Đặt[TEX] z=x+yi[/TEX] với [TEX]x,y[/TEX] thuộc[TEX] R[/TEX]
[TEX]\|z+2+i\|=\|z-1\|\Leftrightarrow{(x+2)^2+(y+1)^2=(x-1)^2+y^2[/TEX][TEX]\Leftrightarrow{y=-(3x+2)[/TEX]
[TEX]\|z+1-2i\|^2=(x+1)^2+(y-2)^2=(x+1)^2+(3x+4)^2=10x^2+26x+17=10[(x+\frac{13}{10})^2+\frac{1}{100}][/TEX][TEX]\ge{\frac{1}{10}}[/TEX]
Đẳng thức xảy ra khi[TEX] \left{x=\frac{-13}{10}\\y=\frac{19}{10}[/TEX] Vậy [TEX]z=\frac{-13}{10}+\frac{19}{10}[/TEX]