ចំនួនកុំផ្លិចទម្រង់ពិជគណិត

តារាងចម្លើយមេរៀនទី១ (ចំនួនកុំផ្លិច ១០០ សំណួរ)

ដំណោះស្រាយលម្អិត លំហាត់អនុវត្តន៍មេរៀនទី១

លំហាត់ ១៖ បំប្លែងជាចំនួននិម្មិត
រូបមន្តគន្លឹះ៖ ចំពោះ \(a > 0\) គេបាន \(\sqrt{-a} = \sqrt{a \cdot (-1)} = i\sqrt{a}\) ដែល \(i^2 = -1\)។

1. \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i\)

2. \(-\sqrt{-121} = -\sqrt{121} \cdot i = -11i\)

3. \(\sqrt{-6} = \sqrt{6} \cdot i = i\sqrt{6}\)

4. \(\sqrt{-12} = \sqrt{4 \cdot 3} \cdot i = 2i\sqrt{3}\)

5. \(\sqrt{-144} = \sqrt{144} \cdot i = 12i\)

6. \(-\sqrt{-17} = -\sqrt{17} \cdot i = -i\sqrt{17}\)

7. \(\sqrt{-64} = \sqrt{64} \cdot i = 8i\)

8. \(\sqrt{-169} = \sqrt{169} \cdot i = 13i\)

លំហាត់ ២៖ កំណត់តម្លៃ \(x\) និង \(y\)
ទ្រឹស្តីបទ៖ ចំនួនកុំផ្លិចពីរស្មើគ្នា \(A+Bi = C+Di \iff \begin{cases} A = C \\ B = D \end{cases}\) (\(A, B, C, D \in \mathbb{R}\))។

1. \((2-i)x + (1+3i)y = 4+5i\)
   \(\iff 2x - xi + y + 3yi = 4+5i\)
   \(\iff (2x+y) + (-x+3y)i = 4+5i\)
   តាមសមភាពចំនួនកុំផ្លិច គេទាញបានប្រព័ន្ធសមីការ៖
   \(\begin{cases} 2x + y = 4 \quad{} (1) \\ -x + 3y = 5 \quad{} (2) \end{cases}\)
   ពី \((1)\)៖ \(y = 4-2x\) ជំនួសក្នុង \((2)\)៖ \(-x + 12 - 6x = 5 \implies x = 1\)
   \(\implies y = 4 - 2 = 2\)
   ដូចនេះ \( x = 1,\; y = 2\)

2. \(-4 - \sqrt{-8} = -x - iy\)
   \(\iff -4 - i\sqrt{8} = -x - iy \iff -4 - 2i\sqrt{2} = -x - iy\)
   \(\implies \begin{cases} -x = -4 \\ -y = -2\sqrt{2} \end{cases} \iff \begin{cases} x = 4 \\ y = 2\sqrt{2} \end{cases}\)
   ដូចនេះ \( x = 4,\; y = 2\sqrt{2}\)

3. \((x-1) + (y+3)i = 1 - 3i\)
   \(\implies \begin{cases} x - 1 = 1 \implies x = 2 \\ y + 3 = -3 \implies y = -6 \end{cases}\)
   ដូចនេះ \( x = 2,\; y = -6\)

4. \((x+iy)^2 = 8 - 6i\)
   ពន្លាត៖ \(x^2 - y^2 + 2xyi = 8 - 6i\)
   ម៉្យាងទៀត យកម៉ូឌុលសងខាង៖
   \(|(x+iy)^2| = |8-6i| \implies x^2 + y^2 = \sqrt{8^2 + (-6)^2} = 10\)
   គេបានប្រព័ន្ធសមីការ៖
   \(\begin{cases} x^2 - y^2 = 8 \quad{} (1) \\ 2xy = -6 \implies xy = -3 < 0 \quad{} (2) \\ x^2 + y^2 = 10 \quad{} (3) \end{cases}\)
   បូក \((1) + (3)\)៖ \(2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3\)
   ដក \((3) - (1)\)៖ \(2y^2 = 2 \implies y^2 = 1 \implies y = \pm 1\)
   ដោយ \(xy = -3 < 0 \implies x, y\) មានសញ្ញាផ្ទុយគ្នា៖
   ដូចនេះ \( (x,y) \in \{(3, -1),\; (-3, 1)\}\)

5. \((x+iy)^2 = -5 + 12i\)
   ប្រព័ន្ធសមីការ៖
   \(\begin{cases} x^2 - y^2 = -5 \quad{} (1) \\ xy = 6 > 0 \quad{} (2) \\ x^2 + y^2 = \sqrt{(-5)^2+12^2} = 13 \quad{} (3) \end{cases}\)
   បូក \((1) + (3)\)៖ \(2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2\)
   ដក \((3) - (1)\)៖ \(2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3\)
   ដោយ \(xy > 0 \implies x, y\) មានសញ្ញាដូចគ្នា៖
   ដូចនេះ \( (x,y) \in \{(2, 3),\; (-2, -3)\}\)

6. \((x+iy)^2 = 24 - 10i\)
   ប្រព័ន្ធសមីការ៖
   \(\begin{cases} x^2 - y^2 = 24 \\ xy = -5 < 0 \\ x^2 + y^2 = \sqrt{24^2+(-10)^2} = 26 \end{cases}\)
   \(\implies 2x^2 = 50 \implies x = \pm 5\)
   \(\implies 2y^2 = 2 \implies y = \pm 1\)
   ដោយ \(xy < 0\)៖
   ដូចនេះ \( (x,y) \in \{(5, -1),\; (-5, 1)\}\)

លំហាត់ ៣៖ ដោះស្រាយសមីការ \(x^2 = -a\)
រូបមន្ត៖ ចំពោះ \(a > 0\) សមីការ \(x^2 = -a\) មានឫសកុំផ្លិចពីរគឺ \(x = \pm i\sqrt{a}\)។

1. \(x^2 = -18 \iff x = \pm i\sqrt{18} = \pm 3i\sqrt{2}\)

2. \(x^2 = -24 \iff x = \pm i\sqrt{24} = \pm 2i\sqrt{6}\)

3. \(x^2 = -81 \iff x = \pm i\sqrt{81} = \pm 9i\)

4. \(x^2 = -75 \iff x = \pm i\sqrt{75} = \pm 5i\sqrt{3}\)

5. \(x^2 = -625 \iff x = \pm i\sqrt{625} = \pm 25i\)

6. \(x^2 = -10^{-4} \iff x = \pm i\sqrt{10^{-4}} = \pm 10^{-2}i = \pm 0.01i\)

7. \(x^2 = -\frac{48}{121} \iff x = \pm i\frac{\sqrt{48}}{\sqrt{121}} = \pm \frac{4\sqrt{3}}{11}i\)

8. \(x^2 = -\frac{3}{100} \iff x = \pm i\frac{\sqrt{3}}{\sqrt{100}} = \pm \frac{\sqrt{3}}{10}i\)

លំហាត់ ៤៖ ប្រមាណវិធីលើ \(z_1, z_2\)

1. \(z_1 = 2+5i,\; z_2 = 1-7i\)៖

   • \(z_1 + z_2 = (2+1) + (5-7)i = 3 - 2i\)

   • \(z_1 - z_2 = (2-1) + (5 - (-7))i = 1 + 12i\)

   • \(z_1 z_2 = (2+5i)(1-7i) = 2 - 14i + 5i - 35i^2\)
     \(= 2 - 9i + 35 = 37 - 9i\)

   • \(\frac{z_1}{z_2} = \frac{(2+5i)(1+7i)}{(1-7i)(1+7i)} = \frac{2+14i+5i+35i^2}{1^2+7^2}\)
     \(= \frac{2+19i-35}{50} = -\frac{33}{50} + \frac{19}{50}i\)

2. \(z_1 = \sqrt{2}-i\sqrt{3},\; z_2 = \sqrt{2}+i\sqrt{3}\)៖

   • \(z_1 + z_2 = 2\sqrt{2}\)

   • \(z_1 - z_2 = -2i\sqrt{3}\)

   • \(z_1 z_2 = (\sqrt{2})^2 - (i\sqrt{3})^2 = 2 - 3(-1) = 5\)

   • \(\frac{z_1}{z_2} = \frac{(\sqrt{2}-i\sqrt{3})^2}{5} = \frac{2 - 2i\sqrt{6} - 3}{5} = -\frac{1}{5} - \frac{2\sqrt{6}}{5}i\)

3. \(z_1 = 3+4i,\; z_2 = -1+2i\)៖

   • \(z_1 + z_2 = 2 + 6i\)

   • \(z_1 - z_2 = 4 + 2i\)

   • \(z_1 z_2 = -3 + 6i - 4i - 8 = -11 + 2i\)

   • \(\frac{z_1}{z_2} = \frac{(3+4i)(-1-2i)}{(-1)^2+2^2} = \frac{-3-6i-4i+8}{5} = \frac{5-10i}{5} = 1 - 2i\)

4. \(z_1 = 5-2i,\; z_2 = 2+3i\)៖

   • \(z_1 + z_2 = 7 + i\)

   • \(z_1 - z_2 = 3 - 5i\)

   • \(z_1 z_2 = 10 + 15i - 4i + 6 = 16 + 11i\)

   • \(\frac{z_1}{z_2} = \frac{(5-2i)(2-3i)}{2^2+3^2} = \frac{10-15i-4i-6}{13} = \frac{4-19i}{13} = \frac{4}{13} - \frac{19}{13}i\)

លំហាត់ ៥៖ សរសេរជាទម្រង់ \(a+bi\)

1. \(A = \frac{13+12i}{6i-8} + \frac{(2i+1)^2}{2+i}\)៖
   - ភាគទី១៖ \(\frac{13+12i}{-8+6i} = \frac{(13+12i)(-8-6i)}{(-8)^2+6^2} = \frac{-32-174i}{100} = -\frac{8}{25} - \frac{87}{50}i\)
   - ភាគទី២៖ \(\frac{(1+2i)^2}{2+i} = \frac{-3+4i}{2+i} = \frac{(-3+4i)(2-i)}{2^2+1^2} = \frac{-2+11i}{5} = -\frac{4}{10} + \frac{22}{10}i = -\frac{20}{50} + \frac{110}{50}i\)
   - បូកបញ្ចូលគ្នា៖
   \(A = \left(-\frac{16}{50} - \frac{87}{50}i\right) + \left(-\frac{20}{50} + \frac{110}{50}i\right) = -\frac{36}{50} + \frac{23}{50}i = -\frac{18}{25} + \frac{23}{50}i\)

2. \(B = \frac{(i\sqrt{2}-\sqrt{3})(i\sqrt{3}-\sqrt{2})}{1-i}\)៖
   ភាគយក៖ \(i^2\sqrt{6} - 2i - 3i + \sqrt{6} = -\sqrt{6} - 5i + \sqrt{6} = -5i\)
   \(\implies B = \frac{-5i}{1-i} = \frac{-5i(1+i)}{1^2+(-1)^2} = \frac{-5i + 5}{2} = \frac{5}{2} - \frac{5}{2}i\)

3. \(C = \frac{1+i}{\sqrt{2}-i} + \frac{1-i}{\sqrt{2}+i}\)៖
   តម្រូវភាគបែងរួម៖
   \(C = \frac{(1+i)(\sqrt{2}+i) + (1-i)(\sqrt{2}-i)}{(\sqrt{2})^2 - i^2}\)
   \(= \frac{(\sqrt{2}-1 + i(\sqrt{2}+1)) + (\sqrt{2}-1 - i(\sqrt{2}+1))}{2+1} = \frac{2\sqrt{2}-2}{3}\)

4. \(D = \frac{3+i}{2-i} + \frac{1-2i}{1+i}\)៖
   \(\frac{3+i}{2-i} = \frac{(3+i)(2+i)}{5} = \frac{5+5i}{5} = 1+i\)
   \(\frac{1-2i}{1+i} = \frac{(1-2i)(1-i)}{2} = \frac{-1-3i}{2} = -\frac{1}{2} - \frac{3}{2}i\)
   \(\implies D = (1+i) + \left(-\frac{1}{2} - \frac{3}{2}i\right) = \frac{1}{2} - \frac{1}{2}i\)

5. \(E = \frac{4-3i}{2+3i} + \frac{2+i}{1-i}\)៖
   \(= \frac{(4-3i)(2-3i)}{13} + \frac{(2+i)(1+i)}{2} = \frac{-1-18i}{13} + \frac{1+3i}{2}\)
   \(= \frac{-2-36i + 13+39i}{26} = \frac{11+3i}{26} = \frac{11}{26} + \frac{3}{26}i\)

6. \(F = \frac{5-2i}{1+i} - \frac{3+4i}{2-i}\)៖
   \(= \frac{(5-2i)(1-i)}{2} - \frac{(3+4i)(2+i)}{5} = \frac{3-7i}{2} - \frac{2+11i}{5}\)
   \(= \frac{15-35i - (4+22i)}{10} = \frac{11-57i}{10} = \frac{11}{10} - \frac{57}{10}i\)

លំហាត់ ៦៖ គណនា \(i^n\)
រូបមន្ត៖ \(i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1\)។ សម្រាប់ \(n = 4q + r\) នោះ \(i^n = (i^4)^q \cdot i^r = i^r\)។

1. \(i^5 = i^{4(1)+1} = i\)

2. \(i^{12} = i^{4(3)+0} = 1\)

3. \(i^{23} = i^{4(5)+3} = i^3 = -i\)

4. \(i^{34} = i^{4(8)+2} = i^2 = -1\)

5. \(i^{47} = i^{4(11)+3} = i^3 = -i\)

6. \(i^{56} = i^{4(14)+0} = 1\)

7. \(i^{63} = i^{4(15)+3} = i^3 = -i\)

8. \(i^{78} = i^{4(19)+2} = i^2 = -1\)

9. \(i^{91} = i^{4(22)+3} = i^3 = -i\)

10. \(i^{104} = i^{4(26)+0} = 1\)

លំហាត់ ៧៖ ពន្លាតជាទម្រង់ \(a+bi\)

1. \((2+i)^3 = 2^3 + 3(2^2)(i) + 3(2)(i^2) + i^3\)
   \(= 8 + 12i - 6 - i = 2 + 11i\)

2. \((1-i)^4 = [(1-i)^2]^2 = (1-2i-1)^2 = (-2i)^2 = 4i^2 = -4\)

3. \((3+2i)^3 = 3^3 + 3(3^2)(2i) + 3(3)(2i)^2 + (2i)^3\)
   \(= 27 + 54i - 36 - 8i = -9 + 46i\)

4. \((2-3i)^4 = [(2-3i)^2]^2 = (4 - 12i - 9)^2 = (-5 - 12i)^2\)
   \(= 25 + 120i - 144 = -119 + 120i\)

5. \((1+3i)^3 = 1 + 9i - 27 - 27i = -26 - 18i\)

6. \((4-i)^4 = [(4-i)^2]^2 = (15-8i)^2 = 225 - 240i - 64 = 161 - 240i\)

7. \((2-i)^3 = 8 - 12i - 6 + i = 2 - 11i\)

8. \((3-2i)^4 = [(3-2i)^2]^2 = (5-12i)^2 = -119 - 120i\)

លំហាត់ ៨៖ កំណត់ \(x\) ឱ្យកន្សោមជាចំនួនពិត
ចំនួនកុំផ្លិច \(Z\) ជាចំនួនពិត \(\iff \text{Im}(Z) = 0\)។

1. \(Z = \frac{2+i}{1-i} + \frac{x+3i}{1+2i} = \frac{1+3i}{2} + \frac{x+6 + (3-2x)i}{5}\)
   \(\text{Im}(Z) = \frac{3}{2} + \frac{3-2x}{5} = \frac{15 + 6 - 4x}{10} = \frac{21-4x}{10}\)
   \(Z \in \mathbb{R} \iff 21 - 4x = 0 \iff x = \frac{21}{4}\)

2. \(Z = \frac{1-2i}{1+i} + \frac{x-i}{2-i} = \frac{-1-3i}{2} + \frac{2x+1 + (x-2)i}{5}\)
   \(\text{Im}(Z) = -\frac{3}{2} + \frac{x-2}{5} = \frac{-15 + 2x - 4}{10} = \frac{2x-19}{10}\)
   \(Z \in \mathbb{R} \iff 2x - 19 = 0 \iff x = \frac{19}{2}\)

3. \(Z = \frac{3+4i}{2-i} + \frac{x+2i}{1+i} = \frac{2+11i}{5} + \frac{x+2 + (2-x)i}{2}\)
   \(\text{Im}(Z) = \frac{11}{5} + \frac{2-x}{2} = \frac{22 + 10 - 5x}{10} = \frac{32-5x}{10}\)
   \(Z \in \mathbb{R} \iff 32 - 5x = 0 \iff x = \frac{32}{5}\)

4. \(Z = \frac{5-3i}{3+2i} + \frac{x-i}{2-i} = \frac{9-19i}{13} + \frac{2x+1 + (x-2)i}{5}\)
   \(\text{Im}(Z) = -\frac{19}{13} + \frac{x-2}{5} = \frac{-95 + 13x - 26}{65} = \frac{13x - 121}{65}\)
   \(Z \in \mathbb{R} \iff 13x - 121 = 0 \iff x = \frac{121}{13}\)

លំហាត់ ៩៖ សំណុំចំណុចក្នុងប្លង់កុំផ្លិច
ចំណុចរូបភាព \(M(a, b)\) តំណាងឱ្យចំនួនកុំផ្លិច \(z = a+bi\)៖

• \(z_1 = -1-3i \implies A(-1, -3)\)

• \(z_2 = -1+3i \implies B(-1, 3)\)

• \(z_3 = 2 = 2+0i \implies C(2, 0)\)

• \(z_4 = 4i = 0+4i \implies D(0, 4)\)

• \(z_5 = 2-3i \implies E(2, -3)\)

• \(z_6 = 1+i \implies F(1, 1)\)

លំហាត់ ១០៖ រូបភាពវ៉ិចទ័រ
គេឱ្យ \(z_1 = 3+2i \implies \vec{u}(3, 2)\) និង \(z_2 = 2+4i \implies \vec{v}(2, 4)\)។

1. \(z_1+z_2 = 5+6i \implies \vec{w}_1(5, 6)\)

2. \(z_1-z_2 = 1-2i \implies \vec{w}_2(1, -2)\)

3. \(2z_2 = 4+8i \implies \vec{w}_3(4, 8)\)

4. \(z_2+2z_1 = (2+4i) + (6+4i) = 8+8i \implies \vec{w}_4(8, 8)\)

5. \(-z_1 = -3-2i \implies \vec{w}_5(-3, -2)\)

6. \(-2z_1+z_2 = (-6-4i) + (2+4i) = -4 \implies \vec{w}_6(-4, 0)\)

7. \(z_1+3z_2 = (3+2i) + (6+12i) = 9+14i \implies \vec{w}_7(9, 14)\)

8. \(2z_1-3z_2 = (6+4i) - (6+12i) = -8i \implies \vec{w}_8(0, -8)\)

លំហាត់ ១១៖ រូបភាពវ៉ិចទ័រ \(z_1, z_2\)
គេឱ្យ \(z_1 = 2+i \implies (2, 1)\), \(z_2 = -2+i \implies (-2, 1)\)៖

1. \(z_1+z_2 = 2i \implies (0, 2)\)

2. \(z_1-z_2 = 4 \implies (4, 0)\)

3. \(-2z_2 = 4-2i \implies (4, -2)\)

4. \(3z_1+z_2 = (6+3i) + (-2+i) = 4+4i \implies (4, 4)\)

លំហាត់ ១២៖ ផ្ទៀងផ្ទាត់លក្ខណៈកុំផ្លិចឆ្លាស់

1. \(z = 2-i,\; w = 1+3i\)៖
   \(z+w = 3+2i \implies \overline{z+w} = 3-2i\)
   \(\bar{z} = 2+i,\; \bar{w} = 1-3i \implies \bar{z}+\bar{w} = 3-2i\)
   ដូចនេះ \( \overline{z+w} = \bar{z}+\bar{w}\) (ផ្ទៀងផ្ទាត់)

2. \(z = 3+i,\; w = 2-2i\)៖
   \(z-w = 1+3i \implies \overline{z-w} = 1-3i\)
   \(\bar{z} = 3-i,\; \bar{w} = 2+2i \implies \bar{z}-\bar{w} = 1-3i\)
   ដូចនេះ \( \overline{z-w} = \bar{z}-\bar{w}\) (ផ្ទៀងផ្ទាត់)

3. \(z = 1+i,\; w = 2-i\)៖
   \(zw = (1+i)(2-i) = 3+i \implies \overline{zw} = 3-i\)
   \(\bar{z}\bar{w} = (1-i)(2+i) = 3-i\)
   ដូចនេះ \( \overline{zw} = \bar{z}\bar{w}\) (ផ្ទៀងផ្ទាត់)

4. \(z = 3-2i,\; w = 1+2i\)៖
   \(\frac{z}{w} = \frac{(3-2i)(1-2i)}{5} = \frac{-1-8i}{5} \implies \overline{\left(\frac{z}{w}\right)} = -\frac{1}{5} + \frac{8}{5}i\)
   \(\frac{\bar{z}}{\bar{w}} = \frac{3+2i}{1-2i} = \frac{(3+2i)(1+2i)}{5} = \frac{-1+8i}{5}\)
   ដូចនេះ \( \overline{\left(\frac{z}{w}\right)} = \frac{\bar{z}}{\bar{w}}\) (ផ្ទៀងផ្ទាត់)

លំហាត់ ១៣៖ គណនា \(z\cdot \bar{z}\)
រូបមន្ត៖ ចំពោះ \(z = a+bi\) គេបាន \(z\cdot\bar{z} = (a+bi)(a-bi) = a^2+b^2\)។

1. \(z = 3-4i \implies z\bar{z} = 3^2+(-4)^2 = 25\)

2. \(z = 1+2i \implies z\bar{z} = 1^2+2^2 = 5\)

3. \(z = -2+3i \implies z\bar{z} = (-2)^2+3^2 = 13\)

4. \(z = 2-5i \implies z\bar{z} = 2^2+(-5)^2 = 29\)

5. \(z = 4+i \implies z\bar{z} = 4^2+1^2 = 17\)

6. \(z = -3-4i \implies z\bar{z} = (-3)^2+(-4)^2 = 25\)

លំហាត់ ១៤៖ សមីការដឺក្រេទីពីរមេគុណពិត

1. \(x^2 + 4x + 13 = 0\)៖
   \(\Delta' = 2^2 - 13 = -9 = (3i)^2\)
   \(\implies x = -2 \pm 3i\)

2. \(x^2 - 4x + 13 = 0\)៖
   \(\Delta' = (-2)^2 - 13 = -9 = (3i)^2\)
   \(\implies x = 2 \pm 3i\)

3. \(2x^2 + 3x + 4 = 0\)៖
   \(\Delta = 3^2 - 4(2)(4) = 9 - 32 = -23 = (i\sqrt{23})^2\)
   \(\implies x = \frac{-3 \pm i\sqrt{23}}{4}\)

4. \(x^2 + x + 1 = 0\)៖
   \(\Delta = 1 - 4 = -3 = (i\sqrt{3})^2\)
   \(\implies x = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}\)

5. \(x^2 - 2x + 2 = 0\)៖
   \(\Delta' = 1 - 2 = -1 = i^2 \implies x = 1 \pm i\)

6. \(x^2 + 6x + 13 = 0\)៖
   \(\Delta' = 9 - 13 = -4 = (2i)^2 \implies x = -3 \pm 2i\)

លំហាត់ ១៥៖ កំណត់តម្លៃ \(a\) និង \(b\)

• ករណី \((a+bi)(1-2i) = 7+i\)៖
  \(a+bi = \frac{7+i}{1-2i} = \frac{(7+i)(1+2i)}{5} = \frac{7+14i+i-2}{5} = \frac{5+15i}{5} = 1+3i\)
  \(\implies a = 1,\; b = 3\)

• ករណី \((a+bi)^2 = -3-4i\)៖
  \(a^2-b^2 = -3\), \(2ab = -4 \implies ab = -2 < 0\), \(a^2+b^2 = \sqrt{(-3)^2+(-4)^2} = 5\)
  \(\implies 2a^2 = 2 \implies a = \pm 1\) និង \(2b^2 = 8 \implies b = \pm 2\) (\(a, b\) សញ្ញាផ្ទុយគ្នា)
  ដូចនេះ \( (a,b) = (1, -2) \) ឬ \( (-1, 2)\)

លំហាត់ ១៦៖ កំណត់តម្លៃ \(a\) និង \(b\)

• ករណី \((a+bi)(3+i) = 5+5i\)៖
  \(a+bi = \frac{5+5i}{3+i} = \frac{(5+5i)(3-i)}{10} = \frac{15-5i+15i+5}{10} = \frac{20+10i}{10} = 2+i\)
  \(\implies a = 2,\; b = 1\)

• ករណី \(\frac{a+bi}{2-i} = 1+3i\)៖
  \(a+bi = (1+3i)(2-i) = 2 - i + 6i + 3 = 5+5i\)
  \(\implies a = 5,\; b = 5\)

លំហាត់ ១៧៖ ប្រមាណវិធីចំនួនកុំផ្លិច

1. \((4+3i) + (-6+2i) = -2 + 5i\)

2. \((1-4i) - (3-2i) = -2 - 2i\)

3. \((3-i)(2+5i) = 6 + 15i - 2i + 5 = 11 + 13i\)

4. \((4+i)(4-i) = 4^2 + 1^2 = 17\)

5. \(\frac{7+i}{1+i} = \frac{(7+i)(1-i)}{2} = \frac{7-7i+i+1}{2} = \frac{8-6i}{2} = 4 - 3i\)

6. \(\frac{5-i}{2+3i} = \frac{(5-i)(2-3i)}{13} = \frac{10-15i-2i-3}{13} = \frac{7}{13} - \frac{17}{13}i\)

លំហាត់ ១៨៖ រូបភាពវ៉ិចទ័រ \(z_1, z_2\)
\(z_1 = 3-i \implies \vec{u}(3, -1)\), \(z_2 = 1+2i \implies \vec{v}(1, 2)\)៖

• \(z_1+z_2 = 4+i \implies\) តំណាងដោយវ៉ិចទ័រ \((4, 1)\) (អង្កត់ទ្រូងប្រលេឡូក្រាមសង់លើ \(\vec{u}, \vec{v}\))

• \(z_1-z_2 = 2-3i \implies\) តំណាងដោយវ៉ិចទ័រ \((2, -3)\) (\(=\vec{u}+(-\vec{v})\))

លំហាត់ ១៩៖ រូបភាពវ៉ិចទ័រ \(3z\) និង \(-2z\)

• ចំពោះ \(z = 1-i \implies 3z = 3-3i\) តំណាងដោយ \((3, -3)\) (ទិសដៅដូច \(\overrightarrow{OM}(z)\) ប្រវែង ៣ ដង)

• ចំពោះ \(z = -1+3i \implies -2z = 2-6i\) តំណាងដោយ \((2, -6)\) (ទិសដៅផ្ទុយ ប្រវែង ២ ដង)

លំហាត់ ២០៖ សំណុំចំណុចក្នុងប្លង់កុំផ្លិច
- \(2-i \implies A(2, -1)\)
- \(-3+2i \implies B(-3, 2)\)
- \(-1-2i \implies C(-1, -2)\)
- \(4 = 4+0i \implies D(4, 0)\) (លើអ័ក្សពិត)

លំហាត់ ២១៖ សរសេរជាទម្រង់ \(a+bi\)

1. \((3+i)(2-4i) = 6 - 12i + 2i + 4 = 10 - 10i\)

2. \(\frac{5+2i}{2-i} = \frac{(5+2i)(2+i)}{5} = \frac{10+5i+4i-2}{5} = \frac{8}{5} + \frac{9}{5}i\)

3. \(\frac{3-4i}{1+3i} = \frac{(3-4i)(1-3i)}{10} = \frac{3-9i-4i-12}{10} = -\frac{9}{10} - \frac{13}{10}i\)

លំហាត់ ២២៖ លក្ខណៈឆ្លាស់
គេឱ្យ \(z_1 = 2-i,\; z_2 = 3+4i\)៖

• \(z_1+z_2 = 5+3i \implies \overline{z_1+z_2} = 5-3i\)

• \(z_1-z_2 = -1-5i \implies \overline{z_1-z_2} = -1+5i\)

• \(z_1 z_2 = (2-i)(3+4i) = 10+5i \implies \overline{z_1 z_2} = 10-5i\)

• \(\frac{z_1}{z_2} = \frac{(2-i)(3-4i)}{25} = \frac{2-11i}{25} \implies \overline{\left(\frac{z_1}{z_2}\right)} = \frac{2+11i}{25}\)

លំហាត់ ២៣៖ កំណត់ \(x\) ឱ្យកន្សោមជាចំនួនពិត
\(Z = \frac{1+i}{2-i} + \frac{x-i}{1+i} = \frac{1+3i}{5} + \frac{x-1 - (x+1)i}{2}\)
\(\text{Im}(Z) = \frac{3}{5} - \frac{x+1}{2} = \frac{6 - 5(x+1)}{10} = \frac{1-5x}{10}\)
\(Z \in \mathbb{R} \iff 1 - 5x = 0 \iff x = \frac{1}{5}\)

លំហាត់ ២៤៖ ប្រមាណវិធី \(z_1, z_2\)

1. \(z_1 = 3-4i, z_2 = 2+i\)៖
   \(z_1+z_2 = 5-3i,\; z_1-z_2 = 1-5i\)
   \(z_1 z_2 = 6+3i-8i-4i^2 = 10-5i\)
   \(\frac{z_1}{z_2} = \frac{(3-4i)(2-i)}{5} = \frac{6-3i-8i-4}{5} = \frac{2}{5} - \frac{11}{5}i\)

2. \(z_1 = \sqrt{3}+i\sqrt{2}, z_2 = \sqrt{3}-i\sqrt{2}\)៖
   \(z_1+z_2 = 2\sqrt{3},\; z_1-z_2 = 2i\sqrt{2},\; z_1 z_2 = 3+2 = 5\)
   \(\frac{z_1}{z_2} = \frac{(\sqrt{3}+i\sqrt{2})^2}{5} = \frac{3+2i\sqrt{6}-2}{5} = \frac{1}{5} + \frac{2\sqrt{6}}{5}i\)

លំហាត់ ២៥៖ ទម្រង់ពិជគណិត

1. \(\frac{5+i}{2-3i} = \frac{(5+i)(2+3i)}{13} = \frac{7+17i}{13}\) ; \(\frac{(1-i)^2}{1+i} = \frac{-2i(1-i)}{2} = -1-i\)
   \(\implies A = \frac{7+17i}{13} - 1 - i = -\frac{6}{13} + \frac{4}{13}i\)

2. \((2+i)(1-3i) = 5-5i \implies B = \frac{(5-5i)(1-i)}{2} = \frac{-10i}{2} = -5i\)

3. \(\frac{1+2i}{3-i} = \frac{(1+2i)(3+i)}{10} = \frac{1+7i}{10}\) ; \(\frac{1-2i}{3+i} = \frac{(1-2i)(3-i)}{10} = \frac{1-7i}{10}\)
   \(\implies C = \frac{14i}{10} = \frac{7}{5}i\)

លំហាត់ ២៦៖ គណនាស្វ័យគុណ

• \((3+i)^3 = 27 + 27i + 9i^2 + i^3 = 27+27i-9-i = 18+26i\)

• \((2+i)^4 = [(2+i)^2]^2 = (3+4i)^2 = 9+24i-16 = -7+24i\)

លំហាត់ ២៧៖ កំណត់ \(x\) ឱ្យជាចំនួនពិត
\(\frac{1+3i}{2+i} = \frac{(1+3i)(2-i)}{5} = \frac{5+5i}{5} = 1+i\) ; \(\frac{x+2i}{1-i} = \frac{(x+2i)(1+i)}{2} = \frac{(x-2)+(x+2)i}{2}\)
\(\text{Im}(Z) = 1 + \frac{x+2}{2} = \frac{x+4}{2} = 0 \iff x = -4\) (ពេលនោះ \(Z = -2\))។

លំហាត់ ២៨៖ រូបភាពសំណុំចំណុច
\(A(3, 2),\; B(-2, 1),\; C(-3, -2),\; D(1, -3),\; E(-4, 0),\; F(0, 3)\)។

លំហាត់ ២៩៖ វ៉ិចទ័ររូបភាព
\(z_1 = 1+3i, z_2 = 3-i\)៖
\(z_1+z_2 = 4+2i \to (4, 2),\; z_1-z_2 = -2+4i \to (-2, 4),\; 3z_1 = 3+9i \to (3, 9),\; z_1-2z_2 = -5+5i \to (-5, 5)\)។

លំហាត់ ៣០៖ វ៉ិចទ័ររូបភាព
\(z_1 = -1+2i, z_2 = 3+i\)៖
\(z_1+z_2 = 2+3i \to (2, 3),\; z_2-z_1 = 4-i \to (4, -1),\; -3z_1 = 3-6i \to (3, -6)\)។

លំហាត់ ៣១៖ គណនាកន្សោម \(z_4\) ដល់ \(z_9\)
គេឱ្យ \(z_1 = 3+2i,\; z_2 = -1+3i,\; z_3 = 1-i\)៖

1. \(z_4 = z_1 - 3z_2 = (3+2i) - 3(-1+3i) = 6 - 7i\)

2. \(z_5 = i z_1^2 = i(3+2i)^2 = i(5+12i) = -12 + 5i\)

3. \(z_6 = -i z_3 = -i(1-i) = -1 - i\)

4. \(z_7 = z_1 z_2 = (3+2i)(-1+3i) = -3 + 9i - 2i - 6 = -9 + 7i\)

5. \(z_8 = z_2^4 = [(-1+3i)^2]^2 = (-8-6i)^2 = 64 + 96i - 36 = 28 + 96i\)

6. \(z_9 = \frac{z_1}{z_3} = \frac{3+2i}{1-i} = \frac{(3+2i)(1+i)}{2} = \frac{1+5i}{2} = \frac{1}{2} + \frac{5}{2}i\)

លំហាត់ ៣២៖ គណនាកន្សោម
គេឱ្យ \(Z_1 = 1+2i,\; Z_2 = 1,\; Z_3 = 2i\)៖

1. \(A = 2Z_1 - 3Z_2 + Z_3 = 2(1+2i) - 3(1) + 2i = -1 + 6i\)

2. \(B = Z_1^2 - Z_2 Z_3 = (1+2i)^2 - (1)(2i) = (-3+4i) - 2i = -3 + 2i\)

3. \(C = \frac{Z_1 + Z_3}{Z_2 - Z_3} = \frac{1+4i}{1-2i} = \frac{(1+4i)(1+2i)}{5} = \frac{-7+6i}{5} = -\frac{7}{5} + \frac{6}{5}i\)

លំហាត់ ៣៣៖ ដោះស្រាយសមីការក្នុង \(\mathbb{C}\)

1. \(z + (2-i)\bar{z} = 3+4i\)
   តាង \(z = x+iy\) គេបាន៖
   \((x+iy) + (2-i)(x-iy) = 3+4i\)
   \(\iff (x+iy) + (2x - 2iy - ix - y) = 3+4i\)
   \(\iff (3x-y) - (x+y)i = 3+4i\)
   \(\implies \begin{cases} 3x - y = 3 \\ -(x+y) = 4 \implies x+y = -4 \end{cases}\)
   បូកបញ្ចូលគ្នា៖ \(4x = -1 \implies x = -\frac{1}{4}\)
   \(y = -4 - x = -4 + \frac{1}{4} = -\frac{15}{4}\)
   ដូចនេះ \( z = -\frac{1}{4} - \frac{15}{4}i\)

2. \(\frac{z-1}{z+1} = i \iff z - 1 = i(z + 1) \iff z(1-i) = 1+i\)
   \(\implies z = \frac{1+i}{1-i} = \frac{(1+i)^2}{2} = \frac{2i}{2} = i\)
   ដូចនេះ \( z = i\)

3. \(z^2 - (3+i)z + 4 + 3i = 0\)
   \(\Delta = (3+i)^2 - 4(4+3i) = 8+6i - 16 - 12i = -8 - 6i\)
   រកឫសការេ \(\delta = u+iv\) នៃ \(\Delta\)៖
   \(u^2 - v^2 = -8\), \(2uv = -6 \implies uv = -3 < 0\), \(u^2+v^2 = \sqrt{64+36} = 10\)
   \(\implies u^2 = 1,\; v^2 = 9 \implies \delta = 1-3i\) (ឬ \(-1+3i\))
   ឫសសមីការ៖ \(z = \frac{(3+i) \pm (1-3i)}{2}\)
   ដូចនេះ \( z_1 = 2-i,\; z_2 = 1+2i\)

4. \(\begin{cases} z_1 + z_2 = 4-2i \\ z_1 z_2 = 2-4i \end{cases}\)
   \(z_1, z_2\) ជាឫសនៃសមីការ \(Z^2 - (4-2i)Z + (2-4i) = 0\)៖
   \(\Delta' = (2-i)^2 - (2-4i) = (3-4i) - (2-4i) = 1\)
   \(\implies Z = (2-i) \pm 1 \implies Z = 3-i\) ឬ \(Z = 1-i\)
   ដូចនេះ \( (z_1, z_2) = (3-i,\; 1-i)\) ឬ \((1-i,\; 3-i)\)