Question 767477
<font face="Times New Roman" size="+2">


If *[tex \LARGE a\ +\ bi] is a zero of a polynomial function, then the conjugate, *[tex \LARGE a\ -\ bi] must also be a zero.  That is because complex roots always come in conjugate pairs.


So if *[tex \LARGE 0\ +\ 3i] is a zero, *[tex \LARGE 0\ -\ 3i] must also be a zero.  That means that both *[tex \LARGE x\ -\ 3i] and *[tex \LARGE x\ +\ 3i] must both be factors of the polynomial.  The product of a pair of conjugates is the difference of two squares so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ (x\ -\ 3i)(x\ +\ 3i)\ =\ x^2\ -\ (-9)\ =\ x^2\ +\ 9]


Divide the original polynomial by *[tex \LARGE x^2\ +\ 9] using polynomial long division.  The quotient will be the remaining factor from which solving for the remaining zero is trivial.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>