Cho số phức z thỏa mãn |z+i|=|z+2-i| . Khi đó |z| đạt giá trị nhỏ nhất bằng

* Hướng dẫn giải

Chọn B

$$\begin{gathered} \mathrm{Đặt} \mathrm{z}=\mathrm{a}+\mathrm{bi} (\mathrm{a};\mathrm{b}\in \mathrm{ℝ}) \\ \Rightarrow \left|\mathrm{a}+\mathrm{bi}+\mathrm{i}\right|=\left|\mathrm{a}+2+\mathrm{bi}-\mathrm{i}\right| \\ \iff \sqrt{{\mathrm{a}}^2+{(\mathrm{b}+1)}^2}=\sqrt{{(\mathrm{a}+2)}^2+{(\mathrm{b}-1)}^2} \\ \iff 2\mathrm{b}+1=4\mathrm{a}+4-2\mathrm{b}+1 \\ \iff 4\mathrm{a}-4\mathrm{b}=-4 \\ \iff \mathrm{a}-\mathrm{b}=-1 \\ \iff \mathrm{a}=\mathrm{b}-1 \\ \mathrm{Khi} \mathrm{đó} \left|\mathrm{z}\right|=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{{(\mathrm{b}-1)}^2+{\mathrm{b}}^2}=\sqrt{2{\mathrm{b}}^2-2\mathrm{b}+1} \\ \mathrm{Ta} \mathrm{thấy} \mathrm{hàm} \mathrm{f}\left(\mathrm{b}\right)=2{\mathrm{b}}^2-2\mathrm{b}+1 \mathrm{đạt} \mathrm{GTNN} \mathrm{bằng} \frac{1}{2} \mathrm{khi} \mathrm{b}=\frac{1}{2} \\ \Rightarrow {\left|\mathrm{z}\right|}_{\min }=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\mathrm{khi} \mathrm{z}=\frac{-1}{2}+\frac{1}{2}\mathrm{i} \end{gathered}$$