3. Let $a, b \in \mathbf{C}, a^{2}+b^{2}=7, a^{3}+b^{3}=10$, and let $m, n$ be the maximum and minimum real values that $a+b$ can take, respectively. Then $m+n=$ $\qquad$
3. -1 .

Let $p=a+b$.
$$
\begin{array}{l}
\text { By } 7=a^{2}+b^{2}=p^{2}-2 a b \\
\Rightarrow a b=\frac{p^{2}-7}{2} \text {. }
\end{array}
$$

Then $10=a^{3}+b^{3}=p^{3}-3 a b p$
$$
\begin{array}{l}
=p^{3}-\frac{3 p\left(p^{2}-7\right)}{2} \\
\Rightarrow p^{3}-21 p+20=0 \\
\Rightarrow(p-1)(p-4)(p+5)=0 .
\end{array}
$$

Thus, $m=4, n=-5$.
It is easy to verify,
when $a=\frac{4+\sqrt{2} \mathrm{i}}{2}, b=\frac{4-\sqrt{2}}{2} \mathrm{i}$, $m=4$;
when $a=\frac{-5+\sqrt{11} \mathrm{i}}{2}, b=\frac{-5-\sqrt{11} \mathrm{i}}{2}$, $n=-5$.
Therefore, $m+n=4-5=-1$.