លំហាត់ស្រាវជ្រាវ និង វិញ្ញាសាពង្រឹងសមត្ថភាព

តារាងចម្លើយ វិញ្ញាសាត្រៀមប្រឡងប្រជែង (ផ្នែកទី១, ៤៤ សំណួរ)

ដំណោះស្រាយលម្អិត លំហាត់ដែលធ្លាប់ចេញប្រឡង

លំហាត់ ១៖ [2002]

1. \(z = 1-i\sqrt{3}\) មាន \(r = \sqrt{1+3} = 2\)
   \(z = 2\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 2\left[\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right]\)
   តាមរូបមន្តដឺម័រ៖ \(z^{2002} = 2^{2002}\left[\cos\left(-\frac{2002\pi}{3}\right)+i\sin\left(-\frac{2002\pi}{3}\right)\right]\)
   ដោយ \(-\frac{2002\pi}{3} = -668\pi + \frac{2\pi}{3}\)
   ដូចនេះ \( z^{2002} = 2^{2002}\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)

2. \(z_1^2 = (a+ib)^2 = a^2-b^2+2abi\) និង \(iz_1 = -b+ai\)
   \(u = z_1^2+iz_1-\frac{1}{2} = \left(a^2-b^2-b-\frac{1}{2}\right) + (2ab+a)i\)
   ដូចនេះ \( x = a^2-b^2-b-\frac{1}{2},\; y = a(2b+1)\)

លំហាត់ ២៖ [2003]

1. \(Z = \frac{2(-1+i\sqrt{3})(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})} = \frac{2(-1+i\sqrt{3}+i\sqrt{3}+3)}{4} = \frac{2(2+2i\sqrt{3})}{4}\)
   ដូចនេះ \( Z = 1+i\sqrt{3} = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)

2. \(Z' = 1+i = \sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)
   \(\frac{Z'}{Z} = \frac{\sqrt{2}}{2}\left[\cos\left(\frac{\pi}{4}-\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}-\frac{\pi}{3}\right)\right]\)
   ដូចនេះ \( \left|\frac{Z'}{Z}\right| = \frac{\sqrt{2}}{2},\; \arg\left(\frac{Z'}{Z}\right) = -\frac{\pi}{12}+2k\pi\)

លំហាត់ ៣៖ [2004]

1. ភាគយក៖ \((3-2i)(1+i) = 3+3i-2i+2 = 5+i\)
   ភាគបែង៖ \(i(1+2i) = -2+i\)
   \(\frac{5+i}{-2+i} = \frac{(5+i)(-2-i)}{(-2)^2+1^2} = \frac{-10-5i-2i+1}{5} = -\frac{9}{5}-\frac{7}{5}i\)
   \(2xi-y = -\frac{9}{5}-\frac{7}{5}i \implies \begin{cases} -y = -\frac{9}{5} \\ 2x = -\frac{7}{5} \end{cases}\)
   ដូចនេះ \( x = -\frac{7}{10},\; y = \frac{9}{5}\)

2. \(1+Z = 1+\cos\frac{2\pi}{9}+i\sin\frac{2\pi}{9} = 2\cos^2\frac{\pi}{9}+2i\sin\frac{\pi}{9}\cos\frac{\pi}{9}\)
   \(= 2\cos\frac{\pi}{9}\left(\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}\right)\) ដែល \(2\cos\frac{\pi}{9} > 0\)
   ដូចនេះ \( (1+Z)^4 = 16\cos^4\frac{\pi}{9}\left(\cos\frac{4\pi}{9}+i\sin\frac{4\pi}{9}\right)\)

លំហាត់ ៤៖ [2005]

1. ភាគយក៖ \(2\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)^2 = 2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\)
   ភាគបែង៖ \(1+i\sqrt{3} = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)
   \(Z_1 = \cos\left(\frac{\pi}{6}-\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{6}-\frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\)
   ដូចនេះ \( Z_1 = \frac{\sqrt{3}}{2}-\frac{1}{2}i\)

2. \(2\overline{Z}_1 = \sqrt{3}+i\)
   \(Z_2 = x-ix+(1-y)+(1-y)i = (x+1-y)+(1-y-x)i\)
   \(Z_2+y-1 = x+(1-x-y)i\)
   \(2\overline{Z}_1 - (Z_2+y-1) = 0 \iff \sqrt{3}+i = x+(1-x-y)i\)
   \(\implies \begin{cases} x = \sqrt{3} \\ 1-x-y = 1 \end{cases}\)
   ដូចនេះ \( x = \sqrt{3},\; y = -\sqrt{3}\)

លំហាត់ ៥៖ [2006]

1. \(x_1 = 1+i\sqrt{3}\) ជាឫស \(\iff x_1^2+ax_1+b = 0\)
   \(x_1^2 = 1+2i\sqrt{3}-3 = -2+2i\sqrt{3}\)
   \(\implies (-2+a+b)+(2\sqrt{3}+a\sqrt{3})i = 0\)
   \(\implies \begin{cases} a+b-2 = 0 \\ \sqrt{3}(2+a) = 0 \end{cases}\)
   ដូចនេះ \( a = -2,\; b = 4\)

2. ផលបូកឫស \(x_1+x_2 = -a = 2 \implies x_2 = 2-x_1 = 1-i\sqrt{3}\)
   \(x_1 = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right),\; x_2 = 2\left[\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right]\)
   \(\frac{x_1}{x_2} = \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\)
   ដូចនេះ \( Z = \left(\frac{x_1}{x_2}\right)^2 = \cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\)

លំហាត់ ៦៖ [2007]

1. \(A^2 = (\sqrt{3}-1)^2-(\sqrt{3}+1)^2+2i(\sqrt{3}-1)(\sqrt{3}+1)\)
   \(= (4-2\sqrt{3})-(4+2\sqrt{3})+2i(3-1) = -4\sqrt{3}+4i\)
   \(|A^2| = \sqrt{48+16} = 8\), \(\cos\theta = -\frac{\sqrt{3}}{2},\; \sin\theta = \frac{1}{2} \implies \theta = \frac{5\pi}{6}\)
   ដូចនេះ \( A^2 = 8\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)\)

2. \(B = \frac{(x+iy)(1-i)}{2} = \frac{x+y}{2}+\frac{y-x}{2}i \implies 2\overline{B} = (x+y)+(x-y)i\)
   \(2\overline{B}-A^2 = 0 \iff (x+y)+(x-y)i = -4\sqrt{3}+4i\)
   \(\implies \begin{cases} x+y = -4\sqrt{3} \\ x-y = 4 \end{cases}\)
   ដូចនេះ \( x = 2-2\sqrt{3},\; y = -2-2\sqrt{3}\)

លំហាត់ ៧៖ [2008]

1. \(|W| = \sqrt{\cos^2\alpha+\sin^2\alpha} = 1\) និង \(|Z| = \sqrt{x^2+y^2}\)
   \(|Z| = |W| \iff x^2+y^2 = 1\)

2. ដោយ \(|Z| = 1\) នោះ \(Z\cdot\overline{Z} = |Z|^2 = 1\)
   ដូចនេះ \( \frac{1}{Z} = \overline{Z}\)

3. \(Z = 1 \iff x = 1,\; y = 0\) (ផ្ទៀងផ្ទាត់ \(x^2+y^2 = 1\))
   \(W = 1 \iff \cos\alpha = 1,\; \sin\alpha = 0 \iff \alpha = 2k\pi,\; k\in\mathbb{Z}\)

លំហាត់ ៨៖ [2009]
\((1+i)^2 = 2i = 2\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\) និង \(1-i\sqrt{3} = 2\left[\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right]\)
\(A = \frac{2\cdot 2}{2}\left[\cos\left(\frac{\pi}{2}+\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{\pi}{3}\right)\right]\)
ដូចនេះ \( A = 2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right) = -\sqrt{3}+i\)

លំហាត់ ៩៖ [2010]

1. \(\Delta' = (\sqrt{2})^2-4 = -2 = (i\sqrt{2})^2\)
   \(\implies z_1 = \sqrt{2}+i\sqrt{2},\; z_2 = \sqrt{2}-i\sqrt{2}\)
   \(|z_1| = |z_2| = \sqrt{2+2} = 2\)
   \(\arg(z_1) = \frac{\pi}{4},\; \arg(z_2) = -\frac{\pi}{4}\) (រួម \(2k\pi\))

2. \(\frac{\sqrt{2}+i\sqrt{2}}{\sqrt{2}-i\sqrt{2}} = \frac{2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)}{2\left[\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right]} = \cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\)
   ដូចនេះ \( w = \cos\pi+i\sin\pi\; (= -1)\)

លំហាត់ ១០៖ [2011]

1. \(-t^2+2t-4 = 0 \iff t^2-2t+4 = 0\)
   \(\Delta' = 1-4 = -3 = (i\sqrt{3})^2 \implies t = 1\pm i\sqrt{3}\)
   ដូចនេះ \( t_1 = 1-i\sqrt{3},\; t_2 = 1+i\sqrt{3}\)

2. \(t_1 = 2\left[\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right] \implies t_1^3 = 8(\cos(-\pi)+i\sin(-\pi)) = -8\)
   \(t_2 = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)
   \(Z = \frac{4t_2}{t_1^3} = \frac{8\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)}{8(\cos(-\pi)+i\sin(-\pi))}\)
   ដូចនេះ \( Z = \cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\)

លំហាត់ ១១៖ [2012]

1. \(y^2 = \left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4}-i\frac{\sqrt{3}}{2}-\frac{3}{4} = -\frac{1}{2}-i\frac{\sqrt{3}}{2} = x\)
   \(\implies A = x-y^2 = 0\)
   \(x^2 = \frac{1}{4}+i\frac{\sqrt{3}}{2}-\frac{3}{4} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}\)
   \(\implies B = x^2+x+1 = -\frac{1}{2}+i\frac{\sqrt{3}}{2}-\frac{1}{2}-i\frac{\sqrt{3}}{2}+1 = 0\)

2. \(x = \cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right),\; y = \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\)
   \(x^{2013} = \cos(-1342\pi)+i\sin(-1342\pi) = 1\)
   \(y^{2013} = \cos(1342\pi)+i\sin(1342\pi) = 1\)
   ដូចនេះ \( C = x^{2013}+y^{2013} = 2\) ជាចំនួនពិត។

លំហាត់ ១២៖ [2013]

1. \(a^2 = 12-8\sqrt{3}i-4 = 8-8\sqrt{3}i\)
   \(b^2 = 2-4i-2 = -4i\)
   \(4ai = 4i(2\sqrt{3}-2i) = 8+8\sqrt{3}i\)
   \(\sqrt{2}b = -2+2i\)
   ដូចនេះ \( Z = (8-8\sqrt{3}i)-4i+(8+8\sqrt{3}i)+(-2+2i) = 14-2i\)

2. \(a = 4\left[\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right]\)
   \(b = 2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)
   \(ab = 8\left(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12}\right)\)

លំហាត់ ១៣៖ [2014 លើកទី១]

1. \(z_1+z_2 = -2\), \(z_1-z_2 = 2i\sqrt{3}\)
   \(z_1z_2 = (-1)^2-(i\sqrt{3})^2 = 1+3 = 4\)

2. \(z_1 = 2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)
   \(z_2 = 2\left[\cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right)\right]\)

3. \(z_1^3 = 8(\cos 2\pi+i\sin 2\pi) = 8\) និង \(z_2^3 = 8(\cos(-2\pi)+i\sin(-2\pi)) = 8\)
   ដូចនេះ \( z_1^3-8 = 0\) និង \(z_2^3-8 = 0\) ពោលគឺ \(z_1, z_2\) ជាឫសនៃ \(z^3-8 = 0\)។

លំហាត់ ១៤៖ [2014 លើកទី២]

1. 1. \(z_1+z_2 = \sqrt{2}-i\sqrt{2}\), \(z_1+z_3 = \sqrt{2}+i\sqrt{2}\)
      \((z_1+z_2)(z_1+z_3) = (\sqrt{2})^2-(i\sqrt{2})^2 = 2+2 = 4\)

   2. \(|z_1+z_2| = 2,\; \arg(z_1+z_2) = -\frac{\pi}{4}\)
      \(|z_1+z_3| = 2,\; \arg(z_1+z_3) = \frac{\pi}{4}\)
      \(\frac{z_1+z_3}{z_1+z_2} = \cos\frac{\pi}{2}+i\sin\frac{\pi}{2} \implies \left(\frac{z_1+z_3}{z_1+z_2}\right)^2 = \cos\pi+i\sin\pi\)
      មានម៉ូឌុល \(1\) និងអាគុយម៉ង់ \(\pi\)

2. \(i^{4k} = 1,\; i^{4k+1} = i,\; i^{4k+2} = -1,\; i^{4k+3} = -i\)
   \(2015 = 4(503)+3 \implies i^{2015} = -i\)
   \(2014 = 4(503)+2 \implies i^{2014} = -1\)
   ដូចនេះ \( i^{2015}-i^{2014} = 1-i\)

លំហាត់ ១៥៖ [2015]

1. \(z_1+z_2 = 0\), \(z_1-z_2 = -2+2i\sqrt{3}\)
   \(z_1z_2 = (-1+i\sqrt{3})(1-i\sqrt{3}) = -1+i\sqrt{3}+i\sqrt{3}+3 = 2+2i\sqrt{3}\)

2. \(z_1-z_2 = 4\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)
   \(z_1z_2 = 4\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 4\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)

លំហាត់ ១៦៖ [2016]
\(z_1+z_2 = (\sqrt{3}+1-\sqrt{3})+(-1+1-\sqrt{3})i = 1-i\sqrt{3}\)
\(Z = (z_1+z_2)\times z_3 = -\frac{1}{2}(1-i\sqrt{3}) = -\frac{1}{2}+i\frac{\sqrt{3}}{2}\)
\(Z = \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\)
ដូចនេះ \( Z^3 = \cos 2\pi+i\sin 2\pi = 1\)

លំហាត់ ១៧៖ [2017]

1. \(z_1 = 1+i\sqrt{3} = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)

2. \(z_1^3 = 8(\cos\pi+i\sin\pi)\)
   ដូចនេះ \( |z_1^3| = 8,\; \arg(z_1^3) = \pi+2k\pi\)

3. \(z_2 = 6\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right) = 3\sqrt{2}-3\sqrt{2}i\)
   \(z_1z_2 = (1+i\sqrt{3})(3\sqrt{2}-3\sqrt{2}i) = 3\sqrt{2}-3\sqrt{2}i+3\sqrt{6}i+3\sqrt{6}\)
   ដូចនេះ \( z_1z_2 = (3\sqrt{2}+3\sqrt{6})+(3\sqrt{6}-3\sqrt{2})i\)

លំហាត់ ១៨៖ [2018]

1. \(z_1z_2 = (3+3i\sqrt{3})(\sqrt{3}+i) = 3\sqrt{3}+3i+9i-3\sqrt{3} = 12i\)
   \(\frac{z_1}{z_2} = \frac{(3+3i\sqrt{3})(\sqrt{3}-i)}{4} = \frac{3\sqrt{3}-3i+9i+3\sqrt{3}}{4} = \frac{3\sqrt{3}}{2}+\frac{3}{2}i\)

2. \(z_1z_2 = 12\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)
   \(\frac{z_1}{z_2} = 3\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right) \implies \left(\frac{z_1}{z_2}\right)^2 = 9\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)

3. \(\left(\frac{z_1}{z_2}\right)^3 = 27\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) = 27i\)

លំហាត់ ១៩៖ [2019]

1. \(\Delta' = (-4)^2-64 = -48 = (4i\sqrt{3})^2\)
   ដូចនេះ \( z = 4+4i\sqrt{3}\) ឬ \(z = 4-4i\sqrt{3}\)

2. \(\overline{z_2} = 4+4i\sqrt{3} = z_1 \implies 2z_1+\overline{z_2} = 3z_1 = 12+12i\sqrt{3}\)
   \(= 24\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 24\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)
   \((2z_1+\overline{z_2})^3 = 24^3(\cos\pi+i\sin\pi)\)
   ដូចនេះ \( (2z_1+\overline{z_2})^3 = -13824\)

លំហាត់ ២០៖ [2021]

1. \(\overline{z_1} = \sqrt{2}-i\sqrt{2}\)

2. \(|z_1| = \sqrt{2+2} = 2\), \(\cos\theta = \frac{\sqrt{2}}{2},\; \sin\theta = \frac{\sqrt{2}}{2} \implies \arg(z_1) = \frac{\pi}{4}\)
   \(z_1 = 2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)

3. សមីការ \(z^2 = 2(z\sqrt{2}-2) \iff z^2-2\sqrt{2}z+4 = 0\)
   ជំនួស \(z = \overline{z_1}\)៖ \((\sqrt{2}-i\sqrt{2})^2 = 2-4i-2 = -4i\)
   \(2\sqrt{2}\,\overline{z_1} = 4-4i\)
   \(\implies \overline{z_1}^2-2\sqrt{2}\,\overline{z_1}+4 = -4i-4+4i+4 = 0\)
   ដូចនេះ \( \overline{z_1}\) ជាឫសនៃសមីការ។

លំហាត់ ២១៖ [2022]

1. \((2+i)^2 = 3+4i\)
   \((3+2i)(3+4i) = 9+12i+6i-8 = 1+18i\)
   \(z_1 = i(1+18i) = -18+i \implies \overline{z_1} = -18-i\)

2. \(z_2 = z_1+19 = 1+i = \sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)
   \(z_2^4 = 4(\cos\pi+i\sin\pi) = -4\)
   \(z_2^6 = 8\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right) = -8i\)

លំហាត់ ២២៖ [2023]
\(\Delta' = 2^2-16 = -12 = (2i\sqrt{3})^2 \implies z = -2\pm 2i\sqrt{3}\)
\(z_1 = -2+2i\sqrt{3} = 4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)
\(z_2 = -2-2i\sqrt{3} = 4\left[\cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right)\right]\)

លំហាត់ ២៣៖ [2024]

1. តាង \(z = x+iy\) នោះ \(\overline{z} = x-iy\)
   \(2z-(1+i)\overline{z} = 2x+2iy-(x-iy+ix+y) = (x-y)+(3y-x)i\)
   \(\implies \begin{cases} x-y = -1 \\ -x+3y = 5 \end{cases} \implies y = 2,\; x = 1\)
   ដូចនេះ \( z = 1+2i\)
   \(2025 = 4(506)+1 \implies i^{2025} = i\)

2. \(z_1+z_2 = -4+3+i = -1+i\)
   ដូចនេះ \( z_1+z_2 = \sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)

លំហាត់ ២៤៖ [2025]
\(z_1 = \sqrt{3}-i\)៖ \(|z_1| = 2,\; \arg(z_1) = -\frac{\pi}{6}\)
\(z_1 = 2\left[\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right]\)
\(z_2 = 2(1-i)\)៖ \(|z_2| = 2\sqrt{2},\; \arg(z_2) = -\frac{\pi}{4}\)
\(z_2 = 2\sqrt{2}\left[\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right]\)
\(z_1^7 = 128\left[\cos\left(-\frac{7\pi}{6}\right)+i\sin\left(-\frac{7\pi}{6}\right)\right]\)
\(z_2^6 = (2\sqrt{2})^6\left[\cos\left(-\frac{3\pi}{2}\right)+i\sin\left(-\frac{3\pi}{2}\right)\right]\)
\(= 512\left[\cos\left(-\frac{3\pi}{2}\right)+i\sin\left(-\frac{3\pi}{2}\right)\right]\)
\(Z = \frac{128}{512}\left[\cos\left(-\frac{7\pi}{6}+\frac{3\pi}{2}\right)+i\sin\left(-\frac{7\pi}{6}+\frac{3\pi}{2}\right)\right]\)
\(Z = \frac{1}{4}\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\) (\(|Z| = \frac{1}{4},\; \arg(Z) = \frac{\pi}{3}\))
ដូចនេះ \( Z = \frac{1}{8}+\frac{\sqrt{3}}{8}i\)

លំហាត់ ២៥៖ [2026]

1. \(1+i = \sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)
   \(1-i = \sqrt{2}\left[\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right]\)

2. \(S_2 = (1+i)^2+(1-i)^2 = 2i-2i = 0\)
   \(S_4 = (2i)^2+(-2i)^2 = -4-4 = -8\)
   \(S_6 = (2i)^3+(-2i)^3 = -8i+8i = 0\)
   តាមដឺម័រ៖ \(S_n = (\sqrt{2})^n\left[\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}+\cos\frac{n\pi}{4}-i\sin\frac{n\pi}{4}\right]\)
   \(\implies S_n = 2^{\frac{n}{2}+1}\cos\frac{n\pi}{4}\) (ជាចំនួនពិត)។
   \(\bullet\) ពន្លាតតាមទ្វេធាញូតុន \((1+i)^n = \sum \binom{n}{k}i^k\) និង \((1-i)^n = \sum \binom{n}{k}(-i)^k\)។ តួដែល \(k\) សេសលុបគ្នា ហើយតួដែល \(k\) គូជាចំនួនគត់
   \(\implies S_n = 2\sum_{k\text{ គូ}}\binom{n}{k}i^k\) ជាចំនួនគត់ជានិច្ច។ សំណើនេះ ពិត។
   \(\bullet\) \(S_n = 0 \iff \cos\frac{n\pi}{4} = 0 \iff \frac{n\pi}{4} = \frac{\pi}{2}+k\pi \iff n = 4k+2\)
   ដូចនេះ មាន ឧទាហរណ៍ \(n = 2, 6, 10, \dots\) ដែល \(S_n = 0\)។

ដំណោះស្រាយលម្អិត លំហាត់ស្រាវជ្រាវ និង កម្រិតខ្ពស់

លំហាត់ ១៖ ទម្រង់ពិជគណិត \(x+iy\)

1. \((9-3i) + (-4+5i) = (9-4) + (-3+5)i = 5 + 2i\)

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

3. \((4-i)(3+2i) = 12 + 8i - 3i - 2i^2 = 14 + 5i\)

4. \((3+2i)^3 = 27 + 54i - 36 - 8i = -9 + 46i\)

5. \(\frac{1+i}{\sqrt{2}-\sqrt{2}i} = \frac{(1+i)(\sqrt{2}+\sqrt{2}i)}{4} = \frac{2\sqrt{2}i}{4} = \frac{\sqrt{2}}{2}i\)

6. \(\frac{1+i}{1+3i} = \frac{(1+i)(1-3i)}{10} = \frac{4-2i}{10} = \frac{2}{5} - \frac{1}{5}i\)

7. \(\frac{(1+i)(1-2i)}{2-4i} = \frac{3-i}{2(1-2i)} = \frac{(3-i)(1+2i)}{2(1^2+2^2)} = \frac{5+5i}{10} = \frac{1}{2} + \frac{1}{2}i\)

លំហាត់ ២៖ សមីការលីនេអ៊ែរក្នុង \(\mathbb{C}\)

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

2. \((3-4i)(z-1) = 10-5i\)
   \(\implies z-1 = \frac{10-5i}{3-4i} = \frac{(10-5i)(3+4i)}{25} = \frac{50+25i}{25} = 2+i\)
   \(\implies z = 3+i\)

3. \((2+i)(z-7+3i) = 15-10i\)
   \(z-7+3i = \frac{15-10i}{2+i} = \frac{(15-10i)(2-i)}{5} = \frac{20-35i}{5} = 4-7i\)
   \(\implies z = 4-7i + 7-3i = 11-10i\)

4. \((3+5i)(z+2-5i) = 6+3i\)
   \(z+2-5i = \frac{(6+3i)(3-5i)}{34} = \frac{33-21i}{34}\)
   \(\implies z = \frac{33-21i}{34} - 2 + 5i = -\frac{35}{34} + \frac{149}{34}i\)

លំហាត់ ៣៖ កំណត់ចំនួនពិត \(a, b\)

1. \(\frac{a}{3+i} + \frac{b}{1+2i} = 1-i\)
   \(\iff \frac{a(3-i)}{10} + \frac{b(1-2i)}{5} = 1-i\)
   \(\iff (3a+2b) - (a+4b)i = 10 - 10i\)
   \(\implies \begin{cases} 3a+2b = 10 \\ a+4b = 10 \end{cases} \implies a=2,\; b=2\)

2. \(\frac{2-3i}{2a+bi} = 3+2i \iff 2a+bi = \frac{2-3i}{3+2i} = \frac{(2-3i)(3-2i)}{13} = -i\)
   \(\implies 2a=0,\; b=-1 \implies a=0,\; b=-1\)

3. \((a+bi)^2 = 8+6i \iff a^2-b^2 = 8,\; 2ab = 6,\; a^2+b^2 = 10\)
   \(\implies 2a^2 = 18 \implies a = \pm 3 \implies b = \pm 1\)
   ដូចនេះ \( (a,b) = (3, 1) \) ឬ \( (-3, -1)\)

លំហាត់ ៤៖ សមីការដឺក្រេទីពីរមេគុណពិត

1. \(z^2+2z+2=0 \iff (z+1)^2 = -1 = i^2 \implies z = -1 \pm i\)

2. \(4z^2-4z+17=0 \iff (2z-1)^2 = -16 \implies 2z-1 = \pm 4i \implies z = \frac{1}{2} \pm 2i\)

3. \(z^2+6z+34=0 \iff (z+3)^2 = -25 \implies z = -3 \pm 5i\)

4. \(z^2-2\sqrt{3}z+4=0 \iff (z-\sqrt{3})^2 = -1 \implies z = \sqrt{3} \pm i\)

លំហាត់ ៥៖ សមីការមេគុណកុំផ្លិច

1. \(z^2-4iz-4=0 \iff (z-2i)^2 = 0 \implies z = 2i\) (ឫសឌុប)

2. \(z^2+iz+6=0\)៖ \(\Delta = i^2 - 24 = -25 = (5i)^2\)
   \(\implies z = \frac{-i \pm 5i}{2} \implies z_1 = 2i,\; z_2 = -3i\)

3. \(z^2+8iz-17=0\)៖ \(\Delta' = (4i)^2 - (-17) = -16+17 = 1\)
   \(\implies z = -4i \pm 1 \implies z = 1-4i \) ឬ \( -1-4i\)

4. \(z^2-(3+2i)z+5+i=0\)៖
   \(\Delta = (3+2i)^2 - 4(5+i) = 5+12i - 20 - 4i = -15+8i\)
   \(\sqrt{\Delta} = \pm(1+4i) \implies z = \frac{3+2i \pm (1+4i)}{2}\)
   \(\implies z_1 = 2+3i,\; z_2 = 1-i\)

លំហាត់ ៦៖ ឫសការេនៃចំនួនកុំផ្លិច

1. \(Z_1 = 3+4i = 4+4i-1 = (2+i)^2 \implies \pm(2+i)\)

2. \(Z_2 = -8-6i = 1-6i-9 = (1-3i)^2 \implies \pm(1-3i)\)

3. \(Z_3 = -15+8i = 1+8i-16 = (1+4i)^2 \implies \pm(1+4i)\)

4. \(Z_4 = 5-12i = 9-12i-4 = (3-2i)^2 \implies \pm(3-2i)\)

លំហាត់ ៧៖ ប្រព័ន្ធសមីការ

1. \(\begin{cases} z_1+z_2 = 3+i \quad{}(1) \\ 2z_1-iz_2 = 5 \quad{}(2) \end{cases}\)
   គុណ \((1)\) នឹង \(i\) រួចបូកនឹង \((2)\)៖
   \((2+i)z_1 = 5 + i(3+i) = 4+3i \implies z_1 = \frac{4+3i}{2+i} = \frac{11+2i}{5}\)
   \(z_2 = (3+i) - z_1 = \frac{4+3i}{5}\)

2. \(\begin{cases} (1+i)z_1+z_2 = 2 \quad{}(1) \\ z_1-iz_2 = 1+i \quad{}(2) \end{cases}\)
   ពី \((1)\)៖ \(z_2 = 2-(1+i)z_1\) ជំនួសក្នុង \((2)\)៖
   \(z_1 - 2i + i(1+i)z_1 = 1+i \iff z_1(1+i-1) = 1+3i\)
   \(\iff iz_1 = 1+3i \implies z_1 = \frac{1+3i}{i} = 3-i\)
   \(z_2 = 2-(1+i)(3-i) = 2-(4+2i) = -2-2i\)
   ដូចនេះ \( z_1 = 3-i,\; z_2 = -2-2i\)

លំហាត់ ៨៖ ស្រាយបញ្ជាក់លក្ខណៈ

1. \(\overline{A} = \overline{z^2-\bar{z}^2} = \bar{z}^2-z^2 = -A \implies A\) ជានិម្មិតសុទ្ធ។

2. \(\overline{B} = \overline{z^n+\bar{z}^n} = \bar{z}^n+z^n = B \implies B \in \mathbb{R}\) (ចំនួនពិត)។

3. \(\overline{C} = \overline{\left(\frac{z-\bar{z}}{1+|z|^2}\right)} = \frac{\bar{z}-z}{1+|z|^2} = -C \implies C\) ជានិម្មិតសុទ្ធ។

លំហាត់ ៩៖ ភាពពិត និងនិម្មិតនៃ \(X, Y\)
តាង \(w = 2-\sqrt{5}i \implies \bar{w} = 2+\sqrt{5}i\)។
\(X = w^n+\bar{w}^n \implies \overline{X} = \bar{w}^n+w^n = X \implies X \in \mathbb{R}\)។
\(Y = w^n-\bar{w}^n \implies \overline{Y} = \bar{w}^n-w^n = -Y \implies Y\) ជានិម្មិតសុទ្ធ។

លំហាត់ ១០៖ គណនាម៉ូឌុល

1. \(|z_1| = \frac{|2+i||1-3i|}{|3+4i|} = \frac{\sqrt{5}\sqrt{10}}{5} = \sqrt{2}\)

2. \(|z_2| = \frac{|1+i\sqrt{3}|^4}{|1-i|^6} = \frac{2^4}{(\sqrt{2})^6} = \frac{16}{8} = 2\)

3. \(|z_3| = |1+\cos\alpha+i\sin\alpha| = \left|2\cos\frac{\alpha}{2}e^{i\alpha/2}\right| = 2\left|\cos\frac{\alpha}{2}\right|\)

លំហាត់ ១១៖ ទម្រង់ត្រីកោណមាត្រ

1. \(4 = 4(\cos 0+i\sin 0),\; -6 = 6(\cos\pi+i\sin\pi)\)

2. \(2i = 2\cis\left(\frac{\pi}{2}\right),\; -3i = 3\cis\left(-\frac{\pi}{2}\right)\)

3. \(1+i = \sqrt{2}\cis\left(\frac{\pi}{4}\right),\; 2-2i = 2\sqrt{2}\cis\left(-\frac{\pi}{4}\right)\)

4. \(-\sqrt{3}+i = 2\cis\left(\frac{5\pi}{6}\right),\; 2\sqrt{3}+2i = 4\cis\left(\frac{\pi}{6}\right)\)

លំហាត់ ១២៖ ទម្រង់ពិជគណិត

1. \(2\cis\left(\frac{\pi}{2}\right) = 2i\)

2. \(8\cis\left(\frac{\pi}{4}\right) = 4\sqrt{2}+4i\sqrt{2}\)

3. \(4\cis\left(\frac{\pi}{6}\right) = 2\sqrt{3}+2i\)

4. \(\sqrt{2}\cis\left(-\frac{\pi}{4}\right) = 1-i\)

5. \(\sqrt{3}\cis\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}+\frac{3}{2}i\)

6. \(5\cis\pi = -5\)

លំហាត់ ១៣៖ គណនាតម្លៃពិត \(\frac{\pi}{12}\)
\(z_1 = 2e^{i\pi/3},\; z_2 = \sqrt{2}e^{i\pi/4}\)
\(\frac{z_1}{z_2} = \sqrt{2}e^{i\pi/12} = \sqrt{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)
ម៉្យាងទៀត \(z_2 = 1+i\) នោះ \(\frac{z_1}{z_2} = \frac{1+i\sqrt{3}}{1+i} = \frac{(1+i\sqrt{3})(1-i)}{2} = \frac{1+\sqrt{3}}{2} + i\frac{\sqrt{3}-1}{2}\)
ផ្ទឹមនឹង \(\sqrt{2}\cos\frac{\pi}{12} + i\sqrt{2}\sin\frac{\pi}{12}\)៖
\(\implies \cos\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4},\; \sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

លំហាត់ ១៤៖ រូបមន្តដឺម័រ

1. \((\sqrt{2}\cis\frac{\pi}{5})^{10} = (\sqrt{2})^{10}\cis(2\pi) = 32\)

2. \((\cis\frac{\pi}{12})^{36} = \cis(3\pi) = -1\)

3. \((\sqrt{2}\cis\frac{\pi}{8})^{12} = 64\cis(\frac{3\pi}{2}) = -64i\)

4. \((2\cis\frac{5\pi}{6})^6 = 64\cis(5\pi) = -64\)

លំហាត់ ១៥៖ គណនាជាទម្រង់ \(x+iy\)

1. \((1+i)^{16} = (\sqrt{2}e^{i\pi/4})^{16} = 2^8 e^{i 4\pi} = 256\)

2. \((1-i\sqrt{3})^{11} = (2e^{-i\pi/3})^{11} = 2048 e^{-i 11\pi/3} = 2048 e^{i\pi/3} = 1024 + 1024i\sqrt{3}\)

3. \((\sqrt{3}+i)^{10} = (2e^{i\pi/6})^{10} = 1024 e^{i 5\pi/3} = 512 - 512i\sqrt{3}\)

4. \((\frac{1+i}{\sqrt{3}-i})^{12} = (\frac{\sqrt{2}e^{i\pi/4}}{2e^{-i\pi/6}})^{12} = (\frac{\sqrt{2}}{2})^{12}e^{i 5\pi} = -\frac{1}{64}\)

លំហាត់ ១៦៖ លក្ខខណ្ឌលើ \(n\) នៃ \(z^n\)
\(z = 2-2\sqrt{3}i = 4e^{-i\pi/3} \implies z^n = 4^n e^{-i\frac{n\pi}{3}}\)

1. \(z^n \in \mathbb{R} \iff \sin\left(-\frac{n\pi}{3}\right) = 0 \iff \frac{n\pi}{3} = k\pi \iff n = 3k\) (\(k \in \mathbb{N}\))

2. \(z^n \in i\mathbb{R} \iff \cos\left(-\frac{n\pi}{3}\right) = 0 \iff \frac{n\pi}{3} = \frac{\pi}{2}+k\pi \iff 2n = 3(2k+1)\) (មិនអាចកើតមានព្រោះ \(2n\) គូ ឯ \(3(2k+1)\) សេស)។ គ្មានតម្លៃ \(n\) ទេ។

លំហាត់ ១៧៖ \(Z = \frac{-1+i}{1+i\sqrt{3}}\)

1. \(Z = \frac{(-1+i)(1-i\sqrt{3})}{4} = \frac{\sqrt{3}-1 + i(\sqrt{3}+1)}{4}\)
   \(\implies \text{Re}(Z) = \frac{\sqrt{3}-1}{4},\; \text{Im}(Z) = \frac{\sqrt{3}+1}{4}\)

2. \(Z = \frac{\sqrt{2}e^{i 3\pi/4}}{2e^{i\pi/3}} = \frac{\sqrt{2}}{2}e^{i 5\pi/12} = \frac{\sqrt{2}}{2}\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)\)

3. \(\cos\frac{5\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}\)
   \(\sin\frac{5\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}\)

លំហាត់ ១៨៖ \(u = (1+i)(1+i\sqrt{3})\)

1. \(u = 1 + i\sqrt{3} + i - \sqrt{3} = (1-\sqrt{3}) + i(1+\sqrt{3})\)

2. \(1+i = \sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\), \(1+i\sqrt{3} = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)
   \(\implies u = 2\sqrt{2}\left(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12}\right)\) (ព្រោះ \(\frac{\pi}{4}+\frac{\pi}{3} = \frac{7\pi}{12}\))

3. ផ្ទឹមទម្រង់ទាំងពីរ៖ \(2\sqrt{2}\cos\frac{7\pi}{12} = 1-\sqrt{3}\) និង \(2\sqrt{2}\sin\frac{7\pi}{12} = 1+\sqrt{3}\)
   \(\implies \cos\frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{2}-\sqrt{6}}{4},\; \sin\frac{7\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}\)

4. \(u^6 = (2\sqrt{2})^6\left(\cos\frac{7\pi}{2}+i\sin\frac{7\pi}{2}\right) = 512\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right) = -512i\)

លំហាត់ ១៩៖ គណនាតម្លៃពិត \(\frac{\pi}{8}\)
\(z = \sqrt{1+\sqrt{2}} + i\sqrt{\sqrt{2}-1}\)
\(z^2 = (1+\sqrt{2}) - (\sqrt{2}-1) + 2i\sqrt{(\sqrt{2}+1)(\sqrt{2}-1)} = 2 + 2i\)
\(Z = z^4 = (z^2)^2 = (2+2i)^2 = 8i = 8e^{i\pi/2}\)
\(\implies |z|^4 = 8 \implies |z| = \sqrt[4]{8} = 2^{3/4}\)
\(4\arg(z) = \frac{\pi}{2} \implies \arg(z) = \frac{\pi}{8}\)
\(\cos\frac{\pi}{8} = \frac{\sqrt{1+\sqrt{2}}}{|z|} = \frac{\sqrt{2+\sqrt{2}}}{2},\; \sin\frac{\pi}{8} = \frac{\sqrt{2-\sqrt{2}}}{2}\)

លំហាត់ ២០៖ \(z_3 = z_1 \bar{z}_2\)
\(z_1 = \sqrt{2}e^{i 3\pi/4},\; z_2 = 2\sqrt{2}e^{-i\pi/6} \implies \bar{z}_2 = 2\sqrt{2}e^{i\pi/6}\)
ផ្ទៀងផ្ទាត់៖ \(z_1\bar{z}_2 = (-1+i)(\sqrt{6}+i\sqrt{2}) = -\sqrt{6}-i\sqrt{2}+i\sqrt{6}-\sqrt{2} = (-\sqrt{6}-\sqrt{2})+i(\sqrt{6}-\sqrt{2}) = z_3\)
\(z_3 = z_1 \bar{z}_2 = 4e^{i 11\pi/12} \implies |z_3|=4,\; \arg(z_3)=\frac{11\pi}{12}\)
ផ្ទឹម \(z_3 = 4\left(\cos\frac{11\pi}{12}+i\sin\frac{11\pi}{12}\right)\) ជាមួយទម្រង់ពិជគណិតខាងលើ៖
\(\cos\frac{11\pi}{12} = -\frac{\sqrt{6}+\sqrt{2}}{4},\; \sin\frac{11\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\)
\(z_3^{18} = 4^{18}e^{i\frac{33\pi}{2}} = 4^{18}e^{i\frac{\pi}{2}} = 4^{18}i\)

លំហាត់ ២១៖ ឫសការេនៃ \(4\sqrt{3}+4i\)
\(u = 8e^{i\pi/6} \implies \pm\sqrt{8}e^{i\pi/12} = \pm 2\sqrt{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)
តាមពិជគណិត៖ \((x+iy)^2 = 4\sqrt{3}+4i \iff x^2-y^2 = 4\sqrt{3},\; xy = 2,\; x^2+y^2 = 8\)
\(\implies x^2 = 4+2\sqrt{3} = (\sqrt{3}+1)^2,\; y^2 = (\sqrt{3}-1)^2\) (\(x, y\) សញ្ញាដូចគ្នា)
\(\implies x+iy = \pm\left[(\sqrt{3}+1)+i(\sqrt{3}-1)\right]\)
ផ្ទឹមឫស \(x>0\) ជាមួយ \(2\sqrt{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)៖
\(\implies \cos\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4},\; \sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

លំហាត់ ២២៖ ឫសគូបក្នុង \(\mathbb{C}\)

1. \(-27 = 27e^{i\pi} \implies w_k = 3e^{i(\frac{\pi}{3}+\frac{2k\pi}{3})},\; k\in\{0,1,2\}\)
   \(w_0 = \frac{3}{2}+i\frac{3\sqrt{3}}{2},\; w_1 = -3,\; w_2 = \frac{3}{2}-i\frac{3\sqrt{3}}{2}\)

2. \(8i = 8e^{i\pi/2} \implies w_k = 2e^{i(\frac{\pi}{6}+\frac{2k\pi}{3})}\)
   \(w_0 = \sqrt{3}+i,\; w_1 = -\sqrt{3}+i,\; w_2 = -2i\)

3. \(-1+i = \sqrt{2}e^{i 3\pi/4} \implies w_k = \sqrt[6]{2}e^{i(\frac{\pi}{4}+\frac{2k\pi}{3})}\)

លំហាត់ ២៣៖ សមីការ \(z^5-1=0\)
ឫសទាំង ៥៖ \(z_k = \cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}\), \(k=0,1,2,3,4\) (\(z_0=1\), \(z_4=\bar{z}_1\), \(z_3=\bar{z}_2\))។
\(\omega = e^{i 2\pi/5}\)។ ផលបូក \(1+\omega+\omega^2+\omega^3+\omega^4 = \frac{\omega^5-1}{\omega-1} = 0\)។
យកផ្នែកពិត៖ \(1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0 \implies \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}\)។
ម៉្យាងទៀត \(2\cos\frac{2\pi}{5}\cos\frac{4\pi}{5} = \cos\frac{6\pi}{5}+\cos\frac{2\pi}{5} = -\frac{1}{2}\)
\(\implies X = \cos\frac{2\pi}{5}\) ជាឫសវិជ្ជមាននៃ \(4X^2+2X-1=0\)
ដូចនេះ \( \cos\frac{2\pi}{5} = \frac{\sqrt{5}-1}{4},\; \sin\frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)

លំហាត់ ២៤៖ ឫសគូបមិនពិត \(j\)

1. \(j = e^{i 2\pi/3} \implies j^2 = e^{i 4\pi/3} = e^{-i 2\pi/3} = \bar{j}\)
   \(1+j+j^2 = \frac{j^3-1}{j-1} = 0\)

2. \(A = (1-j+j^2)(1+j-j^2) = (-2j)(-2j^2) = 4j^3 = 4\)

3. ដោយ \(j^3=1\) និង \(j+j^2=-1\)៖
   \((a+bj+cj^2)(a+bj^2+cj) = a^2+b^2j^3+c^2j^3+ab(j+j^2)+bc(j^2+j^4)+ca(j^2+j)\)
   \(= a^2+b^2+c^2-ab-bc-ca\) (ព្រោះ \(j^4=j\))
   \(\implies (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc\)

លំហាត់ ២៥៖ ពន្លាតដឺម័រដឺក្រេ ៣
\((\cos\theta+i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta\)
\(= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)\)
ផ្ទឹមជាមួយ \(\cos 3\theta + i\sin 3\theta\)៖
\(\cos 3\theta = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) = 4\cos^3\theta - 3\cos\theta\)
\(\sin 3\theta = 3(1-\sin^2\theta)\sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta\)

លំហាត់ ២៦៖ ពន្លាតដឺម័រដឺក្រេ ៤
\((\cos\theta+i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta\)
\(\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1\)
\(\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta\)
\(\tan 4\theta = \frac{\sin 4\theta}{\cos 4\theta} = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}\)

លំហាត់ ២៧៖ កន្សោម \(z^n \pm 1/z^n\)

1. \(z = e^{i\theta} \implies z^n + z^{-n} = e^{i n\theta} + e^{-i n\theta} = 2\cos n\theta\)
   \(z^n - z^{-n} = 2i\sin n\theta\)

2. \((z+z^{-1})^3 = (2\cos\theta)^3 = 8\cos^3\theta\)
   ម៉្យាងទៀត \((z+z^{-1})^3 = (z^3+z^{-3}) + 3(z+z^{-1}) = 2\cos 3\theta + 6\cos\theta\)
   \(\implies \cos^3\theta = \frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta\)

3. \(\cos^3\left(\frac{13\pi}{12}\right) = \frac{1}{4}\cos\left(\frac{13\pi}{4}\right) + \frac{3}{4}\cos\left(\frac{13\pi}{12}\right) = -\frac{\sqrt{2}}{8} - \frac{3(\sqrt{6}+\sqrt{2})}{16} = -\frac{5\sqrt{2}+3\sqrt{6}}{16}\)

លំហាត់ ២៨៖ រូបមន្តអយល័រ និងអាំងតេក្រាល

1. \(\cos^4\theta = \left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^4 = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}\)

2. \(\sin^4\theta = \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^4 = \frac{1}{8}\cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8}\)

3. \(I = \int_{0}^{\pi/2} \sin^4\theta\,d\theta = \int_{0}^{\pi/2} \left(\frac{1}{8}\cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8}\right)d\theta\)
   \(= \left[\frac{\sin 4\theta}{32} - \frac{\sin 2\theta}{4} + \frac{3\theta}{8}\right]_{0}^{\pi/2} = \frac{3\pi}{16}\)

លំហាត់ ២៩៖ សមីការ និងទាញរក \(u\)
(ក) \(z^2-2z+5=0\)៖ \(\Delta' = 1-5 = -4 = (2i)^2 \implies z = 1\pm 2i\)
(ខ) តាង \(X = \frac{u+i}{u-1}\) នោះ \(X^2-2X+5=0 \implies X = 1\pm 2i\)
- \(\frac{u+i}{u-1} = 1+2i \iff u+i = (1+2i)u-1-2i \iff -2iu = -1-3i\)
\(\implies u = \frac{1+3i}{2i} = \frac{(1+3i)(-i)}{2} = \frac{3}{2}-\frac{1}{2}i\)
- \(\frac{u+i}{u-1} = 1-2i \iff u+i = (1-2i)u-1+2i \iff 2iu = -1+i\)
\(\implies u = \frac{-1+i}{2i} = \frac{(-1+i)(-i)}{2} = \frac{1}{2}+\frac{1}{2}i\)
ដូចនេះ \( u \in \left\{\frac{3}{2}-\frac{1}{2}i,\; \frac{1}{2}+\frac{1}{2}i\right\}\) (ទាំងពីរខុសពី \(1\))

លំហាត់ ៣០៖ សមីការ \((z+1)^4 = (z-1)^4\)

1. យកម៉ូឌុល៖ \(|z+1|^4 = |z-1|^4 \implies \left|\frac{z+1}{z-1}\right| = 1\)

2. \(|z-(-1)| = |z-1| \implies M(z)\) ស្មើចម្ងាយពី \(-1\) និង \(1\)
   \(\implies M\) ស្ថិតនៅលើអ័ក្សនិម្មិត \(\implies z = iy\) (\(y \in \mathbb{R}\))

3. \(\frac{z+1}{z-1} = e^{i\frac{2k\pi}{4}} = e^{i\frac{k\pi}{2}} \in \{i, -1, -i\}\) (ដោយសារ \(\ne 1\))
   \(\frac{z+1}{z-1} = i \implies z = -i\); \(\frac{z+1}{z-1} = -1 \implies z = 0\); \(\frac{z+1}{z-1} = -i \implies z = i\)
   ដូចនេះ \( z \in \{0,\; i,\; -i\}\)

លំហាត់ ៣១៖ សមីការផលបូកធរណីមាត្រ
តាង \(u = \frac{z+i}{z-i}\) នោះសមីការក្លាយជា៖
\(u^3+u^2+u+1 = 0 \iff \frac{u^4-1}{u-1} = 0 \implies u^4 = 1\) (\(u \ne 1\))
\(\implies u \in \{i, -1, -i\}\)
- \(u = -1 \implies \frac{z+i}{z-i} = -1 \implies z+i = -z+i \implies z = 0\)
- \(u = i \implies z+i = iz+1 \implies z(1-i) = 1-i \implies z = 1\)
- \(u = -i \implies z = -1\)
ដូចនេះ \( z \in \{-1,\; 0,\; 1\}\)

លំហាត់ ៣២៖ ផលបូក Lagrange

1. \(C_n = \sum_{k=0}^n e^{ikx} = \frac{e^{i(n+1)x}-1}{e^{ix}-1} = \frac{e^{i\frac{(n+1)x}{2}}(e^{i\frac{(n+1)x}{2}}-e^{-i\frac{(n+1)x}{2}})}{e^{i\frac{x}{2}}(e^{i\frac{x}{2}}-e^{-i\frac{x}{2}})}\)
   \(= e^{i\frac{nx}{2}}\frac{2i\sin\frac{(n+1)x}{2}}{2i\sin\frac{x}{2}} = \frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}\left(\cos\frac{nx}{2}+i\sin\frac{nx}{2}\right)\)

2. ផ្ទឹមផ្នែកពិត និងនិម្មិត៖
   \(S_n = \frac{\sin\frac{(n+1)x}{2}\cos\frac{nx}{2}}{\sin\frac{x}{2}},\quad{} T_n = \frac{\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}}{\sin\frac{x}{2}}\)

លំហាត់ ៣៣៖ \(2\bar{B} = A\)

1. \(A = (1+i\sqrt{3})^2 = 1 + 2i\sqrt{3} - 3 = -2+2i\sqrt{3}\)
   \(|A| = 4\), \(\cos\theta = -\frac{1}{2},\ \sin\theta = \frac{\sqrt{3}}{2} \implies A = 4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)

2. \(B = (x+iy)(1-i) = x - ix + iy + y = (x+y) + (y-x)i\)
   \(\implies 2\bar{B} = 2(x+y) + 2(x-y)i\)
   \(2\bar{B} = A \iff \begin{cases} 2(x+y) = -2 \\ 2(x-y) = 2\sqrt{3} \end{cases} \iff \begin{cases} x+y = -1 \\ x-y = \sqrt{3} \end{cases}\)
   ដូចនេះ \( x = \frac{\sqrt{3}-1}{2},\; y = -\frac{\sqrt{3}+1}{2}\)

លំហាត់ ៣៤៖ \(|z-2i| = |z+2|\)

1. \(x^2+(y-2)^2 = (x+2)^2+y^2 \iff -4y+4 = 4x+4 \iff y = -x\)
   បកស្រាយ៖ \(M(z)\) ស្មើចម្ងាយពីចំណុច \((0,2)\) និង \((-2,0)\) ដូច្នេះវាស្ថិតលើមេដ្យាទ័រ \(y=-x\) នៃអង្កត់នោះ។

2. \(|z| = 2\sqrt{2} \iff x^2+y^2 = 8 \iff 2x^2 = 8 \iff x = \pm 2\)
   \(\implies z = 2-2i\) ឬ \(z = -2+2i\)

3. \(2-2i = 2\sqrt{2}\left[\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right]\)
   \(-2+2i = 2\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)

លំហាត់ ៣៥៖ សំណុំចំណុច

1. \(|z-(2-3i)| = 4\)៖ រង្វង់ផ្ចិត \(I(2, -3)\) កាំ \(R = 4\)។

2. \(|z-3i| = |z-(-2+i)|\)៖ បន្ទាត់មេដ្យាទ័រនៃអង្កត់ភ្ជាប់ចំណុច \((0, 3)\) និង \((-2, 1)\)។

3. \(|z-(1-2i)| \le 3\)៖ ថាសបិទ (ផ្ទៃក្នុង និងបរិមណ្ឌលរង្វង់) ផ្ចិត \((1, -2)\) កាំ \(R=3\)។

លំហាត់ ៣៦៖ សមីការដេកាត

1. \(\arg(z-i) = \frac{\pi}{2} \iff x = 0\) និង \(y - 1 > 0 \iff y > 1\) (កន្លះបន្ទាត់ឈរលើអ័ក្សអរដោនេខាងលើចំណុច \((0,1)\))។

2. \(|z+2| = 2|z-2| \iff (x+2)^2+y^2 = 4[(x-2)^2+y^2]\)
   \(\iff 3x^2+3y^2-20x+12 = 0 \iff \left(x-\frac{10}{3}\right)^2 + y^2 = \frac{64}{9}\)
   ជារង្វង់ Apollonius ផ្ចិត \(\left(\frac{10}{3}, 0\right)\) កាំ \(R = \frac{8}{3}\)។

លំហាត់ ៣៧៖ ត្រីកោណសម័ង្ស
\(|z| = |1/z| \implies |z|^2 = 1 \implies |z|=1\) ដូច្នេះតាង \(z = \cos\theta+i\sin\theta\)។
\(|z+1|^2 = (1+\cos\theta)^2+\sin^2\theta = 2+2\cos\theta = 1 \iff \cos\theta = -\frac{1}{2}\)
\(\implies z = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2} = \cos\frac{2\pi}{3} \pm i\sin\frac{2\pi}{3}\)។
ដោយ \(|z|=1\) នោះ \(-\frac{1}{z} = -\bar{z} = \frac{1}{2} \pm i\frac{\sqrt{3}}{2} = z+1\)។
ចំណុច \(O(0), J(-1), A(z)\)៖
\(OJ = 1,\quad{} OA = |z| = 1,\quad{} JA = |z-(-1)| = |z+1| = 1\)
ប្រវែងជ្រុងទាំងបីស្មើគ្នា \(\implies \triangle OJA\) ជាត្រីកោណសម័ង្ស។

លំហាត់ ៣៨៖ \(z = 1+\cos 2\theta+i\sin 2\theta\)

1. \(z = 2\cos^2\theta + 2i\sin\theta\cos\theta = 2\cos\theta(\cos\theta+i\sin\theta)\)
   ដោយ \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \implies \cos\theta > 0 \implies |z| = 2\cos\theta\)

2. \(\frac{1}{z} = \frac{e^{-i\theta}}{2\cos\theta} = \frac{\cos\theta-i\sin\theta}{2\cos\theta} = \frac{1}{2} - i\frac{\tan\theta}{2}\)
   \(\implies \text{Re}\left(\frac{1}{z}\right) = \frac{1}{2}\) (ជាចំនួនថេរមិនអាស្រ័យនឹង \(\theta\))

លំហាត់ ៣៩៖ \(z = 2\cos\theta+i(1-2\sin\theta)\)

1. \(z-i = 2\cos\theta - 2i\sin\theta = 2e^{-i\theta} \implies |z-i| = 2\)

2. \(z+2-i = 2(1+\cos\theta) - 2i\sin\theta = 4\cos\frac{\theta}{2}e^{-i\theta/2}\)
   \(\frac{1}{z+2-i} = \frac{e^{i\theta/2}}{4\cos\frac{\theta}{2}} = \frac{1}{4} + i\frac{\tan\frac{\theta}{2}}{4} \implies \text{Re} = \frac{1}{4}\) (ថេរ)

លំហាត់ ៤០៖ ធរណីមាត្រវ៉ិចទ័រ

1. \(\overrightarrow{AB} \parallel \overrightarrow{AC} \iff \arg\left(\frac{z_C-z_A}{z_B-z_A}\right) \equiv 0 \pmod\pi \iff \frac{z_C-z_A}{z_B-z_A} \in \mathbb{R}\)

2. \(AB \perp AC \iff \arg \equiv \pm\frac{\pi}{2} \pmod\pi \iff \frac{z_C-z_A}{z_B-z_A} \in i\mathbb{R}\)

3. អនុវត្ត៖ \(\frac{z_C-z_A}{z_B-z_A} = \frac{(4+5i)-(1+2i)}{(-2-i)-(1+2i)} = \frac{3+3i}{-3-3i} = -1 \in \mathbb{R}\)
   ដូចនេះ \( A, B, C\) រត់ត្រង់ជួរគ្នា!

លំហាត់ ៤១៖ លក្ខខណ្ឌត្រីកោណសម័ង្ស
\(\triangle ABC\) សម័ង្ស \(\iff \frac{z_3-z_1}{z_2-z_1} = e^{\pm i\pi/3} \iff \left(\frac{z_3-z_1}{z_2-z_1}\right)^2 - \frac{z_3-z_1}{z_2-z_1} + 1 = 0\)
\(\iff (z_3-z_1)^2 - (z_3-z_1)(z_2-z_1) + (z_2-z_1)^2 = 0\)
ពន្លាតនិងបង្រួមគេបាន៖
\(z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1\) (ពិត)

លំហាត់ ៤២៖ សំណុំចំណុចនៃ \(Z = \frac{z-2i}{z+1}\)
\(Z = \frac{(x+iy-2i)(x+1-iy)}{(x+1)^2+y^2} = \frac{(x^2+x+y^2-2y) + i(y-2x-2)}{(x+1)^2+y^2}\)

1. \(Z \in \mathbb{R} \iff y - 2x - 2 = 0 \iff y = 2x + 2\) (បន្ទាត់ ដកចំណុច \((-1, 0)\))

2. \(Z \in i\mathbb{R} \iff x^2+x+y^2-2y = 0 \iff \left(x+\frac{1}{2}\right)^2 + (y-1)^2 = \frac{5}{4}\) (រង្វង់ផ្ចិត \((-\frac{1}{2}, 1)\) កាំ \(\frac{\sqrt{5}}{2}\) ដក \((-1, 0)\))

លំហាត់ ៤៣៖ ពហុធា \(P(z)\)
\(P(i) = i^4 - 2i^3 + 3i^2 - 2i + 2 = 1 + 2i - 3 - 2i + 2 = 0\)
ដោយមេគុណជាចំនួនពិត នោះ \(-i\) ក៏ជាឫស \(\implies z^2+1\) ជាកត្តា។
ចែកពហុធា៖ \(P(z) = (z^2+1)(z^2-2z+2)\)
\(z^2-2z+2 = 0 \iff (z-1)^2 = -1 \implies z = 1 \pm i\)
ដូចនេះ \( z \in \{i,\; -i,\; 1+i,\; 1-i\}\)

លំហាត់ ៤៤៖ សមីការ \((z+i)^n = (z-i)^n\)

1. \(|z+i| = |z-i| \implies z\) ស្មើចម្ងាយពី \(-i\) និង \(i \implies z \in \mathbb{R}\)

2. \(\frac{z+i}{z-i} = e^{i\frac{2k\pi}{n}} \implies z+i = e^{i\frac{2k\pi}{n}}(z-i)\)
   \(\implies z = i\frac{e^{i\frac{2k\pi}{n}}+1}{e^{i\frac{2k\pi}{n}}-1} = i\frac{2\cos\frac{k\pi}{n}e^{i\frac{k\pi}{n}}}{2i\sin\frac{k\pi}{n}e^{i\frac{k\pi}{n}}} = \cot\left(\frac{k\pi}{n}\right)\)
   ចំពោះ \(k = 1, 2, \dots, n-1\)

3. ករណី \(n=5\)៖ ឫសទាំង ៤ គឺ \(\cot\frac{\pi}{5}, \cot\frac{2\pi}{5}, -\cot\frac{2\pi}{5}, -\cot\frac{\pi}{5}\)
   ដោយ \(\tan\frac{\pi}{5} = \sqrt{5-2\sqrt{5}}\) និង \(\tan\frac{2\pi}{5} = \sqrt{5+2\sqrt{5}}\) គេបាន
   \(\cot\frac{\pi}{5} = \frac{1}{\sqrt{5-2\sqrt{5}}} = \sqrt{\frac{5+2\sqrt{5}}{5}},\; \cot\frac{2\pi}{5} = \frac{1}{\sqrt{5+2\sqrt{5}}} = \sqrt{\frac{5-2\sqrt{5}}{5}}\)

លំហាត់ ៤៥៖ ផលគុណស៊ីនុស
\(z^{2n+1}-1 = (z-1)\prod_{k=1}^{2n}(z - e^{i\frac{2k\pi}{2n+1}})\)
ចែកនឹង \(z-1\) រួចផ្គុំគូកត្តាឆ្លាស់ \(k\) និង \(2n+1-k\)៖
\(\prod_{k=1}^n \left|1 - e^{i\frac{2k\pi}{2n+1}}\right|^2 = 2n+1\)
ដោយ \(|1-e^{i 2\theta}| = 2\sin\theta \implies \prod_{k=1}^n \left(2\sin\frac{k\pi}{2n+1}\right)^2 = 2n+1\)
\(\implies \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^n}\) (រូបមន្តផលគុណស៊ីនុសដ៏ល្បី)

លំហាត់ ៤៦៖ វិសមភាពត្រីកោណ និងបរមា
(ក) \(|z_1+z_2|^2 = |z_1|^2+|z_2|^2+2\text{Re}(z_1\bar{z}_2) \le |z_1|^2+|z_2|^2+2|z_1||z_2| = (|z_1|+|z_2|)^2\)
សមភាពកើតមាន \(\iff \text{Re}(z_1\bar{z}_2)=|z_1\bar{z}_2| \iff z_1\bar{z}_2\ge 0\) (ពោលគឺ \(z_1, z_2\) មានទិសដៅដូចគ្នា ឬមួយស្មើសូន្យ)។
(ខ) \(|z_1| = |(z_1-z_2)+z_2| \le |z_1-z_2|+|z_2|\) និងដូចគ្នា \(|z_2| \le |z_1-z_2|+|z_1|\)
\(\implies ||z_1|-|z_2|| \le |z_1-z_2|\)
(គ) យក \(z_1=z,\ z_2=\frac{3}{z}\)៖ \(||z| - |3/z|| \le |z - 3/z| = 2\)
\(\implies -2 \le |z| - \frac{3}{|z|} \le 2\)
- \(|z| - \frac{3}{|z|} \le 2 \implies |z|^2 - 2|z| - 3 \le 0 \implies |z| \le 3\)
- \(|z| - \frac{3}{|z|} \ge -2 \implies |z|^2 + 2|z| - 3 \ge 0 \implies |z| \ge 1\)
ដូចនេះ \( 1 \le |z| \le 3\)។ តម្លៃទាំងពីរត្រូវបានសម្រេច៖ \(z=1\) ឱ្យ \(|1-3|=2\) និង \(z=3\) ឱ្យ \(|3-1|=2\) \(\implies |z|_{\min} = 1,\; |z|_{\max} = 3\)

លំហាត់ ៤៧៖ \(u+v+w=0\) លើរង្វង់ឯកតា

1. \(\frac{1}{u}+\frac{1}{v}+\frac{1}{w} = \bar{u}+\bar{v}+\bar{w} = \overline{u+v+w} = 0\)

2. \(uv+vw+wu = uvw\left(\frac{1}{w}+\frac{1}{u}+\frac{1}{v}\right) = 0\)

3. \((u+v+w)^2 = u^2+v^2+w^2+2(uv+vw+wu) \implies u^2+v^2+w^2 = 0\)
   \(u, v, w\) ជាឫសនៃ \(Z^3 - 0Z^2 + 0Z - uvw = 0 \iff Z^3 = uvw\)
   \(\implies u^3 = v^3 = w^3 = uvw\)
   មុំរវាងពួកវាគឺ \(\frac{2\pi}{3} \implies\) បង្កើតបានជាត្រីកោណសម័ង្ស។

លំហាត់ ៤៨៖ ប្រព័ន្ធសមីការ
\(\sum z_k = 1,\; \prod z_k = 1,\; |z_k|=1\)
\(\sum z_1 z_2 = z_1 z_2 z_3 \sum \frac{1}{z_k} = 1 \cdot \overline{\sum z_k} = 1\)
\(z_1, z_2, z_3\) ជាឫសនៃពហុធា៖
\(Z^3 - Z^2 + Z - 1 = 0 \iff (Z-1)(Z^2+1) = 0\)
\(\implies Z \in \{1,\; i,\; -i\}\)
ដូចនេះ \( \{z_1, z_2, z_3\} = \{1,\; i,\; -i\}\) (៦ ចម្លើយតាមបម្លាស់ប្តូរ)

លំហាត់ ៤៩៖ ទ្រឹស្តីបទណាប៉ូឡេអុង
តាង \(\rho = e^{i\pi/3}\) និង \(\omega = e^{-i\pi/3} = \bar{\rho}\)។ គេមាន \(\rho+\omega = 1\), \(\rho\omega = 1\), \(\rho^2 = \rho - 1\)។
ដោយ \(ABC\) តាមទិសដៅវិជ្ជមាន ត្រីកោណ \(BCD\) នៅក្រៅ \(\implies z_D - z_B = \omega(z_C - z_B)\)
\(\implies z_D = (1-\omega)z_B + \omega z_C = \rho z_B + \omega z_C\)
\(P = \frac{z_B+z_C+z_D}{3} = \frac{(1+\rho)z_B + (1+\omega)z_C}{3}\)
ដូចគ្នា \(Q = \frac{(1+\rho)z_C + (1+\omega)z_A}{3}\), \(R = \frac{(1+\rho)z_A + (1+\omega)z_B}{3}\)
\(3(Q-P) = (1+\omega)z_A - (1+\rho)z_B + (\rho-\omega)z_C\)
\(3(R-P) = (1+\rho)z_A + (\omega-\rho)z_B - (1+\omega)z_C\)
គុណ \(3(Q-P)\) នឹង \(\rho\)៖ \(\rho(1+\omega) = \rho+1\), \(-\rho(1+\rho) = 1-2\rho = \omega-\rho\), \(\rho(\rho-\omega) = \rho-2 = -(1+\omega)\)
\(\implies R-P = \rho(Q-P) = e^{i\pi/3}(Q-P) \implies PQ = PR\) និង \((\overrightarrow{PQ},\overrightarrow{PR}) = \frac{\pi}{3}\)
ដូចនេះ \( \triangle PQR\) ជាត្រីកោណសម័ង្ស (ទ្រឹស្តីបទណាប៉ូឡេអុង)។

លំហាត់ ៥០៖ ផលបូកមេគុណទ្វេធា

1. \(Z_n = \sum_{k=0}^n \binom{n}{k} e^{ikx} = (1+e^{ix})^n\)

2. \(1+e^{ix} = e^{ix/2}(e^{-ix/2}+e^{ix/2}) = 2\cos\left(\frac{x}{2}\right)e^{ix/2}\)

3. \(Z_n = \left[2\cos\left(\frac{x}{2}\right)e^{ix/2}\right]^n = 2^n\cos^n\left(\frac{x}{2}\right)e^{i\frac{nx}{2}}\)
   \(\implies C_n = 2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{nx}{2}\right)\)
   \(\implies S_n = 2^n\cos^n\left(\frac{x}{2}\right)\sin\left(\frac{nx}{2}\right)\)

4. អនុវត្ត \(x = \frac{\pi}{3}\)៖
   \(\sum_{k=0}^n \binom{n}{k}\cos\left(\frac{k\pi}{3}\right) = 2^n\cos^n\left(\frac{\pi}{6}\right)\cos\left(\frac{n\pi}{6}\right)\)
   \(= 2^n\left(\frac{\sqrt{3}}{2}\right)^n\cos\left(\frac{n\pi}{6}\right) = 3^{n/2}\cos\left(\frac{n\pi}{6}\right)\)

ចម្លើយសង្ខេប លំហាត់ប្រតិបត្តិក្នុងមេរៀន

មេរៀនទី១

1. ក. \(9i\) ខ. \(5i\sqrt{2}\) គ. \(14i\) ឃ. \(4i\sqrt{3}\)

2. ក. \(-i\) ខ. \(-1\) គ. \(1\) ឃ. \(0\)

3. ក. \(4+8i\); ផ្នែកពិត \(4\), ផ្នែកនិម្មិត \(8\); \(\overline{z}=4-8i\); \(-z=-4-8i\)
   ខ. \(-2-5i\sqrt{3}\); ផ្នែកពិត \(-2\), ផ្នែកនិម្មិត \(-5\sqrt{3}\); \(\overline{z}=-2+5i\sqrt{3}\); \(-z=2+5i\sqrt{3}\)
   គ. \(6i\); ផ្នែកពិត \(0\), ផ្នែកនិម្មិត \(6\); \(\overline{z}=-6i\); \(-z=-6i\)
   ឃ. \(7\); ផ្នែកពិត \(7\), ផ្នែកនិម្មិត \(0\); \(\overline{z}=7\); \(-z=-7\)

4. ក. \(x=3,\ y=4\) ខ. \(x=2,\ y=3\) គ. \(x=2,\ y=3\) ឃ. \(x=2,\ y=1\)

5. ក. \(3+2i\) ខ. \(-3-4i\) គ. \(16-11i\) ឃ. \(3-4i\) ង. \(2+3i\) ច. \(i\)

6. ក. \(\overline{z}=1-2i\), \(\overline{w}=3+i\), \(zw=5+5i\), \(\overline{zw}=\overline{z}\,\overline{w}=5-5i\)
   ខ. \(z+\overline{z}=2\), \(z\overline{z}=5\) (ជាចំនួនពិត) គ. \(u=2-i\)

7. ក. \(x=\pm 3i\) ខ. \(x=2\pm 3i\) គ. \(x=-\dfrac{1}{2}\pm\dfrac{3}{2}i\) ឃ. \(x^2-2x+5=0\)

8. \(A(3,2)\), \(B(-2,1)\), \(C(-1,-3)\), \(D(0,2)\), \(E(-3,0)\); អាភិចនៃ \(F\) គឺ \(z_F=4-i\)

9. \(z_1+z_2=4+4i\) (អង្កត់ទ្រូងនៃប្រលេឡូក្រាម \(OM_1MM_2\))

10. \(z_1-z_2=2-i\)

11. \(2z=2+4i\), \(-z=-1-2i\); វ៉ិចទ័រទាំងបីស្ថិតលើបន្ទាត់តែមួយកាត់តាម \(O\) (កូលីនេអ៊ែរ)

មេរៀនទី២

1. ក. \(10\) ខ. \(13\) គ. \(5\sqrt{2}\) ឃ. \(\sqrt{5}\)

2. ក. \(r=2,\ \varphi=\dfrac{\pi}{6}\) ខ. \(r=2\sqrt{2},\ \varphi=-\dfrac{3\pi}{4}\) គ. \(r=3,\ \varphi=-\dfrac{\pi}{2}\) ឃ. \(r=5,\ \varphi=\pi\)

3. ក. \(\sqrt{2}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)\) ខ. \(2\left[\cos\left(-\dfrac{2\pi}{3}\right)+i\sin\left(-\dfrac{2\pi}{3}\right)\right]\)
   គ. \(4\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)\) ឃ. \(4\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\)

4. \(z_1z_2=2\sqrt{2}\left(\cos\dfrac{11\pi}{12}+i\sin\dfrac{11\pi}{12}\right)\)

5. \(\dfrac{z_1}{z_2}=\dfrac{\sqrt{2}}{2}\left[\cos\left(-\dfrac{5\pi}{12}\right)+i\sin\left(-\dfrac{5\pi}{12}\right)\right]\)

6. ក. \(z'=-1+3i\) ខ. \(z'=1+i\sqrt{3}\)

មេរៀនទី៣

1. ក. \(4i\) ខ. \(16\) គ. \(-64\)

2. ក. \(\dfrac{z_C-z_A}{z_B-z_A}=i\) ខ. ត្រីកោណ \(ABC\) ជាត្រីកោណកែងសមបាតត្រង់ \(A\)

3. ក. \(z\in\left\{3,\ \frac{3}{2}\pm\frac{\sqrt{3}}{2}i\right\}\) ខ. \((z^3+1)^2=0 \implies z\in\left\{-1,\ \frac{1}{2}\pm\frac{\sqrt{3}}{2}i\right\}\) (ឫសឌុប)
   គ. \(z\in\left\{\pm i,\ \pm\frac{\sqrt{3}}{2}\pm\frac{1}{2}i\right\}\) ឃ. \(z\in\left\{\pm 2i,\ \pm\sqrt{3}\pm i\right\}\)