9th Putnam 1949 Problem A5 Let p(z) ≡ z 6 + 6z + 10. How many roots lie in each quadrant of the complex plane?
There are no real roots, since p(z) has a minimum (for real value of z) at z = -1 of 5. Similarly there are no purely imaginary roots, since p(i y) has imaginary part 6iy (and z = 0 is not a root). Also a + ib is a root iff a - ib is a root, so either there is one root in each of the 1st and 4th quadrants and two in each of the 2nd and 3rd, or vice versa. The argument principle states that the change in arg p(z) as we move (anti-clockwise) round a closed contour which passes through no zeros and contains k zeros is 2 π k. Take a contour in the first quadrant from z = 0 to z = X, then around the arc |z| = X to iX, then back along the imaginary axis to z = 0. Moving along the real axis, arg p(z) does not change. Moving along the arc the change will be the same as that in arg z 6 for X sufficiently large and hence will be 3π. As we move back down the imaginary axis, the imaginary part remains positive, so p(z) is confined to the upper half plane, and we end up at p(z) = 10, so the change must be -π. Thus the change for a complete circuit is 2π, and so there is just one root in the first quadrant. 9th Putnam 1949 © John Scholes jscholes@kalva.demon.co.uk 11 Mar 2002