Lời giải:
$|z|(-3+i)=z-4+3i+2i(z+2)$
$\Leftrightarrow \sqrt{a^2+b^2}(-3+i) = a+bi-4+3i+2i(a+bi+2)$
$\Leftrightarrow -3\sqrt{a^2+b^2}+\sqrt{a^2+b^2}i=(a-4-2b)+i(2a+b+7)$
\(\Rightarrow \left\{\begin{matrix} -3\sqrt{a^2+b^2}=a-4-2b\\ \sqrt{a^2+b^2}=2a+b+7(*)\end{matrix}\right.\)
$\Rightarrow a-4-2b=-3(2a+b+7)$
$\Leftrightarrow 7a+b+17=0$
$\Leftrightarrow b=-(7a+17)$
Thay vào $(*)$:
$a^2+b^2=(2a+b+7)^2$
$\Leftrightarrow a^2+b^2=4a^2+b^2+49+4ab+28a+14b$
$\Leftrightarrow 3a^2+49+4ab+28a+14b=0$
$\Leftrightarrow 3a^2+49-4a(7a+17)+28a-14(7a+17)=0$
$\Leftrightarrow -25a^2-138a-189=0$
$\Leftrightarrow $-(a+3)(25a+63)=0$
$\Leftrightarrow a=-3$ hoặc $a=\frac{-63}{25}$
Nếu $a=-3$ thì $b=4\Rightarrow a+b=1$
Nếu $a=\frac{-63}{25}$ thì $b=\frac{16}{25}$
$\Rightarrow a+b=\frac{-47}{25}$
