3. Given the complex number $z$ satisfies $z^{2}+2 z=\bar{z} \neq z$ ( $\bar{z}$ denotes the conjugate of $z$), then the product of all possible values of $z$ is $\qquad$ .
Answer: 3.
Solution: Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$. From $z^{2}+2 z=\bar{z}$, we have
$$
a^{2}-b^{2}+2 a b \mathrm{i}+2 a+2 b \mathrm{i}=a-b \mathrm{i},
$$

By comparing the real and imaginary parts, we get $a^{2}-b^{2}+a=0, 2 a b+3 b=0$. Since $\bar{z} \neq z$, we know $b \neq 0$, thus $2 a+3=0$, which means $a=-\frac{3}{2}$, and then $b= \pm \sqrt{a^{2}+a}= \pm \frac{\sqrt{3}}{2}$.
Therefore, the product of the values of $z$ that satisfy the condition is $\left(-\frac{3}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)\left(-\frac{3}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)=3$.