Real and Complex (Posted on 2007-01-23)

	Submitted by Dennis
Rating: 4.0000 (1 votes)
Solution:	(Hide)
If z=0 then by (b) 1=0 so z is not zero. Similarly, if z=1 then by (c) r=2 so z cannot be one. Now z^6-z^5-z+1=(z-1)^2(z^4+z^3+z^2+z+1)=0 --> z^4+z^3+z^2+z+1=0 --> (z+1/z)+(z^2+1/(z^2))=-1. Also z^2-rz+1=0 --> r=z+1/z --> (z+1/z)^2=z^2+1/(z^2)+2=r^2 --> z^2+1/(z^2)=r^2-2. So r+r^2-2=-1 --> r=(sqrt(5)-1)/2.

	Subject	Author	Date
	Puzzle Solution	K Sengupta	2008-07-21 04:26:11
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