Cho số phức $z$ thỏa mãn $\left( z+1-i \right)\left(...

Gọi $z=x+yi \left( x, y\in \mathbb{R} \right)$.
Ta có $\left( z+1-i \right)\left( \overline{z}+1+i \right)=5\Leftrightarrow \left( z+1-i \right)\left( \overline{z+1-i} \right)=5\Leftrightarrow {{\left| z+1-i \right|}^{2}}=5\Leftrightarrow {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=5$.
$P={{\left| z-2i \right|}^{2}}-{{\left| z+1 \right|}^{2}}={{x}^{2}}+{{\left( y-2 \right)}^{2}}-{{\left( x+1 \right)}^{2}}-{{y}^{2}}=-2x-4y+3=-2\left( x+1 \right)-4\left( y-1 \right)+1$
$\Leftrightarrow 1-P=2\left( x+1 \right)+4\left( y-1 \right)$.
Áp dụng bất đẳng thức Bunhia-cốpski ta được: $\left| 2\left( x+1 \right)+4\left( y-1 \right) \right|\le \sqrt{{{2}^{2}}+{{4}^{2}}}.\sqrt{{{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}}$
$\Leftrightarrow \left| 1-P \right|\le 2\sqrt{5}.\sqrt{5}=10$ $\Leftrightarrow -10\le P-1\le 10\Leftrightarrow -9\le P\le 11$.
${{P}_{\text{max}}}=11$ đạt được khi $\left\{ \begin{aligned}
& 11=-2x-4y+3 \\
& \dfrac{x+1}{2}=\dfrac{y-1}{4} \\
& {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2x+4y=-8 \\
& 4x-2y=-6 \\
& {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-2 \\
& y=-1 \\
\end{aligned} \right.\Rightarrow z=-2-i$.
${{P}_{\text{min}}}=-9$ đạt được khi $\left\{ \begin{aligned}
& -9=-2x-4y+3 \\
& \dfrac{x+1}{2}=\dfrac{y-1}{4} \\
& {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2x+4y=12 \\
& 4x-2y=-6 \\
& {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}=5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=0 \\
& y=3 \\
\end{aligned} \right.\Rightarrow z=3i$.
Vậy ${{P}_{\text{max}}}+{{P}_{\text{max}}}=2$.