Cho số phức z thỏa mãn

A

- Hướng dẫn giải

Đáp án A.

Ta có $1={\left|z-2-3i\right|}^2=\left(z-2-3i\right).\overline{\left(z-2-3i\right)}=\left(z-2-3i\right)\left(\overline{z}-\overline{2+3i}\right)=\left(z-2-3i\right)\left(\overline{z}-2+3i\right)$ 

Lấy môđun hai vế, ta được $\left|z-2-3i\right|.\left|\overline{z}-2+3i\right|=1\iff \left|\overline{z}-2+3i\right|=1 (*)$ 

Đặt $w=\overline{z}+1+i\iff \overline{z}=w-1-i$, khi đó (*) $\iff \left|w-1-2-3i\right|=1\iff \left|w-3+2i\right|=1$. 

$\Rightarrow \left\{\begin{array}{l}{\left|w\right|}_{min}=\sqrt{3^2+2^2}-1=\sqrt{13}-1\\ {\left|w\right|}_{min}=\sqrt{3^2+2^2}-1=\sqrt{13}+1\end{array}\right.\Rightarrow \left\{\begin{array}{l}M=\sqrt{13}+1\\ m=\sqrt{13}-1\end{array}\Rightarrow M^2+m^2={\left(\sqrt{13}+1\right)}^2+{\left(\sqrt{13}-1\right)}^2=28.\right.$