21. The equation
$$
(z-1)\left(z^{2}+2 z+4\right)\left(z^{2}+4 z+6\right)=0
$$

has five complex roots $x_{k}+\mathrm{i} y_{k}\left(x_{k} 、 y_{k} \in \mathbf{R}, k=1\right.$, $2, \cdots, 5)$. Let $\Gamma$ be the ellipse passing through the five points $\left(x_{k}, y_{k}\right)(k=1,2$, $\cdots, 5)$, with eccentricity $e=\sqrt{\frac{m}{n}}\left(m 、 n \in \mathbf{Z}_{+}\right.$, $(m, n)=1)$. Then the value of $m+n$ is $(\quad)$.
(A) 7
(B) 9
(C) 11
(D) 13
(E) 15
21. A.

Solving the equation, we get
$$
z_{1}=1, z_{2,3}=-1 \pm \sqrt{3} \mathrm{i}, z_{4,5}=-2 \pm \sqrt{2} \mathrm{i} \text {. }
$$

Thus, the ellipse $\Gamma$ passes through the points $(1,0)$, $(-1, \pm \sqrt{3})$, and $(-2, \pm \sqrt{2})$.
By symmetry, we know the center of the ellipse is on the $x$-axis.
Let the equation of the ellipse $\Gamma$ be
$$
\frac{(x-h)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text {. }
$$

Substituting the points $(1,0)$, $(-1, \sqrt{3})$, and $(-2, \sqrt{2})$ into the equation, we get
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \frac { ( 1 - h ) ^ { 2 } } { a ^ { 2 } } + \frac { 0 } { b ^ { 2 } } = 1 , } \\
{ \frac { ( - 1 - h ) ^ { 2 } } { a ^ { 2 } } + \frac { 3 } { b ^ { 2 } } = 1 , } \\
{ \frac { ( - 2 - h ) ^ { 2 } } { a ^ { 2 } } + \frac { 2 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2}=\frac{361}{100}, \\
b^{2}=\frac{361}{120}, \\
h=-\frac{9}{10} .
\end{array}\right.\right. \\
\text { Hence } e=\frac{c}{a}=\frac{\sqrt{\frac{361}{100}-\frac{361}{120}}}{\sqrt{\frac{361}{100}}}=\frac{\sqrt{6}}{6}=\sqrt{\frac{1}{6}} \\
\Rightarrow m+n=1+6=7 .
\end{array}
$$