= Solution
The <quadratic formula> gives $z=-b\pm\sqrt{b^2-1}$. For $-1\leq b\leq1$, write this as
$$
\boxed{z_\pm=-b\pm i\sqrt{1-b^2}.}
$$
Each <root of a polynomial> has <modulus> $1$, since $b^2+(1-b^2)=1$. As $b$ increases from $-1$ to $1$, the upper root traverses the upper unit semicircle from $1$ to $-1$, and the lower root traverses the lower semicircle in the opposite angular direction. At $b=-1$ the roots coincide at $1$; at $b=1$ they coincide at $-1$. For intermediate $b$ they are <complex conjugates>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1-quadratic-root-locus.png]
{title=The conjugate roots travel along the two unit semicircles}
{description=Arrows show increasing $b$, with endpoints $b=-1$ and $b=1$ and the midpoint pair $b=0$. The figure is an original plot of the solution formula.}
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