+) Ta có: \(P=\left| \frac{\left( 1+i \right)z+i+2}{z+1} \right|=\left| \frac{\left( z+1 \right)\left( 1+i \right)+1}{z+1} \right|=\frac{\left| \left( z+1 \right)\left( 1+i \right)+1 \right|}{\left| z+1 \right|}\)

+) 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:

\(\frac{\left| \left( z+1 \right)\left( 1+i \right) \right|-\left| 1 \right|}{\left| z+1 \right|}\le P\le \frac{\left| \left( z+1 \right)\left( 1+i \right) \right|+\left| 1 \right|}{\left| z+1 \right|}\) \(\Leftrightarrow \left| 1+i \right|-\frac{1}{\left| z+1 \right|}\le P\le \left| 1+i \right|+\frac{1}{\left| z+1 \right|}\)

\(\Leftrightarrow \sqrt{2}-\frac{1}{\left| z+1 \right|}\le P\le \sqrt{2}+\frac{1}{\left| z+1 \right|}\,\,\,\,\,\,\,\,\,\left( 1 \right)\)

Mà: \(\left| z+1 \right|\ge 1\Rightarrow \left\{ \frac{1}{\left| z+1 \right|}\le 1;\,\,\,\frac{-1}{\left| z+1 \right|}\ge -1\,\,\,(2) \right.\)

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:

\(\begin{array}{l} \bullet \,\,\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right| \Leftrightarrow \left\{ \begin{array}{l} {a_1}{b_2} = {a_2}{b_1}\\ {a_1}{a_2} + {b_1}{b_2} \ge 0 \end{array} \right.;\,\,\,\\ \bullet \,\,\,\left| {{z_1}} \right| - \left| {{z_2}} \right| = \left| {{z_1} + {z_2}} \right| \Leftrightarrow \left\{ \begin{array}{l} {a_1}{b_2} = {a_2}{b_1};\,\,\left| {{z_1}} \right| \ge \left| {{z_2}} \right|\\ {a_1}{a_2} + {b_1}{b_2} \le 0 \end{array} \right. \end{array}\)

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:

\(\bullet \,\,\,P=\sqrt{2}+1\) 

\(\Leftrightarrow \left\{ \begin{align} & \left| \left( z+1 \right)\left( 1+i \right)+1 \right|=\left| \left( z+1 \right)\left( 1+i \right) \right|+\left| 1 \right| \\ & \left| z+1 \right|=1 \\ \end{align} \right.\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} a + b + 1 = 0\\ a + 1 - b \ge 0\\ \sqrt {{{(a + 1)}^2} + {b^2}} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a + 1 = - b\\ - 2b \ge 0\\ 2{b^2} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a = \frac{{\sqrt 2 }}{2} - 1\\ b \le 0\\ b = - \frac{{\sqrt 2 }}{2} \end{array} \right.\\ \Leftrightarrow z = \frac{{\sqrt 2 }}{2} - 1 - \frac{{\sqrt 2 }}{2}i \end{array}\)

\(\bullet \,\,\,P=\sqrt{2}-1\)

\(\Leftrightarrow \left\{ \begin{align} & \left| \left( z+1 \right)\left( 1+i \right)+1 \right|=\left| \left( z+1 \right)\left( 1+i \right) \right|-\left| 1 \right| \\ & \left| z+1 \right|=1 \\ \end{align} \right.\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} a + b + 1 = 0\\ a + 1 - b \le 0\\ \sqrt {{{(a + 1)}^2} + {b^2}} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a + 1 = - b\\ - 2b \le 0\\ 2{b^2} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a = - \frac{{\sqrt 2 }}{2} - 1\\ b \ge 0\\ b = \frac{{\sqrt 2 }}{2} \end{array} \right.\\ \Leftrightarrow z = - \frac{{\sqrt 2 }}{2} - 1 + \frac{{\sqrt 2 }}{2}i \end{array}\)

Vậy:

\(\left\{ \begin{align} & M=\sqrt{2}+1 \\ & m=\sqrt{2}-1 \\ \end{align} \right.\)

\(\,\,\,\,\,\Rightarrow {{M}^{2}}+{{m}^{2}}=6\).