Problem:
Consider all polynomials of a complex variable, P(z)=4z4+az3+bz2+cz+d, where a,b,c, and d are integers, 0β€dβ€cβ€bβ€aβ€4, and the polynomial has a zero z0β with β£z0ββ£=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 z0kβ is equal to a positive real r, then 1=β£z0ββ£k=β£β£β£βz0kββ£β£β£β=β£rβ£=r, so z0kβ=1. Suppose that z0kβ=1. If k=1, then z0β=1, but P(1)=4+a+b+ c+dβ₯4 so z0β=1 is not a zero of the polynomial. If k=2, then z0β=Β±1. If z0β=β1, then 0=P(β1)=(4βa)+(bβc)+d and by assumption 4β₯a, bβ₯c, and dβ₯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 z0β=1 or z0β=Ξ³ where Ξ³ is any of the roots of Ξ³2+Ξ³+1=0. If z0β=Ξ³, then 0=P(Ξ³)=4Ξ³+a+b(β1βΞ³)+cΞ³+d=
(aβb)+d+Ξ³((4βb)+c) and by assumption aβ₯b,dβ₯0,4β₯b, and cβ₯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 z0β=cos(2Ο/3)+isin(2Ο/3) is a zero of this polynomial. If k=4, then z0β=Β±1 or z0β=Β±i. If z0β=Β±i, then 0=P(Β±i)=4βiaβbΒ±ic+d= (4βb)+dβi(aβc) and by assumption 4β₯b,dβ₯0, and 4β₯aβ₯bβ₯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 z0β=β1. The remaining case is that z0kβ is not a positive real number for 1β€kβ€4. In this case,
4z5β(zβ1)P(z)=z4(4βa)+z3(aβb)+z2(bβc)+z(cβd)+d
If z=z0β, then the triangle inequality yields
4=β£β£β£βz04β(4βa)+z03β(aβb)+z02β(bβc)+z0β(cβd)+dβ£β£β£ββ€β£β£β£βz04β(4βa)β£β£β£β+β£β£β£βz03β(aβb)β£β£β£β+β£β£β£βz02β(bβc)β£β£β£β+β£z0β(cβd)β£+β£dβ£=β£z0ββ£4(4βa)+β£z0ββ£3(aβb)+β£z0ββ£2(bβc)+β£z0ββ£(cβd)+d=4βa+aβb+bβc+cβd+d=4β
Thus equality must occur throughout. This means that the vectors v4β=z04β(4β a),v3β=z03β(aβb),v2β=z02β(bβc),v1β=z0β(cβd), and v0β=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 z0β by the other would yield a positive rational number times z0kβ for some 1β€kβ€4. Because not all of the vjβ can be zero, it follows that there is exactly one of them that is nonzero. If v0β=dξ =0 and v1β=v2β=v3β=v4β=0, then 4=a=b=c=d, and P(z)=4z4+4z3+4z2+4z+4 satisfies the given conditions because z0β=cos(2Ο/5)+isin(2Ο/5) is a zero of this polynomial. Finally, if vjβξ =0 for some 1β€jβ€4 and the rest are zero, then 4z05β=vjβ=z0jβn for some positive integer n, and so z05βjβ=41βn 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 0β€bβ€4. The required sum is 20+12+βb=04β(8+2b)=32+40+(2+4+6+8)=92β.
The problems on this page are the property of the MAA's American Mathematics Competitions