1. Let $i$ be the imaginary unit, $a$ and $b$ be positive integers, and $|(a+i)(2+i)|=\left|\frac{b-i}{2-i}\right|$, then $a+b=$
Answer 8.
Analysis From the problem, we get $(2 a-1)^{2}+(a+2)^{2}=\left(\frac{2 b+1}{5}\right)^{2}+\left(\frac{b-2}{5}\right)^{2} \Rightarrow(b+5 a)(b-5 a)=24$.
Since $b+5 a$ and $b-5 a$ are integers with the same parity, we have $\left\{\begin{array}{l}b+5 a=12 \\ b-5 a=2\end{array}\right.$ or $\left\{\begin{array}{l}b+5 a=6 \\ b-5 a=4\end{array}\right.$, solving these gives $a=1, b=7$, hence $a+b=8$.