[tex]Z= \frac{1}{a(1+i)+1-2i}= \frac{1}{(a+1)+i(a-2)} [/tex], amplificam fractia cu conjugatul numitorului si obtinem: [tex]Z= \frac{(a+1)-i(a-2)}{(a+1)^2+(a-2)^2}= \frac{a+1}{2a^2-2a+5}+i \frac{a-2}{2a^2-2a+5} [/tex], prima fractie e partea reala si o egalam cu 2/5, rezulta: [tex] \frac{a+1}{2a^2-2a+5}= \frac{2}{5},deci,4a^2-4a+10=5a+5,sau,4a^2-9a+5=0 [/tex], care are radacinile: 1 si 5/4, deci a∈{1;5/4}.
