TÌM SỐ PHỨC CÓ MÔ-ĐUN NHỎ NHẤT TRONG CÁC SỐ PHỨC Z THOẢ MÃN ĐIỀU KIỆN |Z + 1 – 5I| = |Z + 3 – I|

Gọi $z=x+yi\left(x;y\in R\right)$$z=x+yi\left(x;y\in \mathbb{R}\right)$$z = x + yi\,\,\,\,\,\,\,\left( {x;y \in \mathbb{R}} \right)$.
$\left|z+1-5i\right|=\left|\overline{z}+3-i\right|\iff \sqrt{{(x+1)}^2+{(y-5)}^2}=\sqrt{{(x+3)}^2+{(y+1)}^2}$$\left|z+1-5i\right|=\left|\overline{z}+3-i\right|\iff \sqrt{{(x+1)}^2+{(y-5)}^2}=\sqrt{{(x+3)}^2+{(y+1)}^2}$$\left| {z + 1 - 5i} \right| = \left| {\overline z + 3 - i} \right| \Leftrightarrow \sqrt {{{(x + 1)}^2} + {{(y - 5)}^2}}  = \sqrt {{{(x + 3)}^2} + {{(y + 1)}^2}}$
$\iff x+3y-4=0\iff x=4-3y$$\iff x+3y-4=0\iff x=4-3y$$\Leftrightarrow x + 3y - 4 = 0 \Leftrightarrow x = 4 - 3y$.
$\left|z\right|=\sqrt{x^2+y^2}=\sqrt{{(4-3y)}^2+y^2}=\sqrt{10y^2-24y+16}=\sqrt{10{\left(y-\frac{6}{5}\right)}^2+\frac{8}{5}}\ge \frac{2\sqrt{10}}{5}$$\left|z\right|=\sqrt{x^2+y^2}=\sqrt{{(4-3y)}^2+y^2}=\sqrt{10y^2-24y+16}=\sqrt{10{\left(y-{\displaystyle \frac{6}{5}}\right)}^2+{\displaystyle \frac{8}{5}}}\ge {\displaystyle \frac{2\sqrt{10}}{5}}$$\left| z \right| = \sqrt {{x^2} + {y^2}}  = \sqrt {{{(4 - 3y)}^2} + {y^2}}  = \sqrt {10{y^2} - 24y + 16}  = \sqrt {10{{\left( {y - \dfrac{6}{5}} \right)}^2} + \dfrac{8}{5}}  \ge \dfrac{{2\sqrt {10} }}{5}$.
Đẳng thức xảy ra khi $y=\frac{6}{5}\Rightarrow x=\frac{2}{5}$$y={\displaystyle \frac{6}{5}}\Rightarrow x={\displaystyle \frac{2}{5}}$$y = \dfrac{6}{5} \Rightarrow x = \dfrac{2}{5}$.
Vậy $\left|z\right|$$\left|z\right|$$\left| z \right|$ đạt giá trị nhỏ nhất bằng $\frac{2\sqrt{10}}{5}$${\displaystyle \frac{2\sqrt{10}}{5}}$$\dfrac{{2\sqrt {10} }}{5}$ khi $z=\frac{2}{5}+\frac{6}{5}i$$z={\displaystyle \frac{2}{5}}+{\displaystyle \frac{6}{5}}i$$z = \dfrac{2}{5} + \dfrac{6}{5}i$.
Vậy $z=\frac{2}{5}+\frac{6}{5}i$$z={\displaystyle \frac{2}{5}}+{\displaystyle \frac{6}{5}}i$$z = \dfrac{2}{5} + \dfrac{6}{5}i$ là số phức cần tìm.