Cho số phức $z=x+yi$ thỏa mãn $\left( z-1 \right)\left| z...

Ta có $\left( z-1 \right)\left| z \right|=2i\left( z+1 \right)\Leftrightarrow \left( x-1+yi \right)\sqrt{{{x}^{2}}+{{y}^{2}}}=2i\left( x+1+yi \right)$
$\Leftrightarrow \left( x-1 \right)\sqrt{{{x}^{2}}+{{y}^{2}}}+y\sqrt{{{x}^{2}}+{{y}^{2}}}i=-2y+2\left( x+1 \right)i$
$\Leftrightarrow \left\{ \begin{aligned}
& \left( x-1 \right)\sqrt{{{x}^{2}}+{{y}^{2}}}=-2y \left( 1 \right) \\
& y\sqrt{{{x}^{2}}+{{y}^{2}}}=2\left( x+1 \right) \left( 2 \right) \\
\end{aligned} \right.$
Từ (1) và (2) suy ra $\dfrac{x-1}{y}=\dfrac{-y}{x+1}\Leftrightarrow {{x}^{2}}-1=-{{y}^{2}}\Leftrightarrow {{x}^{2}}+{{y}^{2}}=1$ (*).
Thay vào (2) ta có $y=2x+2$
Suy ra $\left( * \right)\Leftrightarrow {{x}^{2}}+{{\left( 2x+2 \right)}^{2}}=1\Leftrightarrow 5{{x}^{2}}+8x+3=0\Leftrightarrow \left[ \begin{aligned}
& x=-1\Rightarrow y=0 \Rightarrow xy=0 \\
& x=-\dfrac{3}{5}\Rightarrow y=\dfrac{4}{5}\Rightarrow xy=-\dfrac{12}{25} \\
\end{aligned} \right.$.
Cách 2: (PB bổ sung)
+ $\left( z-1 \right)\left| z \right|=2i\left( z+1 \right)\Leftrightarrow z\left( \left| z \right|-2i \right)=\left| z \right|+2i \left( 1 \right)$
+ Modun 2 vế ta được: $\left| z \right|.\left| \left| z \right|-2i \right|=\left| \left| z \right|+2i \right| \Leftrightarrow \left| z \right|.\sqrt{{{\left| z \right|}^{2}}+4}=\sqrt{{{\left| z \right|}^{2}}+4} \Leftrightarrow \left| z \right|=1$
+Thay vào (1) ta có $z\left( 1-2i \right)=1+2i \Leftrightarrow z=\dfrac{1+2i}{1-2i} =\dfrac{-3}{5}+\dfrac{4}{5}i$ $\Rightarrow xy=-\dfrac{12}{25}$.