Số phức, các phép toán với số phức

Ta có

\(\begin{array}{l}\dfrac{{3 - 4i}}{z} = \dfrac{{\left( {2 + 3i} \right)\overline z }}{{{{\left| z \right|}^2}}} + 2 + i\\ \Leftrightarrow \dfrac{{3 - 4i}}{z} = \dfrac{{\left( {2 + 3i} \right)\overline z }}{{z.\overline z }} + 2 + i\\ \Leftrightarrow \dfrac{{3 - 4i}}{z} = \dfrac{{2 + 3i}}{z} + 2 + i\\ \Leftrightarrow 3 - 4i = 2 + 3i + \left( {2 + i} \right).z\\ \Leftrightarrow \left( {2 + i} \right).z = 1 - 7i\\ \Leftrightarrow z = \dfrac{{1 - 7i}}{{2 + i}} =  - 1 - 3i\end{array}\)

Vậy \(\left| z \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 3} \right)}^2}}  = \sqrt {10} .\)

Đặt \(z = a + bi \Rightarrow \overline z  = a - bi\).

Theo bài ra ta có:

\(\begin{array}{l}\,\,\,\,\,\dfrac{{5\left( {\overline z  + i} \right)}}{{z + 1}} = 2 - i\\ \Rightarrow \dfrac{{5\left( {a - bi + i} \right)}}{{a + bi + 1}} = 2 - i\\ \Leftrightarrow 5\left[ {a - \left( {b - 1} \right)i} \right] = \left( {a + 1 + bi} \right)\left( {2 - i} \right)\\ \Leftrightarrow 5a - 5\left( {b - 1} \right)i = 2\left( {a + 1} \right) + b + \left( {2b - a - 1} \right)i\\ \Leftrightarrow \left\{ \begin{array}{l}5a = 2a + 2 + b\\5 - 5b = 2b - a - 1\end{array} \right. \Rightarrow a = b = 1\\ \Rightarrow z = 1 + i \Rightarrow {z^2} = 2i\\ \Rightarrow {\rm{w}} = 1 + z + {z^2} = 1 + 1 + i + 2i = 2 + 3i\end{array}\)

Vậy \(\left| {\rm{w}} \right| = \sqrt {{2^2} + {3^2}}  = \sqrt {13} .\)

Ta có:

\(\begin{array}{l}\left| w \right| = 9 \Rightarrow \left| {{z^2}} \right| = 9 \Leftrightarrow {\left| z \right|^2} = 9\\ \Leftrightarrow \left| z \right| = 3 \Leftrightarrow \left| {\frac{{m + 3i}}{{1 - i}}} \right| = 3\\ \Leftrightarrow \frac{{\left| {m + 3i} \right|}}{{\left| {1 - i} \right|}} = 3 \Leftrightarrow \frac{{\left| {m + 3i} \right|}}{{\sqrt 2 }} = 3\\ \Leftrightarrow \left| {m + 3i} \right| = 3\sqrt 2  \Leftrightarrow \sqrt {{m^2} + 9}  = 3\sqrt 2 \\ \Leftrightarrow {m^2} + 9 = 18 \Leftrightarrow {m^2} = 9\\ \Leftrightarrow m =  \pm 3\end{array}\)

Đặt \(z = x + yi\). Theo giả thiết ta có \(xy = 625.\)

Ta có:

\(\begin{array}{l}\dfrac{z}{{3 + 4i}} = \dfrac{{x + yi}}{{3 + 4i}} = \dfrac{{\left( {x + yi} \right)\left( {3 - 4i} \right)}}{{25}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{3x + 4y + \left( { - 4x + 3y} \right)i}}{{25}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{3x + 4y}}{{25}} + \dfrac{{ - 4x + 3y}}{{25}}i\end{array}\)

Số phức \(\dfrac{z}{{3 + 4i}}\) có phần thực là \(a = \dfrac{{3x + 4y}}{{25}} \Rightarrow \left| a \right| = \dfrac{{\left| {3x + 4y} \right|}}{{25}}\).

Ta có: \(xy = 625 \Leftrightarrow y = \dfrac{{625}}{x}\)\( \Rightarrow \left| a \right| = \dfrac{{\left| {3x +4. \dfrac{{625}}{x}} \right|}}{{25}}\).

Vì \(3x,\,\,\dfrac{{625}}{x}\) cùng dấu nên \(\left| {3x +4 .\dfrac{{625}}{x}} \right| \ge 2\sqrt {3x.4.\dfrac{{625}}{x}}  = 100\sqrt 3 \).

Vậy \(\left| a \right| \ge 4\sqrt 3 \). Dấu “=” xảy ra \( \Leftrightarrow 3x = 4.\dfrac{{625}}{x} \Leftrightarrow x =  \pm \dfrac{{50}}{{\sqrt 3 }}\).

Theo bài ra ta có:

\(\begin{array}{l}\left( {z + \dfrac{5}{{\left| z \right|}}} \right)i = 7 - z \Leftrightarrow zi + \dfrac{{5i}}{{\left| z \right|}} = 7 - z \Leftrightarrow z\left( {i + 1} \right) = 7 - \dfrac{{5i}}{{\left| z \right|}}\\ \Leftrightarrow 2{\left| z \right|^2} = 49 + \dfrac{{25}}{{{{\left| z \right|}^2}}} \Leftrightarrow 2{\left| z \right|^4} - 49{\left| z \right|^2} - 25 = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\left| z \right|^2} = 25\,\,\left( {tm} \right)\\\left| z \right| =  - \dfrac{1}{2}\,\,\left( {ktm} \right)\end{array} \right. \Leftrightarrow \left| z \right| = 5\,\,\left( {Do\,\,\left| z \right| > 0} \right)\end{array}\)

Thay \(\left| z \right| = 5\) vào biểu thức đề bài ta có:

\(\left( {z + 1} \right)i = 7 - z \Leftrightarrow z\left( {i + 1} \right) = 7 - i \Leftrightarrow z = \dfrac{{7 - i}}{{i + 1}} = 3 - 4i\).

Ta có : \(\dfrac{{{z_1}}}{{{z_2}}}\) là số thuần ảo nên ta viết lại \(\dfrac{{{z_1}}}{{{z_2}}} = ki \Leftrightarrow {z_1} = ki{z_2}\)

Khi đó \(\left| {{z_1} - {z_2}} \right| = 10 \Leftrightarrow \left| {ki{z_2} - {z_2}} \right| = 10 \Leftrightarrow \left| {{z_2}\left( { - 1 + ki} \right)} \right| = 10\) \( \Leftrightarrow \left| {{z_2}} \right| = \dfrac{{10}}{{\left| { - 1 + ki} \right|}} = \dfrac{{10}}{{\sqrt {{k^2} + 1} }}\)

\( \Rightarrow \left| {{z_1}} \right| = \left| {ki} \right|.\left| {{z_2}} \right| = \left| k \right|.\dfrac{{10}}{{{k^2} + 1}}\) \( \Rightarrow \left| {{z_1}} \right| + \left| {{z_2}} \right| = \dfrac{{10\left| k \right|}}{{\sqrt {{k^2} + 1} }} + \dfrac{{10}}{{\sqrt {{k^2} + 1} }} = \dfrac{{10\left( {\left| k \right| + 1} \right)}}{{\sqrt {{k^2} + 1} }}\)

Xét \(y = f\left( t \right) = \dfrac{{10\left( {t + 1} \right)}}{{\sqrt {{t^2} + 1} }}\) \( \Rightarrow 10\left( {t + 1} \right) = y\sqrt {{t^2} + 1}  \Leftrightarrow 100{\left( {t + 1} \right)^2} = {y^2}\left( {{t^2} + 1} \right)\)

\( \Leftrightarrow 100\left( {{t^2} + 2t + 1} \right) = {y^2}{t^2} + {y^2} \Leftrightarrow \left( {{y^2} - 100} \right){t^2} - 200t + {y^2} - 100 = 0\)

Phương trình có nghiệm \( \Leftrightarrow \Delta ' = {100^2} - {\left( {{y^2} - 100} \right)^2} = {y^2}\left( {200 - {y^2}} \right) \ge 0\) \( \Leftrightarrow  - 10\sqrt 2  \le y \le 10\sqrt 2 \)

Vậy \(\max y = 10\sqrt 2 \) khi \(t = 1\) hay \(k =  \pm 1\).

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

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

\( \Rightarrow \left| {\dfrac{z}{{w - 1}}} \right| = \sqrt {10{{\left| z \right|}^2} - 8\left| z \right| + 2} \)\( \Rightarrow \left| {w - 1} \right| = \sqrt {\dfrac{{{{\left| z \right|}^2}}}{{10{{\left| z \right|}^2} - 8\left| z \right| + 2}}} \)

Nhận xét: \(T = \left| {w + i} \right| \le \left| {w - 1} \right| + \left| {1 + i} \right| \)\(= \dfrac{1}{{\sqrt {\dfrac{2}{{{{\left| z \right|}^2}}} - \dfrac{8}{{\left| z \right|}} + 10} }} + \sqrt 2  \)\(= \dfrac{1}{{\sqrt {2{{\left( {\dfrac{1}{{\left| z \right|}} - 2} \right)}^2} + 2} }} + \sqrt 2  \le \dfrac{{3\sqrt 2 }}{2}\)

Dấu “=” xảy ra khi và chỉ khỉ

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{\left| z \right| = \dfrac{1}{2}}\\{w - 1 = k\left( {1 + i} \right)}\\{\left( {3 - i} \right)\left| z \right| = \dfrac{z}{{w - 1}} + 1 - i}\end{array}} \right.,\left( {k > 0} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}\left| z \right| = \dfrac{1}{2}\\w - 1 = k\left( {1 + i} \right)\\\left( {3 - i} \right)\dfrac{1}{2} = \dfrac{z}{{k\left( {1 + i} \right)}} + 1 - i\end{array} \right.\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\left| z \right| = \dfrac{1}{2}}\\\begin{array}{l}w - 1 = k\left( {1 + i} \right)\\z = \dfrac{{1 + i}}{2}.\dfrac{{2k}}{{1 - i}}\end{array}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\left| z \right| = \dfrac{1}{2}}\\\begin{array}{l}w - 1 = k\left( {1 + i} \right)\\\left| z \right| = k(dok > 0)\end{array}\end{array}} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left| z \right| = \dfrac{1}{2} = k\\w - 1 = \dfrac{1}{2}\left( {1 + i} \right)\\z = \dfrac{{1 + i}}{2}.\dfrac{{2k}}{{1 - i}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}z = \dfrac{i}{2}\\w = \dfrac{3}{2} + \dfrac{1}{2}i\end{array} \right.\end{array}\)

Vậy, \(\max T = \dfrac{{3\sqrt 2 }}{2}\).