Google
Câu hỏi: Cho số phức $z=a+bi, \left( a, b\in \mathbb{R} \right)$ thỏa $\left( 2z-1 \right)\left( 1+i \right)-\left( \overline{z}+3i \right)\left( 1-i \right)=3-7i$. Tính $P={{a}^{2}}+b$.
A. $5$.
B. $2$.
C. $13$.
D. $7$.
A. $5$.
B. $2$.
C. $13$.
D. $7$.
Ta có $\left( 2z-1 \right)\left( 1+i \right)-\left( \overline{z}+3i \right)\left( 1-i \right)=3-7i$
$\begin{aligned}
& \Leftrightarrow \left( 2a+2bi-1 \right)\left( 1+i \right)-\left( a-bi+3i \right)\left( 1-i \right)=3-7i \\
& \Leftrightarrow a-b-4+\left( 3a+3b-2 \right)i=3-7i \\
& \Leftrightarrow \left\{ \begin{aligned}
& a-b-4=3 \\
& 3a+3b-4=-7 \\
\end{aligned} \right. \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow \left\{ \begin{aligned}
& a-b=7 \\
& 3a+3b=-3 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b= -4 \\
\end{aligned} \right. \\
\end{aligned}$
Vậy $P={{a}^{2}}+b=5$.
$\begin{aligned}
& \Leftrightarrow \left( 2a+2bi-1 \right)\left( 1+i \right)-\left( a-bi+3i \right)\left( 1-i \right)=3-7i \\
& \Leftrightarrow a-b-4+\left( 3a+3b-2 \right)i=3-7i \\
& \Leftrightarrow \left\{ \begin{aligned}
& a-b-4=3 \\
& 3a+3b-4=-7 \\
\end{aligned} \right. \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow \left\{ \begin{aligned}
& a-b=7 \\
& 3a+3b=-3 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& a=3 \\
& b= -4 \\
\end{aligned} \right. \\
\end{aligned}$
Vậy $P={{a}^{2}}+b=5$.
Đáp án A.