Question 1125241
<font face="Times New Roman" size="+2">


Since f is of degree 4, there are exactly 4 zeros, counting multiplicities.  Complex zeros ALWAYS appear in conjugate pairs, i.e. if *[tex \Large a\ +\ bi] is a zero, then *[tex \Large a\ -\ bi] must also be a zero.


Since *[tex \Large 0\ +\ i] is a zero, then *[tex \Large 0\ -\ i] must also be a zero.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>