Ответ: z_1=2+3i;z_2=1-2i\\
\frac{z_1}{z_2}= \frac{2+3i}{1-2i} = \frac{(2+3i)(1+2i)}{1-4i^2} =
\frac{2+4i+3i+6}{1+4} = \frac{8+7i}{5} =az_1+bz_2\\
8+7i=5a(2+3i)+5b(1-2i)\\
8+7i=(10a+5b)+(15a-10b)i\\
\left \{ {{10a+5b=8} \atop {15a-10b=7}} \right. \iff
\left \{ {{20a+10b=14} \atop {15a-10b=7}} \right. \iff
35a=21\\
\iff a= \frac{21}{35} = \frac{3}{5} =0,6\\
10*0,6+5b=8 \iff
5b=8-6=2 \iff
b= \frac{2}{5} =0,4\\
\underline{ \left \{ {{a=0,6} \atop {b=0,4}} \right. }
Источник znanija.site