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Câu hỏi: Số phức $z$ thỏa mãn $(1+z)(3-i)-5 i z-6 i+1=0$. Giá trị $\left| z \right|$ bằng:
A. $\dfrac{\sqrt{10}}{3}.$
B. $\dfrac{\sqrt{13}}{3}$.
C. $\dfrac{1}{2}$
D. $\dfrac{\sqrt{15}}{4}.$
A. $\dfrac{\sqrt{10}}{3}.$
B. $\dfrac{\sqrt{13}}{3}$.
C. $\dfrac{1}{2}$
D. $\dfrac{\sqrt{15}}{4}.$
$\begin{aligned}
& (1+z)(3-i)-5iz-6i+1=0 \\
& \Leftrightarrow \left( 3-6i \right)z=-4+7i \\
& \Leftrightarrow z=\dfrac{-4+7i}{3-6i}=\dfrac{-6}{5}-\dfrac{1}{15}i \\
\end{aligned}$.
$\left| z \right|=\sqrt{{{\left( \dfrac{6}{5} \right)}^{2}}+{{\left( \dfrac{1}{15} \right)}^{2}}}=\dfrac{\sqrt{13}}{3}$
& (1+z)(3-i)-5iz-6i+1=0 \\
& \Leftrightarrow \left( 3-6i \right)z=-4+7i \\
& \Leftrightarrow z=\dfrac{-4+7i}{3-6i}=\dfrac{-6}{5}-\dfrac{1}{15}i \\
\end{aligned}$.
$\left| z \right|=\sqrt{{{\left( \dfrac{6}{5} \right)}^{2}}+{{\left( \dfrac{1}{15} \right)}^{2}}}=\dfrac{\sqrt{13}}{3}$
Đáp án C.