Chọn D
Cách 1:
Ta có
\(z+2+i-\left|z\right|\left(1+i\right)=0\)
\(\Leftrightarrow a+bi+2+i-\sqrt{a^{2} +b^{2} } \left(1+i\right)=0 \)
\(\Leftrightarrow a+2-\sqrt{a^{2} +b^{2} } +\left(b+1-\sqrt{a^{2} +b^{2} } \right)i=0\)
\(\Leftrightarrow \left\{\begin{array}{l} {a+2-\sqrt{a^{2} +b^{2} } =0\, \, \left(1\right)} \\ {b+1-\sqrt{a^{2} +b^{2} } =0\, \, \, \left(2\right)} \end{array}\right. \)
Lấy \((1)
\) trừ \((2)\) ta được \(a-b+1=0\Leftrightarrow b=a+1\)
thế vào \((1)
\) ta được :
\(a+2-\sqrt{a^{2} +\left(a+1\right)^{2} } =0\Leftrightarrow a+2=\sqrt{2a^{2} +2a+1} \)
\(\Leftrightarrow \left\{\begin{array}{l} {a\ge -2} \\ {a^{2} +4a+4=2a^{2} +2a+1} \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {a\ge -2} \\ {a^{2} -2a-3=0} \end{array}\right. \)
\(\Leftrightarrow \left\{\begin{array}{l} {a\ge -2} \\ {\left[\begin{array}{l} {a=3\, \, \, \, \, \, \left(tm\right)} \\ {a=-1\, \, \, \left(tm\right)} \end{array}\right. } \end{array}\right. . \)
Với a=3 \(\Rightarrow b=4 ; a=-1\Rightarrow b=0 .\)
Vì \(\left|z\right|>1\Rightarrow z=3+4i\Rightarrow \left\{\begin{array}{l} {a=3} \\ {b=4} \end{array}\right. \Rightarrow P=a+b=3+4=7 .\)
Cách 2:
Ta có \(z+2+i-\left|z\right|\left(1+i\right)=0\)
\(\Leftrightarrow z=\left(\left|z\right|-2\right)+\left(\left|z\right|-1\right)i{\rm \; }\left(*\right)\)
\(z+2+i-\left|z\right|\left(1+i\right)=0\Leftrightarrow z=\left(\left|z\right|-2\right)+\left(\left|z\right|-1\right)i\)
\(\Leftrightarrow \left|z\right|^{2} =\left(\left|z\right|-2\right)^{2} +\left(\left|z\right|-1\right)^{2} \)
\(\\ {\Leftrightarrow \left|z\right|^{2} -6\left|z\right|+5=0\Leftrightarrow \left[\begin{array}{l} {\left|z\right|=5{\rm \; }\left(N\right)} \\ {\left|z\right|=1{\rm \; }\left(L\right)} \end{array}\right. } \)
Thay \(\left|z\right|=5\) vào \(\left(*\right)\) ta được \(z=3+4i=a+bi\)
\(\Rightarrow \left\{\begin{array}{l} {a=3} \\ {b=4} \end{array}\right. \Rightarrow P=a+b=3+4=7. \)