2005 AMC 12B Problem 22

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}

Since

|z_n| = 1,

\overline{z_n} = \dfrac{1}{z_n},

so

z_{n+1} = i z_n^2

Iterating gives

z_n

as a fixed constant times

z_0^{2^n}

; count roots of the resulting equation

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.

Problem 22 in other years