Problem:
Consider all polynomials of a complex variable, P(z)=4z4+az3+bz2+cz+d, where a,b,c, and d are integers, 0≤d≤c≤b≤a≤4, and the polynomial has a zero z0 with ∣z0∣=1. What is the sum of all values P(1) over all the polynomials with these properties?
Answer Choices:
A. 84
B. 92
C. 100
D. 108
E. 120
Solution:
If z0k is equal to a positive real r, then 1=∣z0∣k=∣∣∣z0k∣∣∣=∣r∣=r, so z0k=1. Suppose that z0k=1. If k=1, then z0=1, but P(1)=4+a+b+ c+d≥4 so z0=1 is not a zero of the polynomial. If k=2, then z0=±1. If z0=−1, then 0=P(−1)=(4−a)+(b−c)+d and by assumption 4≥a, b≥c, and d≥0. Thus a=4,b=c, and d=0. Conversely, if a=4,b=c, and d=0, then P(z)=4z4+4z3+bz2+bz=z(z+1)(4z2+b) satisfies the required conditions. If k=3, then z0=1 or z0=γ where γ is any of the roots of γ2+γ+1=0. If z0=γ, then 0=P(γ)=4γ+a+b(−1−γ)+cγ+d=
(a−b)+d+γ((4−b)+c) and by assumption a≥b,d≥0,4≥b, and c≥0. Thus a=b,d=0,b=4, and c=0. Conversely, if a=b=4 and c=d=0, then P(z)=4z4+4z3+4z2=4z2(z2+z+1) satisfies the given conditions because z0=cos(2π/3)+isin(2π/3) is a zero of this polynomial. If k=4, then z0=±1 or z0=±i. If z0=±i, then 0=P(±i)=4∓ia−b±ic+d= (4−b)+d∓i(a−c) and by assumption 4≥b,d≥0, and 4≥a≥b≥c. Thus b=4,d=0, and a=c=4. Conversely, if a=b=c=4 and d=0, then P(z)=4z4+4z3+4z2+4z=4z(z+1)(z2+1) satisfies the given conditions, but it was already considered in the case when z0=−1. The remaining case is that z0k is not a positive real number for 1≤k≤4. In this case,
4z5−(z−1)P(z)=z4(4−a)+z3(a−b)+z2(b−c)+z(c−d)+d
If z=z0, then the triangle inequality yields
4=∣∣∣z04(4−a)+z03(a−b)+z02(b−c)+z0(c−d)+d∣∣∣≤∣∣∣z04(4−a)∣∣∣+∣∣∣z03(a−b)∣∣∣+∣∣∣z02(b−c)∣∣∣+∣z0(c−d)∣+∣d∣=∣z0∣4(4−a)+∣z0∣3(a−b)+∣z0∣2(b−c)+∣z0∣(c−d)+d=4−a+a−b+b−c+c−d+d=4
Thus equality must occur throughout. This means that the vectors v4=z04(4− a),v3=z03(a−b),v2=z02(b−c),v1=z0(c−d), and v0=d are parallel and they belong to the same quadrant. If two of these vectors are nonzero, then the quotient must be a positive real number; but dividing the vector with the largest exponent of z0 by the other would yield a positive rational number times z0k for some 1≤k≤4. Because not all of the vj can be zero, it follows that there is exactly one of them that is nonzero. If v0=d=0 and v1=v2=v3=v4=0, then 4=a=b=c=d, and P(z)=4z4+4z3+4z2+4z+4 satisfies the given conditions because z0=cos(2π/5)+isin(2π/5) is a zero of this polynomial. Finally, if vj=0 for some 1≤j≤4 and the rest are zero, then 4z05=vj=z0jn for some positive integer n, and so z05−j=41n is a positive real.
Therefore the complete list of polynomials is: 4z4+4z3+4z2+4z+4,4z4+ 4z3+4z2, and 4z4+4z3+bz2+bz with 0≤b≤4. The required sum is 20+12+∑b=04(8+2b)=32+40+(2+4+6+8)=92.
The problems on this page are the property of the MAA's American Mathematics Competitions