Let f(X) be a complex monic quadratic with real roots 1/3 and 2/3 (The polynomial f(X) is of the form X^2 + bX + c where b, c, X are complex numbers.) If |z| = 1, what is the sum of all possible values of f(z) such that f(z) = f(z)?
This is from the Stanford Math Tournament Problem number 6. https://sumo.stanford.edu/pdfs/smt2019/algebra-solutions.pdf
First of all, I would like to ask that if the roots of f(X) are 1/3 and 2/3 then wouldn't f(X) be equivalent to x^2+x+2/9? However, the problem says the coefficients b and c are complex. How can this be possible?
The polynomial f(z) = z$^2$ - z + 2/9.
As |z| = 1, exists real t with z = cos t + sin t.
Assume f(z) = conj f(z). Thus
cos$^2$ t + 2i.cos t sin t - sin$^2$ t - cos t - i.sin t + 2/9 =
cos$^2$ t - 2i.cos t sin t - sin$^2$ t - cos t + i.sin t + 2/9.
Simplifying:
2i.cos t sin t - i.sin t = -2i.cos t sin t + i.sin t;
4.cos t sin t = 2.sin t; sin t (2.cos t - 1) = 0.
The solutions to this equation are 0, 180, 60, 270 degrees.
Find the value of f for each of those roots and add the values together.