2.4. Find all complex numbers $z$ for which

$$
|z-2+i|=5, \quad \operatorname{Re}(z(1+i))=2
$$
2.4. The first equation is equivalent to $(x-2)^{2}+(y+1)^{2}=25$, and the second is $x-y=2$, where $z=x+iy$, $x, y \in R$. By substituting from the second equation into the first, we have $y^{2}+(y+1)^{2}=25$. The solutions are $(5,3)$ and $(-2,-4)$, respectively $z_{1}=5+3i, z_{2}=-2-4i$.