Cho số phức z thỏa mãn:$\left| {z + i + 1} \right| = \left| {\overline z - 2i} \right|$. Tìm giá trị nhỏ nhất của $\left| z \right|$.
A. \( - \frac{1}{2}\).
B. \( - \frac{{\sqrt 2 }}{2}\).
C. \(\frac{1}{2}\).
B. \(\frac{{\sqrt 2 }}{2}\).
A. \( - \frac{1}{2}\).
B. \( - \frac{{\sqrt 2 }}{2}\).
C. \(\frac{1}{2}\).
B. \(\frac{{\sqrt 2 }}{2}\).
Ta có: $\left| {x + yi + i + 1} \right| = \left| {x - yi - 2i} \right|\,\, \Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y + 1} \right)^2} = {x^2} + {\left( {y + 2} \right)^2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 2x - 2y - 2 = 0 \Rightarrow x = 1 + y$
$ \Rightarrow \left| z \right| = \sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {y + 1} \right)}^2} + {y^2}} = \sqrt {2{y^2} + 2y + 1} \ge \frac{{\sqrt 2 }}{2}$
$ \Rightarrow \left| z \right| \ge \frac{{\sqrt 2 }}{2} \Rightarrow {\left| z \right|_{\min }} = \frac{{\sqrt 2 }}{2}$$ \Leftrightarrow x = \frac{1}{2};\,\,\,y = \frac{{ - 1}}{2}$
Vậy chọn đáp án A.
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 2x - 2y - 2 = 0 \Rightarrow x = 1 + y$
$ \Rightarrow \left| z \right| = \sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {y + 1} \right)}^2} + {y^2}} = \sqrt {2{y^2} + 2y + 1} \ge \frac{{\sqrt 2 }}{2}$
$ \Rightarrow \left| z \right| \ge \frac{{\sqrt 2 }}{2} \Rightarrow {\left| z \right|_{\min }} = \frac{{\sqrt 2 }}{2}$$ \Leftrightarrow x = \frac{1}{2};\,\,\,y = \frac{{ - 1}}{2}$
Vậy chọn đáp án A.