Let's say Z is a sum of nth roots of unity and thus an algebraic integer, and D is a rational integer. If z+D is an integer, what can we conclude regarding Z? Can we say Z is an integer? Another related question is: For which nonzero D can we conclude that Z is an integer if Z+D is an integer?
2 Answers
If $D$ is a nonzero rational number and $R$ is a positive number, the complex numbers $z$ with $z+D=R$ form the circle of radius $R$ centred at $D$. The intersection of this with the circle $z = k$ (if nonempty) consists of one or two points satisfying the quadratic $D z^2 + (D^2 + k^2  R^2) z + k^2 D = 0$. So if $z$ with $z+D^2$ rational has degree $> 2$ over the rationals, $z^2$ can't be a rational number.

$\begingroup$ So, what is this telling us regarding Z and Z?!! $\endgroup$– katieSep 17, 2012 at 3:34
Just to give a numerical example, if $\zeta=(1+i\sqrt3)/2$, a 6th root of unity, and $z=1+8\zeta=5+4i\sqrt3$, and $D=6$, then $z$ is a sum of roots of unity, $D$ is a rational integer, $z+D=11+4i\sqrt3=\sqrt{11^2+3\times4^2}=13$ is an integer, but $z=\sqrt{5^2+3\times4^2}=\sqrt{73}$ is not an integer.