5. Given a non-negative real number $\alpha$. Find the smallest real number $\lambda=$
$\lambda(\alpha)$, such that for any complex numbers $z_{1}, z_{2}$ and real number $x \in$ $[0,1]$, if $\left|z_{1}\right| \leqslant \alpha\left|z_{1}-z_{2}\right|$, then
$\left|z_{1}-x z_{2}\right| \leqslant \lambda\left|z_{1}-z_{2}\right|$. (Provided by Li Shenghong)
5. As shown in Figure 4, in the complex plane, points $A, B, C$ correspond to the complex numbers $z_{1}, z_{2}, x z_{2}$, respectively. 
Obviously, point $C$ lies on line segment $OB$. The vectors $\overrightarrow{B A}$ and $\overrightarrow{C A}$ correspond to the complex numbers 
$z_{1}-z_{2}$ and $z_{1}-x z_{2}$, respectively. From 
$$
\left|z_{1}\right| \leqslant \alpha\left|z_{1}-z_{2}\right|,
$$

we get $|\overrightarrow{O A}| \leqslant \alpha|\overrightarrow{B A}|$.
$$
\begin{array}{l}
\text { Therefore, }\left|z_{1}-x z_{2}\right|_{\max }=|\overrightarrow{A C}|_{\max }=\max \{|\overrightarrow{O A}|,|\overrightarrow{B A}|\} \\
=\max \left\{\left|z_{1}\right|,\left|z_{1}-z_{2}\right|\right\} \\
=\max \left\{\alpha\left|z_{1}-z_{2}\right|,\left|z_{1}-z_{2}\right|\right\} . \\
\text { Thus, } \lambda(\alpha)=\max \{\alpha, 1\} .
\end{array}
$$