smc

Since $|z_0|=1$, let $z_0=e^{i{\theta }_0}$, where ${\theta }_0$ is an argument of $z_0$. We will prove by induction that $z_n=e^{i{\theta }_n}$, where ${\theta }_n=2^n({\theta }_0+\frac{\pi }{2})-\frac{\pi }{2}$. Base Case: trivial Inductive Step: Suppose the formula is correct for $z_k$, then $$z_{k+1}=\frac{i{z}_k}{\overline{z_k}}=i{e}^{i{\theta }_k}{e}^{i{\theta }_k}=e^{i(2{\theta }_k+\pi /2)}$$ Since $$2{\theta }_k+\frac{\pi }{2}=2\cdot 2^n({\theta }_0+\frac{\pi }{2})-\pi +\frac{\pi }{2}=2^{n+1}({\theta }_0+\frac{\pi }{2})-\frac{\pi }{2}={\theta }_{n+1}$$ the formula is proven $z_{2005}=1\Rightarrow {\theta }_{2005}=2k\pi$, where $k$ is an integer. Therefore, $$2^{2005}({\theta }_0+\frac{\pi }{2})=(2k+\frac{1}{2})\pi$$ $${\theta }_0=\frac{k}{2^{2004}}\pi +\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi$$ The value of ${\theta }_0$ only matters modulo $2\pi$. Since $\frac{k+2^{2005}}{2^{2004}}\pi \equiv \frac{k}{2^{2004}}\pi \mathrm{mod}2\pi$, k can take values from 0 to $2^{2005}-1$, so the answer is $2^{2005}\Rightarrow \boxed{2^{2005}}$