If $a$ satisfies $a=\sqrt{z+a}$ then $a/z = 1/(a-1)$ which usually does not have modulus $1$.  So for $1/(a-1) = e^{i\theta}$ we get $z = e^{-2i\theta}+e^{-i\theta}$.

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  [1]: http://i44.tinypic.com/2589vub.jpg