Cho số phức

D

- Hướng dẫn giải

Đáp án D.

Đặt

$z=a+bi\Rightarrow a+bi+2+i-\sqrt{a^2+b^2}\left(1+i\right)=0$

$$\begin{gathered} \iff \left\{\begin{array}{l}a+2-\sqrt{a^2+b^2}=0\\ b+1-\sqrt{a^2+b^2}=0\end{array}\right.\iff \left\{\begin{array}{l}a+2=b+1\\ b+1=\sqrt{a^2+b^2}\end{array}\right. \\ \iff \left\{\begin{array}{l}a=b-1\\ b\ge -1\\ b^2+2b+1=a^2+b^2\end{array}\right.\iff \left\{\begin{array}{l}a=b-1\\ b\ge -1\\ 2b+1={\left(b-1\right)}^2\end{array}\right. \\ \iff \left[\begin{array}{l}b=0;a=-1\\ b=4;a=3\end{array}\right.. \end{gathered}$$

Do $\left|z\right|>1\Rightarrow a=3,b=4.$