Cho số phức z=a+bi (a,b thuộc R) thỏa mãn z+2+i - lzl(1+i)=0

* Hướng dẫn giải

Đáp án D

$\begin{array}{l}z+2+i-\left|z\right|(1+i)=0\\ \iff (a+bi)+2+i-\sqrt{a^2+b^2}(1+i)=0\\ \iff a+2-\sqrt{a^2+b^2}+(b+1-\sqrt{a^2+b^2})i=0\\ \Rightarrow \left\{\begin{array}{l}a+2-\sqrt{a^2+b^2}=0\\ b+1-\sqrt{a^2+b^2}=0\end{array}\right.\\ \Rightarrow a-b+1=0\Rightarrow a=b-1\\ \Rightarrow b+1-\sqrt{{(b-1)}^2+b^2}=0\\ \Rightarrow \sqrt{2b^2-2b+1}=b+1\\ \Rightarrow \left\{\begin{array}{l}b\ge -1\\ b^2-4b=0\end{array}\right.\Rightarrow \left[\begin{array}{l}b=0\\ b=4\end{array}\right.\Rightarrow \left[\begin{array}{l}a=-1\left(L\right)\\ a=3\end{array}\right.\\ \Rightarrow P=4+3=7\end{array}$