5. Given the complex number $z_{1}$ satisfies
$\left(z_{1}-2\right)(1+\mathrm{i})=1-\mathrm{i}$ ( $\mathrm{i}$ is the imaginary unit), and the imaginary part of the complex number $z_{2}$ is 2. Then the condition for $z_{1} z_{2}$ to be a real number is
$$
z_{2}=
$$
$\qquad$
$5.4+2 \mathrm{i}$
Since $\left(z_{1}-2\right)(1+\mathrm{i})=1-\mathrm{i}$, therefore, $z_{1}=2-\mathrm{i}$.
Let $z_{2}=a+2 \mathrm{i}(a \in \mathbf{R})$.
Then $z_{1} z_{2}=(2-\mathrm{i})(a+2 \mathrm{i})$
$$
=(2 a+2)+(4-a) \mathrm{i} \text {. }
$$

Since $z_{1} z_{2}$ is a real number, thus, $a=4$.
Hence $z_{2}=4+2 \mathrm{i}$.