Question

A sequence of complex numbers $z_0, z_1, z_2, \ldots$ is defined by the rule $z_{n+1} = \dfrac{i z_n}{\overline{z_n}},$ where $\overline{z_n}$ is the complex conjugate of $z_n$ and $i^2 = -1.$ Suppose that $|z_0| = 1$ and $z_{2005} = 1.$ How many possible values are there for $z_0?$

$1$

$2$

$4$

$2005$

$2^{2005}$

Answer: E

Concepts:complex number roots of unity recursion

Difficulty rating: 2170

Solution:

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.

2005 AMC 12B Problem 22

Answer: E

Difficulty rating: 2170

Solution:

Problem 22 in Other Years