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Câu hỏi: Cho số phức $z=a+bi$ ( với $a,b\in \mathbb{R}$ ) thỏa $\left| z \right|\left( 2+i \right)=z-1+i\left( 2z+3 \right)$. Tính $S=a+b$.
A. $S=-1$.
B. $S=1$.
C. $S=7$.
D. $S=-5$.
A. $S=-1$.
B. $S=1$.
C. $S=7$.
D. $S=-5$.
$\left| z \right|\left( 2+i \right)=z-1+i\left( 2z+3 \right)\Leftrightarrow \left| z \right|\left( 2+i \right)+1-3i=z\left( 1+2i \right)\Leftrightarrow \left( 1+2\left| z \right| \right)+\left( \left| z \right|-3 \right)i=z\left( 1+2i \right)$
Suy ra: ${{\left( 1+2\left| z \right| \right)}^{2}}+{{\left( \left| z \right|-3 \right)}^{2}}=5{{\left| z \right|}^{2}}\Leftrightarrow \left| z \right|=5$
Khi đó, ta có: $5\left( 2+i \right)=z-1+i\left( 2z+3 \right)\Leftrightarrow z\left( 1+2i \right)=11+2i\Leftrightarrow z=\dfrac{11+2i}{1+2i}=3-4i$
Vậy $S=a+b=3-4=-1$.
Suy ra: ${{\left( 1+2\left| z \right| \right)}^{2}}+{{\left( \left| z \right|-3 \right)}^{2}}=5{{\left| z \right|}^{2}}\Leftrightarrow \left| z \right|=5$
Khi đó, ta có: $5\left( 2+i \right)=z-1+i\left( 2z+3 \right)\Leftrightarrow z\left( 1+2i \right)=11+2i\Leftrightarrow z=\dfrac{11+2i}{1+2i}=3-4i$
Vậy $S=a+b=3-4=-1$.
Đáp án A.