2019 AIME II Problem 8

Question

The polynomial

f(z) = az^{2018} + bz^{2017} + cz^{2016}

has real coefficients not exceeding

2019,

and

f\left(\frac{1 + \sqrt{3}i}{2}\right) = 2015 + 2019\sqrt{3}i.

Find the remainder when

f(1)

is divided by

1000.

Answer

\frac{1 + \sqrt{3}i}{2}

is a primitive sixth root of unity, and

2016

is a multiple of

6

Matching imaginary parts gives

a + b = 4038;

the bound on the coefficients then forces both values

Let

\omega = \frac{1 + \sqrt{3}i}{2} = \cos 60^\circ + i\sin 60^\circ,

a primitive sixth root of unity. Since

2016 = 6 \cdot 336,

we get

\omega^{2016} = 1,

\omega^{2017} = \omega,

and

\omega^{2018} = \omega^2 = \frac{-1 + \sqrt{3}i}{2}.

Therefore

\begin{aligned} f(\omega) &= a\omega^2 + b\omega + c \\ &= \left(c + \frac{b - a}{2}\right) \\ &\quad {}+ \frac{(a + b)\sqrt{3}}{2}\,i. \end{aligned}

Matching imaginary parts,

\frac{a + b}{2} = 2019,

so

a + b = 4038.

Since

a \le 2019

and

b \le 2019,

this forces

a = b = 2019.

Matching real parts then gives

c + 0 = 2015,

so

c = 2015.

Hence

f(1) = a + b + c

= 4038 + 2015 = 6053,

whose remainder upon division by

1000

is

53.

Problem 8 in other years