Cho số phức z thỏa mãn |z-1-i|=1. Khi 3|z|

* Hướng dẫn giải

Đáp án B

Đặt $z=a+bi\Rightarrow \left|z-1-i\right|=1\iff {\left(a-1\right)}^2+{\left(b-1\right)}^2=1\iff a^2+b^2=2a+2b-1$ .

Khi đó$$\begin{gathered} 3\left|z\right|+2\left|z-4-4i\right|=3\sqrt{a^2+b^2}+2\sqrt{{\left(a-4\right)}^2+{\left(b-4\right)}^2}=3\sqrt{2a+2b-1}+2\sqrt{\left(a^2+b^2\right)-8a-8b+32} \\ =3\sqrt{2a+2b-1}+2\sqrt{\left(2a+2b-1\right)-8a-8b+32}=3\sqrt{2a+2b-1}+2\sqrt{-6a-6b+31} \\ =\sqrt{3}\sqrt{6a+6b-3}+2\sqrt{-6a-6b+31}\le \sqrt{\left({\sqrt{3}}^2+2^2\right)\left(6a+6b-3-6a-6b+31\right)}=14 \end{gathered}$$

Dấu “=” xảy ra khi$\left\{\begin{array}{l}{a}^2+b^2=2a+2b-1\\ \frac{\sqrt{6a+6b-3}}{\sqrt{3}}=\frac{\sqrt{-6a-6b+31}}{2}\end{array}\right.\iff \left\{\begin{array}{l}a+b=\frac{5}{2}\\ a^2+b^2=4\end{array}\right.\Rightarrow \left|z\right|=\sqrt{a^2+b^2}=2$