Question 24. Let the equation $x^{2}+m x+m=0$ have the complex root $a+b i(b>0)$, as the real number $m$ varies, what kind of figure does the point $(\mathrm{a}, \mathrm{b})$ trace out?
Solve $x^{2}+m x+m=0$.
Equation (1) has a root $a+b i$, and thus also has the conjugate root, $a-bi$, so
$$
\begin{array}{l}
(a+b i)+(a-b i)=-m, \\
(a+b i)(a-b i)=m .
\end{array}
$$

From (2) we get $n_{1}=-2 a$
From (3) we get $a^{2}+b^{2}=m$.
Substituting (3), we get $a^{2}+2 a+b^{2}=0$,
which is $(a+1)^{2}+b^{2}=1$.
$$
\because b>0 \text {, }
$$
$\therefore$ the point $(a, b)$ lies on the upper semicircle with center at $(-1,0)$ and radius 1, but does not include the intersection points of the semicircle with the x-axis $(-2,0)$ and $(0,0)$.