2000 AIME II Problem 9

Attempt Problem 9 of the 2000 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AIME II solutions, or check the answer key.

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9.

Given that

z

is a complex number such that

z + \frac{1}{z} = 2\cos 3^\circ,

find the least integer that is greater than

z^{2000} + \frac{1}{z^{2000}}.

Answer

Answer: 0

Concepts:complex number De Moivre’s Theorem

Difficulty rating: 2330

Small hint:

If

z + \frac{1}{z} = 2\cos\theta,

then

z = \cos\theta \pm i\sin\theta

lies on the unit circle

Big hint:

By de Moivre,

z^{2000} + \frac{1}{z^{2000}} = 2\cos 6000^\circ;

reduce

6000^\circ

modulo

360^\circ

Solution:

From

z + \frac{1}{z} = 2\cos 3^\circ

we get

z^2 - (2\cos 3^\circ) z + 1 = 0,

so

z = \cos 3^\circ \pm i \sin 3^\circ,

a point on the unit circle. By de Moivre’s theorem,

\begin{aligned} z^{2000} + \frac{1}{z^{2000}} &= 2\cos(2000 \cdot 3^\circ) \\ &= 2\cos 6000^\circ. \end{aligned}

Since

6000 = 16 \cdot 360 + 240,

this equals

2\cos 240^\circ = -1.

The least integer greater than

-1

is

0.