Câu hỏi: Cho số phức z có phần thực là số nguyên và z thỏa mãn $\left| x \right|-2\bar{z}=-7+3i+z$. Tính mô-đun của số phức $w=1-z+{{z}^{2}}$ bằng
A. $\left| w \right|=\sqrt{37}$.
B. $\left| w \right|=\sqrt{457}$.
C. $\left| w \right|=\sqrt{425}$.
D. $\left| w \right|=\sqrt{445}$.
A. $\left| w \right|=\sqrt{37}$.
B. $\left| w \right|=\sqrt{457}$.
C. $\left| w \right|=\sqrt{425}$.
D. $\left| w \right|=\sqrt{445}$.
Đặt $z=a+bi\left( a,b\in \mathbb{R} \right)$
Ta có: $\left| z \right|-2\bar{z}=-7+3i+z\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}-2\left( a-bi \right)=-7+3i+a+bi$
$\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}-3a+7+\left( b-3 \right)i=0\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{a}^{2}}+{{b}^{2}}}-3a+7=0 \\
& b-3=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{a}^{2}}+9}=3a-7 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge \dfrac{7}{3} \\
& {{a}^{2}}+9=9{{a}^{2}}-42a+49 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge \dfrac{7}{3} \\
& \left[ \begin{aligned}
& a=4\left( nhan \right) \\
& a=\dfrac{5}{4}\left( loai \right) \\
\end{aligned} \right. \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& a=4 \\
\end{aligned} \right.$;
Vậy $z=4+3i\Rightarrow w=1-z+{{z}^{2}}=4+21i\Rightarrow \left| w \right|=\sqrt{457}$
Ta có: $\left| z \right|-2\bar{z}=-7+3i+z\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}-2\left( a-bi \right)=-7+3i+a+bi$
$\Leftrightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}-3a+7+\left( b-3 \right)i=0\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{a}^{2}}+{{b}^{2}}}-3a+7=0 \\
& b-3=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{a}^{2}}+9}=3a-7 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge \dfrac{7}{3} \\
& {{a}^{2}}+9=9{{a}^{2}}-42a+49 \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge \dfrac{7}{3} \\
& \left[ \begin{aligned}
& a=4\left( nhan \right) \\
& a=\dfrac{5}{4}\left( loai \right) \\
\end{aligned} \right. \\
& b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& a=4 \\
\end{aligned} \right.$;
Vậy $z=4+3i\Rightarrow w=1-z+{{z}^{2}}=4+21i\Rightarrow \left| w \right|=\sqrt{457}$
Đáp án B.