Cho số phức z thỏa mãn 5|z-1|=|z+1-3i|+3|z-1+i|

Gọi A(0;1), $B\left(-1;3\right),C\left(1;-1\right)$ . Ta thấy A  là trung điểm của BC

  $\Rightarrow M{A}^2=\frac{M{B}^2+M{C}^2}{2}-\frac{B{C}^2}{4}\iff M{B}^2+M{C}^2=2M{A}^2+\frac{B{C}^2}{2}=2M{A}^2+10$

Ta lại có : $5\left|z-i\right|=\left|z+1-3i\right|+3\left|z-1+i\right|$

 $\iff 5MA=MB+3MC\le \sqrt{10}.\sqrt{M{B}^2+M{C}^2}$

$\Rightarrow 25M{A}^2\le 10\left(2M{A}^2+10\right)\Rightarrow MC\le 2\sqrt{5}$

Mà $\left|z-2+3i\right|=\left|\left(z-i\right)+\left(-2+4i\right)\right|\le \left|z-i\right|+\left|2-4i\right|\le \left|z-i\right|+2\sqrt{5}\le 4\sqrt{5}$ .

Dấu "="  xảy ra khi $\left\{\begin{array}{l}\left|z-i\right|=2\sqrt{5}\\ \frac{a}{-2}=\frac{b-1}{4}\end{array}\right.$ , với $z=a+bi$ ; $a,b\in ℝ$ .

$\iff \left[\begin{array}{l}z=2-3i\left(loai\right)\\ z=-2+5i\end{array}\right.$