Dễ dàng kiểm tra z=0 không thỏa mãn\[\left( {3 - i} \right)\left| z \right| = \frac{z}{{w - 1}} + 1 - i\]

Ta có: \[\left( {3 - i} \right)\left| z \right| = \frac{z}{{w - 1}} + 1 - i \Leftrightarrow \frac{z}{{w - 1}} = \left( {3 - i} \right)\left| z \right| + i - 1\]

\[ \Leftrightarrow \frac{z}{{w - 1}} = \left( {3\left| z \right| - 1} \right) + \left( {1 - \left| z \right|} \right)i\]

\[ \Rightarrow \left| {\frac{z}{{w - 1}}} \right| = \sqrt {10{{\left| z \right|}^2} - 8\left| z \right| + 2} \Rightarrow \left| {w - 1} \right| = \sqrt {\frac{{{{\left| z \right|}^2}}}{{10{{\left| z \right|}^2} - 8\left| z \right| + 2}}} \]

Nhận xét: \[T = \left| {w + i} \right| \le \left| {w - 1} \right| + \left| {1 + i} \right| = \frac{1}{{\sqrt {\frac{2}{{{{\left| z \right|}^2}}} - \frac{8}{{\left| z \right|}} + 10} }} + \sqrt 2 \]

\[ = \frac{1}{{\sqrt {2{{\left( {\frac{1}{{\left| z \right|}} - 2} \right)}^2} + 2} }} + \sqrt 2 \le \frac{{3\sqrt 2 }}{2}\]

Dấu “=” xảy ra khi và chỉ khỉ

\(\left\{ {\begin{array}{*{20}{c}}{|z| = \frac{1}{2}}\\{w - 1 = k(1 + i)}\\{(3 - i)|z| = \frac{z}{{w - 1}} + 1 - i}\end{array}} \right.(k > 0)\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{|z| = \frac{1}{2}}\\{w - 1 = k(1 + i)}\\{(3 - i)\frac{1}{2} = \frac{z}{{k(1 + i)}} + 1 - i}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{|z| = \frac{1}{2}}\\{w - 1 = k(1 + i)}\\{z = \frac{{1 + i}}{2}.\frac{{2k}}{{1 - i}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{|z| = \frac{1}{2}}\\{w - 1 = k(1 + i)}\\{|z| = k(dok > 0)}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{|z| = \frac{1}{2} = k}\\{w - 1 = \frac{1}{2}(1 + i)}\\{z = \frac{{1 + i}}{2}.\frac{{2k}}{{1 - i}}}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{z = \frac{i}{2}}\\{w = \frac{3}{2} + \frac{1}{2}i}\end{array}} \right.\)

Vậy,\[\max T = \frac{{3\sqrt 2 }}{2}\]

Đáp án cần chọn là: B

Hai số phức z = a + b i , z ′ = a + b ′ i bằng nhau nếu: