Cho số phức z=a+bi (với ) thỏa |z|(2+i)=z-1+i(2z+3)

Chọn A

$\left|z\right|\left(2+i\right)=z-1+i\left(2z+3\right)\iff \left|z\right|\left(2+i\right)+1-3i=z\left(1+2i\right)\iff \left(1+2\left|z\right|\right)+\left(\left|z\right|-3\right)i=z\left(1+2i\right)$

Suy ra: ${\left(1+2\left|z\right|\right)}^2+{\left(\left|z\right|-3\right)}^2=5{\left|z\right|}^2\iff \left|z\right|=5$

Khi đó, ta có: $5\left(2+i\right)=z-1+i\left(2z+3\right)\iff z\left(1+2i\right)=11+2i\iff z=\frac{11+2i}{1+2i}=3-4i$

Vậy S=a+b=3-4=-1.