Problem:
The polynomial f(z)=az2018+bz2017+cz2016 has real coefficients not exceeding 2019, and f(21+3βiβ)=2015+20193βi. Find the remainder when f(1) is divided by 1000.
Solution:
We first compute the first few powers of 21+3βiβ in search of a periodic sequence:
(21+3βiβ)1=21+3βiβ
(21+3βiβ)2=2β1+3βiβ
(21+3βiβ)3=β1
(21+3βiβ)4=2β1β3βiβ
(21+3βiβ)5=21β3βiβ
(21+3βiβ)6=1
So the powers of 21+3βiβ repeat in a cycle of 6.
Let Ο=21+3βiβ. Consider the expression f(Ο)=aΟ2+bΟ+c.
We are given that f(Ο)=2015+20193βi.
Using the identities Ο=21+3βiβ and Ο2=2β1+3βiβ, we compute:
f(Ο)=aβ 2β1+3βiβ+bβ 21+3βiβ+c
=2βa+a3βi+b+b3βiβ+c
=(2βa+bβ+c)+2a+bβ3βi
Now we match real and imaginary parts with 2015+20193βi:
From the real part: 2βa+bβ+c=2015 ββa+b+2c=4030(1)
From the imaginary part: 2a+bβ=2019 βa+b=4038(2)
From equation (2): b=4038βa. Substitute into equation (1):
βa+(4038βa)+2c=4030
4038β2a+2c=4030ββ2a+2c=β8βc=aβ4
Now substitute into f(1)=a+b+c:
f(1)=a+(4038βa)+(aβ4)=4038+aβ4=a+4034
We are told all coefficients are at most 2019, so a=2019, b=2019, and c=2015.
Thus, f(1)=2019+2019+2015=6053.
The remainder when 6053 is divided by 1000 is 53β.