i)\right)\left(x\ -\ (2\ -\ i)\right)\left(x\ -\ 6i\right)\left(x\ +\ 6i\right)]


All you need to do is multiply the factors to obtain a general form representation of your polynomial.


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