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Câu hỏi: Cho số phức $z=a+bi(a,b\in \mathbb{R})$ thoả mãn $\left( 1+i \right)z+2\overline{z}=3+2i$. Tính $P=a+~b$
A. $P=1$
B. $P=-\dfrac{1}{2}$
C. $P=\dfrac{1}{2}$
D. $P=-1$
A. $P=1$
B. $P=-\dfrac{1}{2}$
C. $P=\dfrac{1}{2}$
D. $P=-1$
Lời giải
$\left( 1+i \right)z+2z=3+2i\Leftrightarrow \left( 1+i \right)\left( a+bi \right)+2\left( a-bi \right)=3+2i\Leftrightarrow \left( 3a-b \right)+\left( a-b \right)i=3+2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 3a-b=3 \\
& a-b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2} \\
& b=-\dfrac{3}{2} \\
\end{aligned} \right.$
Suy ra: $P=a+b=-1.~$
$\left( 1+i \right)z+2z=3+2i\Leftrightarrow \left( 1+i \right)\left( a+bi \right)+2\left( a-bi \right)=3+2i\Leftrightarrow \left( 3a-b \right)+\left( a-b \right)i=3+2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 3a-b=3 \\
& a-b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2} \\
& b=-\dfrac{3}{2} \\
\end{aligned} \right.$
Suy ra: $P=a+b=-1.~$
Đáp án D.