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Câu hỏi: Cho số phức $z$ thỏa mãn $\left( 1+2i \right)z+\overline{z}=i$. Tìm số phức $z$.
A. $z=\dfrac{1}{2}+\dfrac{1}{2}i.$
B. $z=1+2i.$
C. $z=2-i.$
D. $z=\dfrac{1}{2}-\dfrac{1}{2}i.$
A. $z=\dfrac{1}{2}+\dfrac{1}{2}i.$
B. $z=1+2i.$
C. $z=2-i.$
D. $z=\dfrac{1}{2}-\dfrac{1}{2}i.$
Gọi số phức $z=a+bi ,\left( a,b\notin \mathbb{R} \right)$.
Ta có: $\left( 1+2i \right)z+\overline{z}=i\Leftrightarrow \left( 1+2i \right)\left( a+bi \right)+\left( a-bi \right)=i$
$\begin{aligned}
& \Leftrightarrow a-2b+\left( 2a+b \right)i+a-bi=i \\
& \Leftrightarrow 2a-2b+\left( 2a-1 \right)i=0 \\
& \Leftrightarrow \left\{ \begin{aligned}
& 2a-2b=0 \\
& 2a-1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2} \\
& b=\dfrac{1}{2} \\
\end{aligned} \right. \\
\end{aligned}$
Vậy $z=\dfrac{1}{2}+\dfrac{1}{2}i$.
Ta có: $\left( 1+2i \right)z+\overline{z}=i\Leftrightarrow \left( 1+2i \right)\left( a+bi \right)+\left( a-bi \right)=i$
$\begin{aligned}
& \Leftrightarrow a-2b+\left( 2a+b \right)i+a-bi=i \\
& \Leftrightarrow 2a-2b+\left( 2a-1 \right)i=0 \\
& \Leftrightarrow \left\{ \begin{aligned}
& 2a-2b=0 \\
& 2a-1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2} \\
& b=\dfrac{1}{2} \\
\end{aligned} \right. \\
\end{aligned}$
Vậy $z=\dfrac{1}{2}+\dfrac{1}{2}i$.
Đáp án A.