Summer Problem Solving Marathon Solution #38

Let x + iy be an arbitrary root of the polynomial. Then the reciprocal of that root is 1/(x + iy). We then convert the denominator to a real number by multiplying by the complex conjugate x - iy, to obtain:

(x - iy)/((x + iy)(x - iy)) = (x - iy)/(x² + y²)

But because the root is on the circle centered at 0 + 0i with radius 1, we have, by the Pythagorean theorem, x² + y² = 1. So the reciprocal of the root is x - iy.

But this is the complex conjugate of the original root. Complex roots of polynomials with real-valued coefficients always comes in pairs by complex conjugates, so the reciprocal of the original root is another root of the polynomial.

Thus the reciprocals of the roots of the polynomial are just the roots of the polynomial. We thus want the sum of the roots of the polynomial. But the sum of the roots of any polynomial with a leading coefficient of 1 is the negative of the coefficient of the next-to-largest power of the variable. Hence the sum of the roots, which is also the sum of the reciprocals of the roots, is -a.