Trong mặt phẳng $Oxy$, cho số phức $z$ thỏa mãn $|z-1+2 i|=3$. Tập...

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$w=z(1+i)\Rightarrow x+yi=\left( x'+y'i \right)\left( 1+i \right)\Leftrightarrow x+yi=x'-y'+\left( x'+y' \right)i\Leftrightarrow \left\{ \begin{aligned}
& x=x'-y' \\
& y=x'+y' \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x'=\dfrac{x+y}{2} \\
& y'=\dfrac{-x+y}{2} \\
\end{aligned} \right.|z-1+2i|=3\Leftrightarrow \left| \dfrac{x+y}{2}-1+\left( \dfrac{y-x}{2}+2 \right)i \right|=3\Leftrightarrow {{\left( \dfrac{x+y}{2}-1 \right)}^{2}}+{{\left( \dfrac{y-x}{2}+2 \right)}^{2}}=9{{\left( x+y-2 \right)}^{2}}+{{\left( y-x+4 \right)}^{2}}=36\Leftrightarrow {{x}^{2}}+{{y}^{2}}-6x+2y-8=0\Leftrightarrow {{\left( x-3 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=18w=z(1+i)I(3 ;-1)R=3 \sqrt{2}$.