This solution is from Tiffany of Island School. Andrei of Tudor
Vianu National College, Romania also sent in a good solution.
In the equation x2 + px + q = 0 if we fix q and vary p and
observe the complex solutions which occur (for p2 - 4q < 0) then, as
p changes, the complex roots of the equation also change. If these
roots are plotted on an Argand diagram then you will see that the
roots lie on a circle.
To prove that the complex roots do lie on a circle we use the fact
that the product of the roots of the quadratic equation x2 + px + q = 0 is q.
This equation has roots z1 and z2 given by
-p ±
______ Öp2 - 4q
2
so
z1 =
-p +
______ Öp2 - 4q
2
= u + iv
and
z2 =
-p -
______ Öp2 - 4q
2
= u - iv
where
u+iv and u-iv are complex conjugates.
The product of the complex conjugate roots is given by
z1z2 = (u+iv)(u-iv) = u2 + v2 = q.
Hence as the quadratic changes keeping q fixed and varying p the
locus of the complex roots in the (u,v) plane is the circle with
radius Öq centre at the origin.