(55) Given the complex number $z_{1}$ satisfies $\left(z_{1}-2\right)(1+\mathrm{i})=1-\mathrm{i}$ (where $\mathrm{i}$ is the imaginary unit), and the imaginary part of the complex number $z_{2}$ is 2, then the condition for $z_{1} \cdot z_{2}$ to be a real number is $z_{2}=$ $\qquad$ .
(5) $4+2 \mathrm{i}$ Hint: Since
$$
\left(z_{1}-2\right)(1+\mathrm{i})=1-\mathrm{i},
$$

so $z_{1}=2-\mathrm{i}$. Let $z_{2}=a+2 \mathrm{i}(a \in \mathbf{R})$,
$$
\begin{aligned}
z_{1} \cdot z_{2} & =(2-\mathrm{i})(a+2 \mathrm{i}) \\
& =(2 a+2)+(4-a) \mathrm{i} .
\end{aligned}
$$

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