Problem:
Let P(z)=z3+az2+bz+c, where a,b, and c are real. There exists a complex number w such that the three roots of P(z) are w+3i,w+9i, and 2w−4, where i2=−1. Find ∣a+b+c∣.
Solution:
Let w=x+yi, where x and y are real. Then because a is real and the sum of the three roots is −a, it follows that Im((w+3i)+(w+9i)+(2w−4))=0. Thus y+3+y+9+2y=0, and y=−3. Therefore the three roots are x,x+6i, and 2x−4−6i. Because the coefficients of P(z) are real, the non-real roots must occur in conjugate pairs, and so x=2x−4 and x=4. Thus P(z)=(z−4)(z−(4+6i))(z−(4−6i)) and 1+a+b+c=P(1)= (−3)(−3−6i)(−3+6i)=−135. Thus ∣a+b+c∣=∣−135−1∣=136. Such a polynomial exists: P(z)=z3−12z2+84z−208 has the zeros 4,4±6i, which satisfy the conditions of the problem for w=4−9i.
The problems on this page are the property of the MAA's American Mathematics Competitions