Question

Consider all polynomials of a complex variable, $P(z) = 4z^4 + az^3 + bz^2 + cz + d,$ where $a,$ $b,$ $c,$ and $d$ are integers, $0 \le d \le c \le b \le a \le 4,$ and the polynomial has a zero $z_0$ with $|z_0| = 1.$ What is the sum of all values $P(1)$ over all the polynomials with these properties?

$84$

$92$

$100$

$108$

$120$

Solution:

Because $|z_0|=1,$ applying the triangle inequality to the identity $4z_0^5-(z_0-1)P(z_0)=z_0^4(4-a)+z_0^3(a-b)+z_0^2(b-c)+z_0(c-d)+d$ forces equality throughout, so all but one of the coefficient differences vanish.

Working through the cases (including $z_0=-1$ and $z_0=\gamma$ a primitive cube root of unity), the polynomials are exactly $4z^4+4z^3+4z^2+4z+4,$ $4z^4+4z^3+4z^2,$ and $4z^4+4z^3+bz^2+bz$ for $0\le b\le4.$

Their values at $1$ are $20,$ $12,$ and $8+2b;$ summing gives $20+12+\sum_{b=0}^{4}(8+2b)=32+40+20=92.$

Thus, the correct answer is B.

2012 AMC 12B Problem 23

Solution:

Problem 23 in Other Years