Minh thu thoi chu minh ko chac a!
\(z=x+yi\left(x,y\in R\right)\)
\(\left|x+yi-x+yi+1-i\right|=\sqrt{5}=\left|1+\left(2y-1\right)i\right|\)
\(\Leftrightarrow1+\left(2y-1\right)^2=5\Leftrightarrow2y-1=\pm2\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vi \(\left(2-z\right)\left(i+\overline{z}\right)\) la so thuan ao
\(\Rightarrow\left(2-x-yi\right)\left(i+x-yi\right)=2i+2x-2yi-xi-x^2+yxi+y-xyi-y^2=\left(2-2y-x\right)i+2x+y-\left(x^2+y^2\right)\) la so thuan ao
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-\left(x^2+y^2\right)=0\\2-2y-x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+y^2+y=0\\x\ne2-2y\end{matrix}\right.\)
Voi \(y=\dfrac{3}{2}\Rightarrow x^2-2x+\dfrac{9}{4}+\dfrac{3}{2}=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Voi \(y=-\dfrac{1}{2}\Rightarrow x^2-2x+\dfrac{1}{4}-\dfrac{1}{2}=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2+\sqrt{5}}{2}\\x=\dfrac{2-\sqrt{5}}{2}\end{matrix}\right.\)
=> 4 so phuc z thoa man
