ស្វ័យគុណទី n និង ឫសទី n នៃចំនួនកុំផ្លិច

តារាងចម្លើយមេរៀនទី៣ (ស្វ័យគុណ និង ឫសទី \(n\))

ដំណោះស្រាយលម្អិត លំហាត់អនុវត្តន៍មេរៀនទី៣ និងលំហាត់រួម

\(\blacksquare\) ក. លំហាត់អនុវត្តន៍មេរៀនទី៣ (១១ លំហាត់)

លំហាត់ ១៖ គណនាស្វ័យគុណតាមដឺម័រ
រូបមន្តដឺម័រ៖ \([r(\cos\theta+i\sin\theta)]^n = r^n(\cos n\theta+i\sin n\theta) = r^n e^{i n\theta}\)។

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

2. \((1-i)^{10} = (\sqrt{2}e^{-i\pi/4})^{10} = 2^5 e^{-i 5\pi/2}\)
   \(= 32(0-i) = -32i\)

3. \((3+3i)^5 = (3\sqrt{2}e^{i\pi/4})^5 = 972\sqrt{2}e^{i 5\pi/4}\)
   \(= 972\sqrt{2}\left(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right) = -972 - 972i\)

4. \((-4\sqrt{3}+4i)^5 = (8e^{i 5\pi/6})^5 = 32768 e^{i 25\pi/6}\)
   \(= 32768 e^{i\pi/6} = 32768\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\)
   \(= 16384\sqrt{3} + 16384i\)

5. \(\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)^{15} = (e^{i 3\pi/4})^{15} = e^{i 45\pi/4}\)
   \(= e^{i(10\pi+5\pi/4)} = e^{i 5\pi/4} = -\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\)

6. \((\sqrt{3}+i)^7 = (2e^{i\pi/6})^7 = 128 e^{i 7\pi/6}\)
   \(= 128\left(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\right) = -64\sqrt{3} - 64i\)

7. \((-1+i)^{50} = (\sqrt{2}e^{i 3\pi/4})^{50} = 2^{25} e^{i 75\pi/2}\)
   \(= 2^{25} e^{i(36\pi + 3\pi/2)} = 2^{25}(0-i) = -2^{25}i\)

8. \(\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)^{25} = (e^{i\pi/4})^{25} = e^{i 25\pi/4}\)
   \(= e^{i\pi/4} = \frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\)

លំហាត់ ២៖ គណនាកន្សោម \(u, v\)
គេមាន \(1+i\sqrt{3} = 2e^{i\pi/3}\) និង \(1-i\sqrt{3} = 2e^{-i\pi/3}\)៖

1. \(u = (1+i\sqrt{3})^5 + (1-i\sqrt{3})^5\)
   \(= 2^5 e^{i 5\pi/3} + 2^5 e^{-i 5\pi/3} = 32\left(2\cos\frac{5\pi}{3}\right)\)
   \(= 64 \cdot \frac{1}{2} = 32\)

2. \(v = (1+i\sqrt{3})^5 - (1-i\sqrt{3})^5\)
   \(= 2^5 e^{i 5\pi/3} - 2^5 e^{-i 5\pi/3} = 32\left(2i\sin\frac{5\pi}{3}\right)\)
   \(= 64i\left(-\frac{\sqrt{3}}{2}\right) = -32i\sqrt{3}\)

លំហាត់ ៣៖ ទម្រង់ត្រីកោណមាត្រ

1. \(\frac{(1+i)^5}{(1-i)^3} = \frac{(\sqrt{2}e^{i\pi/4})^5}{(\sqrt{2}e^{-i\pi/4})^3} = \frac{4\sqrt{2}e^{i 5\pi/4}}{2\sqrt{2}e^{-i 3\pi/4}}\)
   \(= 2e^{i 2\pi} = 2(\cos 0+i\sin 0)\)

2. \(\left(\frac{1+i\sqrt{3}}{1-i}\right)^{20} = \left(\frac{2e^{i\pi/3}}{\sqrt{2}e^{-i\pi/4}}\right)^{20} = (\sqrt{2}e^{i 7\pi/12})^{20}\)
   \(= 2^{10}e^{i 35\pi/3} = 1024 e^{-i\pi/3}\)
   \(= 1024\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right)\)

3. \((1+i)^8(1-i\sqrt{3})^{-6} = \frac{(1+i)^8}{(1-i\sqrt{3})^6}\)
   \(= \frac{(\sqrt{2}e^{i\pi/4})^8}{(2e^{-i\pi/3})^6} = \frac{16e^{i 2\pi}}{64e^{-i 2\pi}} = \frac{1}{4}e^{i 4\pi} = \frac{1}{4}(\cos 0+i\sin 0)\)

លំហាត់ ៤៖ បង្ហាញសមភាព

1. \(1+i = \sqrt{2}\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right) = 2^{1/2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)
   តាមដឺម័រ៖ \((1+i)^n = (2^{1/2})^n\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)\)
   \(= 2^{n/2}\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)\) (ពិត)

2. \(\sqrt{3}-i = 2\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right) = 2\left(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})\right)\)
   \(\implies (\sqrt{3}-i)^n = 2^n\left(\cos\frac{n\pi}{6}-i\sin\frac{n\pi}{6}\right)\) (ពិត)

លំហាត់ ៥៖ ផលបូកស្វ័យគុណ \(z^k\)
\(z = \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} = e^{i\pi/6}\)៖

1. ចំណុចរូបភាព \(z^k = e^{i k\pi/6}\) (\(k=1,\dots,7\)) ចែករង្វង់ត្រីកោណមាត្រជាធ្នូមានមុំ \(\frac{\pi}{6} = 30^\circ\) ស្មើៗគ្នា។

2. \(S = 1+z+z^2+\dots+z^{11} = \frac{z^{12}-1}{z-1}\)
   ដោយ \(z^{12} = (e^{i\pi/6})^{12} = e^{i 2\pi} = 1\)
   \(\implies S = \frac{1-1}{z-1} = 0\)

លំហាត់ ៦៖ គណនាឫសទី ២
រូបមន្ត៖ ឫសទី \(n\) នៃ \(r e^{i\theta}\) គឺ \(w_k = \sqrt[n]{r}e^{i\frac{\theta+2k\pi}{n}}\) (\(k=0,\dots,n-1\))។

1. \(Z = 9\cis\left(\frac{\pi}{4}\right)\)៖
   \(w_0 = 3\cis\left(\frac{\pi}{8}\right),\; w_1 = 3\cis\left(\frac{9\pi}{8}\right) = -3\cis\left(\frac{\pi}{8}\right)\)

2. \(Z = 1-i = \sqrt{2}e^{-i\pi/4}\)៖
   \(w_0 = \sqrt[4]{2}e^{-i\pi/8},\; w_1 = -\sqrt[4]{2}e^{-i\pi/8}\)

3. \(Z = 4+4i = 4\sqrt{2}e^{i\pi/4}\)៖
   \(w_0 = 2\sqrt[4]{2}e^{i\pi/8},\; w_1 = -2\sqrt[4]{2}e^{i\pi/8}\)

លំហាត់ ៧៖ គណនាឫសទី ៣

1. \(Z = 8\cis\left(\frac{3\pi}{4}\right)\)៖
   \(w_k = 2\cis\left(\frac{\pi}{4}+\frac{2k\pi}{3}\right)\) (\(k\in\{0,1,2\}\))
   \(w_0 = 2\cis\left(\frac{\pi}{4}\right) = \sqrt{2}+i\sqrt{2}\)
   \(w_1 = 2\cis\left(\frac{11\pi}{12}\right),\; w_2 = 2\cis\left(\frac{19\pi}{12}\right)\)

2. \(Z = -2+2i = 2\sqrt{2}e^{i 3\pi/4}\)៖
   \(w_k = \sqrt{2}e^{i(\frac{\pi}{4}+\frac{2k\pi}{3})},\; k=0,1,2\)

3. \(Z = -1-i\sqrt{3} = 2e^{-i 2\pi/3}\)៖
   \(w_k = \sqrt[3]{2}e^{i(-\frac{2\pi}{9}+\frac{2k\pi}{3})},\; k=0,1,2\)

លំហាត់ ៨៖ គណនាឫសទី ៤

1. \(Z = 1+i\sqrt{3} = 2e^{i\pi/3} \implies w_k = \sqrt[4]{2}e^{i(\frac{\pi}{12}+\frac{k\pi}{2})},\; k=0,1,2,3\)

2. \(Z = 2\sqrt{3}-2i = 4e^{-i\pi/6} \implies w_k = \sqrt{2}e^{i(-\frac{\pi}{24}+\frac{k\pi}{2})},\; k=0,1,2,3\)

3. \(Z = -2-2i\sqrt{3} = 4e^{-i 2\pi/3} \implies w_k = \sqrt{2}e^{i(-\frac{\pi}{6}+\frac{k\pi}{2})},\; k=0,1,2,3\)

លំហាត់ ៩៖ ឫសទី ៥ នៃ \(-\sqrt{3}-i\)
\(-\sqrt{3}-i = 2e^{i 7\pi/6}\)
\(\implies w_k = \sqrt[5]{2}e^{i(\frac{7\pi}{30}+\frac{2k\pi}{5})},\; k=0,1,2,3,4\)។
កំពូលទាំង ៥ បង្កើតបានជាបញ្ចកោណនិយ័តចារឹកក្នុងរង្វង់ផ្ចិត \(O\) កាំ \(R = \sqrt[5]{2}\)។

លំហាត់ ១០៖ ឫសទី ៦ នៃ \(1\)
\(1 = e^{i 0} \implies w_k = e^{i\frac{2k\pi}{6}} = e^{i\frac{k\pi}{3}},\; k=0,1,2,3,4,5\)។
ឫសទាំង ៦ គឺ៖ \(1,\; \frac{1}{2}+i\frac{\sqrt{3}}{2},\; -\frac{1}{2}+i\frac{\sqrt{3}}{2},\; -1,\; -\frac{1}{2}-i\frac{\sqrt{3}}{2},\; \frac{1}{2}-i\frac{\sqrt{3}}{2}\)។
ពួកវាជាកំពូលនៃឆកោណនិយ័តចារឹកក្នុងរង្វង់ឯកតា។

លំហាត់ ១១៖ ដោះស្រាយសមីការ

1. \(z^4 = 8(-1+i\sqrt{3}) = 16e^{i 2\pi/3} \implies z_k = 2e^{i(\frac{\pi}{6}+\frac{k\pi}{2})},\; k=0,1,2,3\)

2. \(z^3 = -8i = 8e^{-i\pi/2} \implies z_k = 2e^{i(-\frac{\pi}{6}+\frac{2k\pi}{3})},\; k=0,1,2\)

3. \(z^4 = -8i = 8e^{-i\pi/2} \implies z_k = \sqrt[4]{8}e^{i(-\frac{\pi}{8}+\frac{k\pi}{2})},\; k=0,1,2,3\)

4. \(z^3 = -27i = 27e^{-i\pi/2} \implies z_k = 3e^{i(-\frac{\pi}{6}+\frac{2k\pi}{3})},\; k=0,1,2\)

5. \(z^3 = 4\sqrt{3}-4i = 8e^{-i\pi/6} \implies z_k = 2e^{i(-\frac{\pi}{18}+\frac{2k\pi}{3})},\; k=0,1,2\)

6. \(z^5 = -\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}} = e^{i 5\pi/4} \implies z_k = e^{i(\frac{\pi}{4}+\frac{2k\pi}{5})},\; k=0,1,2,3,4\)

\(\blacksquare\) ខ. លំហាត់រួម មេរៀនទី១–៣ (១៦ លំហាត់)

លំហាត់ ១៖ ប្រមាណវិធីលើ \(z_1, z_2\)

1. \(z_1=3+2i, z_2=1-7i\)៖
   \(z_1+z_2 = 4-5i,\; z_1-z_2 = 2+9i\)
   \(z_1 z_2 = (3+2i)(1-7i) = 3 - 21i + 2i + 14 = 17 - 19i\)
   \(\frac{z_1}{z_2} = \frac{(3+2i)(1+7i)}{50} = \frac{3+21i+2i-14}{50} = -\frac{11}{50} + \frac{23}{50}i\)

2. \(z_1=\sqrt{3}-i\sqrt{5}, z_2=\sqrt{3}+i\sqrt{5}\)៖
   \(z_1+z_2 = 2\sqrt{3},\; z_1-z_2 = -2i\sqrt{5}\)
   \(z_1 z_2 = 3 - (-5) = 8\)
   \(\frac{z_1}{z_2} = \frac{(\sqrt{3}-i\sqrt{5})^2}{8} = \frac{3 - 2i\sqrt{15} - 5}{8} = -\frac{1}{4} - \frac{\sqrt{15}}{4}i\)

លំហាត់ ២៖ សរសេរជាទម្រង់ពិជគណិត

1. \(z = \frac{-14+63i}{50} - \frac{6i+1}{1-7i}\)
   \(\frac{1+6i}{1-7i} = \frac{(1+6i)(1+7i)}{50} = \frac{1+7i+6i-42}{50} = \frac{-41+13i}{50}\)
   \(\implies z = \frac{-14+63i - (-41+13i)}{50} = \frac{27+50i}{50} = \frac{27}{50} + i\)

2. \(z = \frac{(1+2i)^2 - (1-i)^2}{(3+2i)^3 - (2+i)^2}\)
   ភាគយក៖ \((-3+4i) - (-2i) = -3+6i = 3(-1+2i)\)
   ភាគបែង៖ \((-9+46i) - (3+4i) = -12+42i = 6(-2+7i)\)
   \(\implies z = \frac{-1+2i}{2(-2+7i)} = \frac{(-1+2i)(-2-7i)}{2(4+49)} = \frac{16+3i}{106}\)

3. \(z = \frac{13+12i}{6i-8} + \frac{(2i+1)^2}{2+i} = -\frac{18}{25} + \frac{23}{50}i\)

4. \(z = \left(\frac{-1+i\sqrt{3}}{2i}\right)^2 = \frac{1 - 2i\sqrt{3} - 3}{-4} = \frac{-2-2i\sqrt{3}}{-4} = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)

លំហាត់ ៣៖ សមីការដឺក្រេទីពីរ និងសមីការបំប្លែង

1. \(z^2 - 6z + 13 = 0\)៖
   \(\Delta' = (-3)^2 - 13 = -4 = (2i)^2\)
   ដូចនេះ \( z = 3 \pm 2i\)

2. តាង \(X = \frac{u-3i}{u+2}\) នោះសមីការក្លាយជា \(X^2-6X+13=0\)
   \(\implies X = 3 \pm 2i\)
   - បើ \(\frac{u-3i}{u+2} = 3+2i \implies u - 3i = (3+2i)(u+2)\)
   \(\iff u(2+2i) = -6 - 7i \implies u = \frac{(-6-7i)(2-2i)}{8} = \frac{-26-2i}{8} = -\frac{13}{4} - \frac{1}{4}i\)
   - បើ \(\frac{u-3i}{u+2} = 3-2i \implies u - 3i = (3-2i)(u+2)\)
   \(\iff u(2-2i) = -6 + i \implies u = \frac{-6+i}{2-2i} = -\frac{7}{4} - \frac{5}{4}i\)

លំហាត់ ៤៖ ស្វែងរក \(z\)
\(|z| = \left|\frac{1}{z}\right| \implies |z| = \frac{1}{|z|} \implies |z|^2 = 1 \implies |z| = 1\)។
ម៉្យាងទៀត \(|1-z| = |z| = 1\)។
តាង \(z = \cos\theta+i\sin\theta\) នោះ៖
\(|1 - \cos\theta - i\sin\theta|^2 = 1 \iff (1-\cos\theta)^2 + \sin^2\theta = 1\)
\(\iff 2 - 2\cos\theta = 1 \iff \cos\theta = \frac{1}{2} \implies \theta = \pm\frac{\pi}{3}\)
ដូចនេះ \( z = \cos\left(\pm\frac{\pi}{3}\right)+i\sin\left(\pm\frac{\pi}{3}\right) = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}\)

លំហាត់ ៥៖ កន្សោម \(z = \frac{i-x}{i+x}\)

1. \(|z| = \frac{|-x+i|}{|x+i|} = \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}} = 1\)
   \(z = \frac{(-x+i)(x-i)}{x^2+1} = \frac{-x^2+1 + 2xi}{x^2+1} = \frac{1-x^2}{1+x^2} + i\frac{2x}{1+x^2}\)

2. \(\text{Re}(z) = \text{Im}(z) \iff 1 - x^2 = 2x \iff x^2 + 2x - 1 = 0\)
   \(\implies x = -1 \pm \sqrt{2}\)
   បើ \(z = \frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \implies \frac{2x}{1+x^2} = \frac{\sqrt{2}}{2} > 0 \implies x = \sqrt{2}-1\)

លំហាត់ ៦៖ ទម្រង់ត្រីកោណមាត្រ

1. \(\sqrt{2}+i\sqrt{6} = 2\sqrt{2}\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 2\sqrt{2}\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\)

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

3. \((\sqrt{3}-i)^{100} = (2e^{-i\pi/6})^{100} = 2^{100}e^{-i 50\pi/3}\)
   \(= 2^{100}e^{-i 2\pi/3} = 2^{100}\left(\cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right)\right)\)

4. \(2+\sqrt{2}+i\sqrt{2} = 2(1+e^{i\pi/4}) = 4\cos\frac{\pi}{8} e^{i\pi/8}\)
   \(= 4\cos\frac{\pi}{8}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)\)

5. \(1+(2+\sqrt{3})i = 2\sqrt{2+\sqrt{3}}\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)\)

6. \(1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7} = 2\cos\frac{\pi}{14}\left(\cos\frac{\pi}{14}+i\sin\frac{\pi}{14}\right)\)

លំហាត់ ៧៖ ដោះស្រាយសមីការ

1. \(z^4 = -8-8i\sqrt{3} = 16e^{-i 2\pi/3}\)
   \(\implies z_k = 2e^{i(-\frac{\pi}{6}+\frac{k\pi}{2})},\; k=0,1,2,3\)

2. \(z^2 + |z|^2 = 0 \iff z^2 = -|z|^2\)
   តាង \(z = x+iy \implies x^2-y^2+2xyi = -(x^2+y^2)\)
   \(\implies 2x^2 = 0 \implies x = 0\) និង \(2xy = 0\) (ផ្ទៀងផ្ទាត់ជានិច្ច)
   ដូចនេះ \( z = iy\) ដែល \(y \in \mathbb{R}\) (\(z\) ជានិម្មិតសុទ្ធ)

លំហាត់ ៨៖ អាគុយម៉ង់វិជ្ជមានតូចបំផុត
\(|z - (-3+i\sqrt{3})| = \sqrt{3}\) ជារង្វង់ផ្ចិត \(\Omega(-3, \sqrt{3})\) កាំ \(R = \sqrt{3}\)។
ចំងាយ \(O\Omega = \sqrt{(-3)^2+(\sqrt{3})^2} = \sqrt{12} = 2\sqrt{3} = 2R\)។
បន្ទាត់ប៉ះ \(OT\) ចេញពី \(O\) (\(T\) ជាចំណុចប៉ះ, \(\Omega T\perp OT\)) បង្កើតមុំ \(\angle TO\Omega\) ដែល \(\sin\angle T O\Omega = \frac{R}{O\Omega} = \frac{1}{2} \implies \angle T O\Omega = \frac{\pi}{6}\)។
មុំ \(\arg(\Omega) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)។
អាគុយម៉ង់វិជ្ជមានតូចបំផុតគឺ \(\theta_{\min} = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}\)។
ប្រវែង \(OT = \sqrt{O\Omega^2-R^2} = \sqrt{12-3} = 3\)។
ដូចនេះ \( z = 3e^{i 2\pi/3} = 3\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = -\frac{3}{2} + i\frac{3\sqrt{3}}{2}\)

លំហាត់ ៩៖ ទម្រង់ត្រីកោណមាត្រ
\(z = a\frac{1+i\tan\alpha}{1-i\tan\alpha} = a\frac{\cos\alpha+i\sin\alpha}{\cos\alpha-i\sin\alpha} = a\frac{e^{i\alpha}}{e^{-i\alpha}} = a e^{i 2\alpha}\)
\(= a(\cos 2\alpha + i\sin 2\alpha)\) (ចំពោះ \(a > 0\))

លំហាត់ ១០៖ ស្វ័យគុណកុំផ្លិច

1. \(\left(\frac{1+i}{\sqrt{2}}\right)^{2020} = (e^{i\pi/4})^{2020} = e^{i 505\pi} = -1\)

2. \(\frac{1+i\sqrt{3}}{\sqrt{3}+i} = \frac{2e^{i\pi/3}}{2e^{i\pi/6}} = e^{i\pi/6}\)
   \(\implies (e^{i\pi/6})^{2021} = e^{i\frac{2021\pi}{6}} = e^{i 5\pi/6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i\)

3. \(\frac{1+i\sqrt{3}}{\sqrt{2}+i\sqrt{2}} = \frac{2e^{i\pi/3}}{2e^{i\pi/4}} = e^{i\pi/12}\)
   \(\implies (e^{i\pi/12})^{2022} = e^{i\frac{2022\pi}{12}} = e^{i\pi/2} = i\)

លំហាត់ ១១៖ តម្លៃពិត \(\cos\frac{5\pi}{12}\) និង \(\sin\frac{5\pi}{12}\)

1. \(z_1 = \sqrt{6}+i\sqrt{2} = 2\sqrt{2}e^{i\pi/6}\)
   \(z_2 = 2+2i = 2\sqrt{2}e^{i\pi/4}\)
   \(z_1 z_2 = (2\sqrt{2}e^{i\pi/6})(2\sqrt{2}e^{i\pi/4})\)
   \(= 8e^{i 5\pi/12} = 8\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)\)

2. ទម្រង់ពិជគណិត៖
   \(z_1 z_2 = (\sqrt{6}+i\sqrt{2})(2+2i)\)
   \(= 2(\sqrt{6}-\sqrt{2}) + 2i(\sqrt{6}+\sqrt{2})\)
   \(\implies \cos\frac{5\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4},\; \sin\frac{5\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}\)

លំហាត់ ១២៖ តម្លៃពិត \(\cos\frac{\pi}{12}\) និង \(\sin\frac{\pi}{12}\)
\(z_1 = \sqrt{3}-i = 2e^{-i\pi/6} \implies z_1^4 = 16e^{-i 2\pi/3} = -8-8i\sqrt{3}\)
\(z_2 = 2-2i = 2\sqrt{2}e^{-i\pi/4} \implies z_2^3 = 16\sqrt{2}e^{-i 3\pi/4} = -16-16i\)
ទម្រង់ពិជគណិត៖ \(A = \frac{-8-8i\sqrt{3}}{-16-16i} = \frac{1+i\sqrt{3}}{2(1+i)} = \frac{(1+i\sqrt{3})(1-i)}{4} = \frac{1+\sqrt{3}}{4} + i\frac{\sqrt{3}-1}{4}\)
ទម្រង់ត្រីកោណមាត្រ៖ \(A = \frac{16e^{-i 2\pi/3}}{16\sqrt{2}e^{-i 3\pi/4}} = \frac{\sqrt{2}}{2}e^{i\pi/12} = \frac{\sqrt{2}}{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)
ផ្ទឹមទម្រង់ទាំងពីរ៖ \(\frac{\sqrt{2}}{2}\cos\frac{\pi}{12} = \frac{1+\sqrt{3}}{4}\) និង \(\frac{\sqrt{2}}{2}\sin\frac{\pi}{12} = \frac{\sqrt{3}-1}{4}\)
\(\implies \cos\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4},\; \sin\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

លំហាត់ ១៣៖ លក្ខណៈ \(z = 1+\cos\varphi+i\sin\varphi\)

1. \(\varphi = \frac{\pi}{3} \implies z = 1+\frac{1}{2}+i\frac{\sqrt{3}}{2} = \sqrt{3}e^{i\pi/6}\)
   \(\frac{1}{z} = \frac{1}{\sqrt{3}}e^{-i\pi/6} = \frac{\sqrt{3}}{3}\left(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6})\right)\)

2. \(z - 1 = \cos\varphi+i\sin\varphi \implies |z-1| = 1\)
   ដូចនេះ ចំណុច \(M(z)\) ស្ថិតនៅលើរង្វង់ផ្ចិត \(I(1, 0)\) កាំ \(R=1\)។

3. \(\frac{1}{z} = \frac{1}{2\cos\frac{\varphi}{2}e^{i\varphi/2}} = \frac{e^{-i\varphi/2}}{2\cos\frac{\varphi}{2}} = \frac{\cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}}{2\cos\frac{\varphi}{2}}\)
   \(= \frac{1}{2} - i\frac{1}{2}\tan\frac{\varphi}{2} \implies \text{Re}\left(\frac{1}{z}\right) = \frac{1}{2}\) (ថេរគ្រប់ \(\varphi\))

លំហាត់ ១៤៖ ឫសការេនៃ \(4\sqrt{3}+4i\)

1. \(|4\sqrt{3}+4i| = \sqrt{48+16} = 8\), \(\cos\theta = \frac{\sqrt{3}}{2},\ \sin\theta = \frac{1}{2} \implies \arg = \frac{\pi}{6}+2k\pi\)
   \(4\sqrt{3}+4i = 8\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\)

2. (i) ទម្រង់ពិជគណិត៖ តាង \(z = x+iy\) នោះ \(z^2 = 4\sqrt{3}+4i\)
   \(\iff \begin{cases} x^2-y^2 = 4\sqrt{3} \\ xy = 2 > 0 \\ x^2+y^2 = |z^2| = 8 \end{cases}\)
   \(\implies x^2 = 4+2\sqrt{3} = (\sqrt{3}+1)^2,\; y^2 = 4-2\sqrt{3} = (\sqrt{3}-1)^2\) (\(x, y\) សញ្ញាដូចគ្នា)
   \(\implies z = \pm\left[(\sqrt{3}+1)+i(\sqrt{3}-1)\right]\)
   (ii) ទម្រង់ត្រីកោណមាត្រ៖
   \(z_k = 2\sqrt{2}\cis\left(\frac{\pi}{12}+k\pi\right),\ k=0,1\)
   \(\implies z = \pm 2\sqrt{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)

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

លំហាត់ ១៥៖ \(a = \sqrt{2+\sqrt{3}}+i\sqrt{2-\sqrt{3}}\)

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

2. \(|a|^2 = 4 \implies |a| = 2\) និង \(2\arg(a) \equiv \frac{\pi}{6} \pmod{2\pi} \implies \arg(a) = \frac{\pi}{12}\) ឬ \(\frac{13\pi}{12}\)
   ដោយ \(\text{Re}(a)>0,\ \text{Im}(a)>0\) នោះ \(\arg(a) = \frac{\pi}{12}\)
   \(\implies a = 2\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)

3. ផ្ទឹមជាមួយ \(a = \sqrt{2+\sqrt{3}}+i\sqrt{2-\sqrt{3}}\)៖
   \(\cos\frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2} = \frac{\sqrt{6}+\sqrt{2}}{4},\; \sin\frac{\pi}{12} = \frac{\sqrt{2-\sqrt{3}}}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

លំហាត់ ១៦៖ \(z = (\sqrt{3}+1)+i(\sqrt{3}-1)\)

1. \(z^2 = [(\sqrt{3}+1)+i(\sqrt{3}-1)]^2 = 4\sqrt{3}+4i = 8e^{i\pi/6}\)
   \(\implies |z|^2 = 8 \implies |z| = 2\sqrt{2}\)
   \(2\arg(z) \equiv \frac{\pi}{6} \pmod{2\pi} \implies \arg(z) = \frac{\pi}{12}\) ឬ \(\frac{13\pi}{12}\)។ ដោយ \(\text{Re}(z)>0,\ \text{Im}(z)>0\) (\(z\) ក្នុងកាដ្រង់ទី១) នោះ \(\arg(z) = \frac{\pi}{12}\)

2. ទម្រង់ពិជគណិត៖ \(z = 2\sqrt{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\)
   \(\implies \cos\frac{\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}\)
   \(\implies \sin\frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

3. សមីការ \((\sqrt{3}+1)\cos x + (\sqrt{3}-1)\sin x = \sqrt{2}\)៖
   ចែកនឹង \(2\sqrt{2}\)៖
   \(\cos\frac{\pi}{12}\cos x + \sin\frac{\pi}{12}\sin x = \frac{1}{2}\)
   \(\iff \cos\left(x - \frac{\pi}{12}\right) = \cos\frac{\pi}{3}\)
   \(\implies x - \frac{\pi}{12} = \pm\frac{\pi}{3} + 2k\pi\)
   \(x_1 = \frac{5\pi}{12} + 2k\pi,\; x_2 = -\frac{\pi}{4} + 2k\pi\) (\(k \in \mathbb{Z}\))