China Girls' Mathematical Olympiad

As shown in the figure, in the complex plane, points $A$, $B$ and $C$ denote the complex numbers $z_1$, $z_2$ and $x{z}_2$, respectively. $C$ is obviously on the segment $OB$. The vectors $\overrightarrow{BA}$ and $\overrightarrow{CA}$ are represented by complex numbers $z_1-z_2$ and $z_1-x{z}_2$, respectively. As $|z_1|\le \alpha |z_1-z_2|$, we have $|\overrightarrow{OA}|\le \alpha |\overrightarrow{BA}|$. Then

$$\begin{array}{rl}|z_1-x{z}_2{|}_{\max }& =|\overrightarrow{AC}{|}_{\max }=\max \{|\overrightarrow{OA}|,|\overrightarrow{BA}|\}\\ & =\max \{|z_1|,|z_1-z_2|\}\\ & =\max \{\alpha |z_1-z_2|,|z_1-z_2|\}.\end{array}$$

Therefore, $\lambda (\alpha )=\max \{\alpha ,1\}$.

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