Cho số phức $z$ thỏa mãn $\left| z+1 \right|\ge 1$. Gọi giá trị...

+) Ta có: $P=\left| \dfrac{\left( 1+i \right)z+i+2}{z+1} \right|=\left| \dfrac{\left( z+1 \right)\left( 1+i \right)+1}{z+1} \right|=\dfrac{\left| \left( z+1 \right)\left( 1+i \right)+1 \right|}{\left| z+1 \right|}$
+) Áp dụng bất đẳng thức: $\left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|\le \left| {{z}_{1}}+{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$, ta có:
$\dfrac{\left| \left( z+1 \right)\left( 1+i \right) \right|-\left| 1 \right|}{\left| z+1 \right|}\le P\le \dfrac{\left| \left( z+1 \right)\left( 1+i \right) \right|+\left| 1 \right|}{\left| z+1 \right|}$ $\Leftrightarrow \left| 1+i \right|-\dfrac{1}{\left| z+1 \right|}\le P\le \left| 1+i \right|+\dfrac{1}{\left| z+1 \right|}$
$\Leftrightarrow \sqrt{2}-\dfrac{1}{\left| z+1 \right|}\le P\le \sqrt{2}+\dfrac{1}{\left| z+1 \right|} \left( 1 \right)$
Mà: $\left| z+1 \right|\ge 1\Rightarrow \left\{ \dfrac{1}{\left| z+1 \right|}\le 1; \dfrac{-1}{\left| z+1 \right|}\ge -1 (2) \right.$
Từ và $\Rightarrow \sqrt{2}-1\le P\le \sqrt{2}+1$
Bây giờ ta xét dấu "=" xảy ra khi nào.
Với ${{z}_{1}}={{a}_{1}}+{{b}_{1}}i; {{z}_{2}}={{a}_{2}}+{{b}_{2}}i ({{a}_{1}},{{b}_{1}},{{a}_{2}},{{b}_{2}}\in \mathbb{R})$, ta có:
$\bullet \left| {{z}_{1}}+{{z}_{2}} \right|=\left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\Leftrightarrow \left\{ \begin{aligned}
& {{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}} \\
& {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}\ge 0 \\
\end{aligned} \right.; \bullet \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|=\left| {{z}_{1}}+{{z}_{2}} \right|\Leftrightarrow \left\{ \begin{aligned}
& {{a}_{1}}{{b}_{2}}={{a}_{2}}{{b}_{1}}; \left| {{z}_{1}} \right|\ge \left| {{z}_{2}} \right| \\
& {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}\le 0 \\
\end{aligned} \right.$.
Giả sử: $z=a+bi (a,b\in \mathbb{R})\Rightarrow \left( z+1 \right)\left( 1+i \right)=\left( a+1-b \right)+\left( a+b+1 \right)i$.
Mà: $1=1+0.i$. Do đó:
$\bullet P=\sqrt{2}+1\Leftrightarrow \left\{ \begin{aligned}
& \left| \left( z+1 \right)\left( 1+i \right)+1 \right|=\left| \left( z+1 \right)\left( 1+i \right) \right|+\left| 1 \right| \\
& \left| z+1 \right|=1 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a+b+1=0 \\
& a+1-b\ge 0 \\
& \sqrt{{{(a+1)}^{2}}+{{b}^{2}}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a+1=-b \\
& -2b\ge 0 \\
& 2{{b}^{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{\sqrt{2}}{2}-1 \\
& b\le 0 \\
& b=-\dfrac{\sqrt{2}}{2} \\
\end{aligned} \right.\Leftrightarrow z=\dfrac{\sqrt{2}}{2}-1-\dfrac{\sqrt{2}}{2}i$.
$\bullet P=\sqrt{2}-1\Leftrightarrow \left\{ \begin{aligned}
& \left| \left( z+1 \right)\left( 1+i \right)+1 \right|=\left| \left( z+1 \right)\left( 1+i \right) \right|-\left| 1 \right| \\
& \left| z+1 \right|=1 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a+b+1=0 \\
& a+1-b\le 0 \\
& \sqrt{{{(a+1)}^{2}}+{{b}^{2}}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a+1=-b \\
& -2b\le 0 \\
& 2{{b}^{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{\sqrt{2}}{2}-1 \\
& b\ge 0 \\
& b=\dfrac{\sqrt{2}}{2} \\
\end{aligned} \right.\Leftrightarrow z=-\dfrac{\sqrt{2}}{2}-1+\dfrac{\sqrt{2}}{2}i$.
Vậy: $\left\{ \begin{aligned}
& M=\sqrt{2}+1 \\
& m=\sqrt{2}-1 \\
\end{aligned} \right. \Rightarrow {{M}^{2}}+{{m}^{2}}=6$.