79 Let the conjugate of the complex number $z$ be $\bar{z}$. If $z \cdot \bar{z}$ satisfies $z \cdot \bar{z}+(1-2 i) z+(1+2 i) \bar{z}=3$, then the minimum value of the sum of the real and imaginary parts of the complex number $z$ is $\qquad$ .
$79-7$. From the given, we have
$$
z \cdot \bar{z}+z+\bar{z}-2 i(z-\bar{z})=3 .
$$

Let $z=x+y i$, where $x, y \in \mathbf{R}$, then
$$
x^{2}+y^{2}+2 x+4 y=3,
$$

which leads to
$$
(x+1)^{2}+(y+2)^{2}=8 .
$$

Also, since
$$
(x+1)^{2}+(y+2)^{2} \geqslant \frac{(x+1+y+2)^{2}}{2},
$$

we have
$$
(x+y+3)^{2} \leqslant 16,
$$

which means
$$
-7 \leqslant x+y \leqslant 1 .
$$

Therefore, the minimum value of $x+y$ is -7, at this point $x=-3, y=-4$.