2005 AMC 12B Problem 22

Because $|z_0| = 1,$ every $|z_n| = 1,$ so $\overline{z_n} = \dfrac{1}{z_n}$ and $$z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2.$$ Iterating, $z_1 = i z_0^2$ and $z_2 = i(i z_0^2)^2 = -i z_0^4.$ Moreover, if $z_n=-i z_0^{2^n},$ then $z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}.$ Thus $z_n=-i z_0^{2^n}$ for every $n\ge2.$ The condition $z_{2005} = 1$ is therefore $z_0^{2^{2005}} = i.$ Every nonzero complex equation $z_0^{N} = i$ has exactly $N$ distinct solutions, all on the unit circle. Here $N = 2^{2005},$ so there are $2^{2005}$ possible values for $z_0.$ Thus, the correct answer is E.