Question

Rezolvare completa, daca se poate :)

Answer 1

$z=a+ib,$  $unde$  $a$ $si$ $b$ ∈ $R$

Conjugatul lui z $=\ \textgreater \ z conjugat =a-ib$

$Astfel$  il inlocuim pe z=>  $2(a+ib)+(a-ib)=\ \textgreater$ $2a+2ib+a-ib=6+i=\ \textgreater \ 3a+ib=6+i=\ \textgreater \ \left \{ {{3a=6} \atop {b=1}} \right. =\ \textgreater \ \left \{ {{a=2} \atop {b=1} \right.$




Si astfel $z=2+i$

Answer 2

[tex]\it z= a+bi,\ \ \ \ \overline{z} = a-bi \\\;\\ 2z+\overline{z} = 6+i \Leftrightarrow 2(a+bi) +a-bi = 6+i \Leftrightarrow \\\;\\ \Leftrightarrow 2a+2bi +a-bi = 6+i \Leftrightarrow 3a+bi=6+i \Leftrightarrow \\\;\\ \Leftrightarrow \begin{cases} \it3a+6 \Longrightarrow a=2 \\\;\\ \it b=1\end{cases}\ Deci\ z = 2+i[/tex]