ចំនួនកុំផ្លិចទម្រង់ត្រីកោណមាត្រ

តារាងចម្លើយមេរៀនទី២ (ទម្រង់ត្រីកោណមាត្រ)

ដំណោះស្រាយលម្អិត លំហាត់អនុវត្តន៍មេរៀនទី២

លំហាត់ ១៖ គណនាម៉ូឌុល
រូបមន្ត៖ ចំពោះ \(z = a+bi\) គេបាន \(|z| = \sqrt{a^2+b^2}\)។

1. \(|-2i| = \sqrt{0^2+(-2)^2} = \sqrt{4} = 2\)

2. \(|\sqrt{3}+i| = \sqrt{(\sqrt{3})^2+1^2} = \sqrt{3+1} = \sqrt{4} = 2\)

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

4. \(|(4+3i) + (4-3i)| = |8| = 8\)

5. \(|i^7| = |(i^4)(i^3)| = |-i| = 1\)

6. \(|i^{100}| = |(i^4)^{25}| = |1| = 1\)

លំហាត់ ២៖ ស្រាយបញ្ជាក់លក្ខណៈកុំផ្លិចឆ្លាស់
តាង \(z_1 = a_1 + b_1i\) និង \(z_2 = a_2 + b_2i\) ដែល \(a_1, b_1, a_2, b_2 \in \mathbb{R}\)។

1. \(\overline{z_1+z_2} = \overline{(a_1+a_2) + (b_1+b_2)i}\)
   \(= (a_1+a_2) - (b_1+b_2)i\)
   \(= (a_1 - b_1i) + (a_2 - b_2i) = \overline{z_1} + \overline{z_2}\) (ពិត)

2. \(z_1 z_2 = (a_1 a_2 - b_1 b_2) + (a_1 b_2 + a_2 b_1)i\)
   \(\implies \overline{z_1 z_2} = (a_1 a_2 - b_1 b_2) - (a_1 b_2 + a_2 b_1)i\)
   ម៉្យាងទៀត \(\overline{z_1} \cdot \overline{z_2} = (a_1 - b_1i)(a_2 - b_2i)\)
   \(= a_1 a_2 - a_1 b_2i - a_2 b_1i + b_1 b_2 i^2\)
   \(= (a_1 a_2 - b_1 b_2) - (a_1 b_2 + a_2 b_1)i = \overline{z_1 z_2}\) (ពិត)

3. \(\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}\) (\(z_2 \ne 0\))៖
   តាង \(w = \frac{z_1}{z_2} \iff z_1 = w z_2\)
   តាមលក្ខណៈ (b)៖ \(\overline{z_1} = \overline{w z_2} = \overline{w} \cdot \overline{z_2}\)
   ដោយ \(z_2 \ne 0 \implies \overline{z_2} \ne 0\) គេទាញបាន៖
   \(\overline{w} = \frac{\overline{z_1}}{\overline{z_2}} \iff \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}\) (ពិត)

លំហាត់ ៣៖ គណនាអាគុយម៉ង់
វិធីសាស្ត្រ៖ រក \(r = |z|\) រួចទាញរក \(\theta\) តាម \(\cos\theta = \frac{a}{r},\; \sin\theta = \frac{b}{r}\)។

1. \(z = 2+2i \implies r = \sqrt{2^2+2^2} = 2\sqrt{2}\)
   \(\cos\theta = \frac{\sqrt{2}}{2},\; \sin\theta = \frac{\sqrt{2}}{2} \implies \arg(z) = \frac{\pi}{4}\)

2. \(z = -3+3i \implies r = 3\sqrt{2}\)
   \(\cos\theta = -\frac{\sqrt{2}}{2},\; \sin\theta = \frac{\sqrt{2}}{2} \implies \arg(z) = \frac{3\pi}{4}\)

3. \(z = -2+2i\sqrt{3} \implies r = \sqrt{4+12} = 4\)
   \(\cos\theta = -\frac{1}{2},\; \sin\theta = \frac{\sqrt{3}}{2} \implies \arg(z) = \frac{2\pi}{3}\)

4. \(z = -1-i \implies r = \sqrt{2}\)
   \(\cos\theta = -\frac{\sqrt{2}}{2},\; \sin\theta = -\frac{\sqrt{2}}{2} \implies \arg(z) = -\frac{3\pi}{4}\) (ឬ \(\frac{5\pi}{4}\))

5. \(z = 4i \implies r = 4,\; \cos\theta = 0,\; \sin\theta = 1 \implies \arg(z) = \frac{\pi}{2}\)

6. \(z = -4i \implies r = 4,\; \cos\theta = 0,\; \sin\theta = -1 \implies \arg(z) = -\frac{\pi}{2}\)

7. \(z = -\sqrt{3}+i \implies r = 2\)
   \(\cos\theta = -\frac{\sqrt{3}}{2},\; \sin\theta = \frac{1}{2} \implies \arg(z) = \frac{5\pi}{6}\)

8. \(z = 8+8i\sqrt{3} \implies r = 16\)
   \(\cos\theta = \frac{1}{2},\; \sin\theta = \frac{\sqrt{3}}{2} \implies \arg(z) = \frac{\pi}{3}\)

លំហាត់ ៤៖ ម៉ូឌុល និងអាគុយម៉ង់
គេឱ្យ \(z_1 = 1+i\) និង \(z_2 = -1+i\sqrt{3}\)៖

1. \(|z_1| = \sqrt{1^2+1^2} = \sqrt{2}\)
   \(\cos\theta_1 = \frac{1}{\sqrt{2}},\; \sin\theta_1 = \frac{1}{\sqrt{2}} \implies \arg(z_1) = \frac{\pi}{4}\)

2. \(|z_2| = \sqrt{(-1)^2+(\sqrt{3})^2} = 2\)
   \(\cos\theta_2 = -\frac{1}{2},\; \sin\theta_2 = \frac{\sqrt{3}}{2} \implies \arg(z_2) = \frac{2\pi}{3}\)

3. ផលគុណ \(z_1 z_2\)៖
   \(|z_1 z_2| = |z_1| \cdot |z_2| = 2\sqrt{2}\)
   \(\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{11\pi}{12}\)

4. ផលចែក \(\frac{z_1}{z_2}\)៖
   \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} = \frac{\sqrt{2}}{2}\)
   \(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) = \frac{\pi}{4} - \frac{2\pi}{3} = -\frac{5\pi}{12}\)

លំហាត់ ៥៖ កំណត់ទម្រង់ត្រីកោណមាត្រ
លក្ខខណ្ឌ៖ ទម្រង់ត្រីកោណមាត្រត្រូវមានទម្រង់ \(r(\cos\theta + i\sin\theta)\) ដែល \(r > 0\) និងមុំដូចគ្នា។

• \(z_1 = 2\left[\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right]\)៖ ជាទម្រង់ត្រីកោណមាត្រ (\(r=2>0\))។

• \(z_2 = -5\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)\)៖ មិនមែនទេ ព្រោះ \(r = -5 < 0\)។

• \(z_3 = -2\left[\cos\left(-\frac{3\pi}{4}\right)+i\sin\left(-\frac{3\pi}{4}\right)\right]\)៖ មិនមែនទេ ព្រោះ \(r = -2 < 0\)។

• \(z_4 = 6\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)\)៖ ជាទម្រង់ត្រីកោណមាត្រ (\(r=6>0\))។

• \(z_5 = \sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)៖ ជាទម្រង់ត្រីកោណមាត្រ (\(r=\sqrt{2}>0\))។

• \(z_6 = -\sqrt{3}\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)៖ មិនមែនទេ ព្រោះ \(r = -\sqrt{3} < 0\)។

លំហាត់ ៦៖ សរសេរជាទម្រង់ត្រីកោណមាត្រ

1. \(z_1 = -5-5i\sqrt{3} = 10\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right) = 10\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)\)

2. \(\frac{-5+i}{2-3i} = \frac{(-5+i)(2+3i)}{13} = \frac{-13-13i}{13} = -1-i\)
   \(= \sqrt{2}\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)\)

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

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

5. \(-20i = 20\left[\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right]\)

6. \(15i = 15\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)

7. \(2\sqrt{3}+2i = 4\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right) = 4\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)\)

8. \(-6i = 6\left[\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right]\)

លំហាត់ ៧៖ ផលចែកត្រីកោណមាត្រ
\(z = \frac{\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}}{\cos\frac{4\pi}{9}+i\sin\frac{4\pi}{9}} = \cos\left(\frac{\pi}{9}-\frac{4\pi}{9}\right) + i\sin\left(\frac{\pi}{9}-\frac{4\pi}{9}\right)\)
\(= \cos\left(-\frac{3\pi}{9}\right) + i\sin\left(-\frac{3\pi}{9}\right) = \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\)

លំហាត់ ៨៖ ទម្រង់ពិជគណិត \(x+iy\)

1. \(|z|=2,\; \arg(z)=\frac{\pi}{3}\)៖
   \(z = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right) = 2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 1+i\sqrt{3}\)

2. \(|z|=4,\; \arg(z)=\frac{3\pi}{4}\)៖
   \(z = 4\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right) = 4\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right) = -2\sqrt{2}+2i\sqrt{2}\)

3. \(|z|=1,\; \arg(z)=-\frac{\pi}{2}\)៖
   \(z = 1\left[\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right] = 0 - i = -i\)

4. \(|z|=6,\; \arg(z)=\frac{7\pi}{6}\)៖
   \(z = 6\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right) = 6\left(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\right) = -3\sqrt{3}-3i\)

លំហាត់ ៩៖ កន្សោមត្រីកោណមាត្រ
\(z = \tan\alpha + i\) (\(0 < \alpha < \frac{\pi}{2}\)), \(w = 4\left(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10}\right)\)៖

1. \(|z| = \sqrt{\tan^2\alpha+1} = \sqrt{\frac{1}{\cos^2\alpha}} = \frac{1}{\cos\alpha}\) (ព្រោះ \(\cos\alpha > 0\))

2. \(|zw| = |z||w| = \frac{1}{\cos\alpha} \cdot 4 = \frac{4}{\cos\alpha}\)

3. \(z = \frac{\sin\alpha}{\cos\alpha}+i = \frac{1}{\cos\alpha}(\sin\alpha+i\cos\alpha)\)
   \(= \frac{1}{\cos\alpha}\left[\cos\left(\frac{\pi}{2}-\alpha\right)+i\sin\left(\frac{\pi}{2}-\alpha\right)\right]\)
   \(\implies \arg(z) = \frac{\pi}{2}-\alpha\)

4. \(\arg\left(\frac{z}{w}\right) = \arg(z) - \arg(w) = \left(\frac{\pi}{2}-\alpha\right) - \frac{\pi}{10} = \frac{2\pi}{5}-\alpha\)

លំហាត់ ១០៖ បំប្លែងត្រីកោណមាត្រ
\(z = \frac{1+\sin\alpha+i\cos\alpha}{1+\sin\alpha-i\cos\alpha} = \frac{1+\cos(\frac{\pi}{2}-\alpha)+i\sin(\frac{\pi}{2}-\alpha)}{1+\cos(\frac{\pi}{2}-\alpha)-i\sin(\frac{\pi}{2}-\alpha)}\)
តាង \(\varphi = \frac{\pi}{4}-\frac{\alpha}{2}\) នោះ \(\frac{\pi}{2}-\alpha = 2\varphi\) (លក្ខខណ្ឌ៖ \(1+\sin\alpha\ne 0\) ពោលគឺ \(\cos\varphi\ne 0\))៖
ភាគយក៖ \(2\cos^2\varphi + 2i\sin\varphi\cos\varphi = 2\cos\varphi e^{i\varphi}\)
ភាគបែង៖ \(2\cos^2\varphi - 2i\sin\varphi\cos\varphi = 2\cos\varphi e^{-i\varphi}\)
\(\implies z = \frac{2\cos\varphi e^{i\varphi}}{2\cos\varphi e^{-i\varphi}} = e^{i 2\varphi} = e^{i(\frac{\pi}{2}-\alpha)}\)
\(= \cos\left(\frac{\pi}{2}-\alpha\right) + i\sin\left(\frac{\pi}{2}-\alpha\right)\)

លំហាត់ ១១៖ រក \(w\) និងរូបភាព

1. \(z = -2+5i,\; zw = 14+23i\)
   \(\implies w = \frac{14+23i}{-2+5i} = \frac{(14+23i)(-2-5i)}{(-2)^2+5^2}\)
   \(= \frac{-28 - 70i - 46i + 115}{29} = \frac{87 - 116i}{29} = 3 - 4i\)

2. រូបភាពក្នុងប្លង់កុំផ្លិច៖
   \(z = -2+5i \implies M(-2, 5)\)
   \(w = 3-4i \implies N(3, -4)\)

លំហាត់ ១២៖ ផលគុណ និងផលចែក

1. \(z_1 = \sqrt{2}\cis\left(\frac{\pi}{2}\right),\; z_2 = 3\sqrt{3}\cis\left(\frac{\pi}{15}\right)\)៖
   \(z_1 z_2 = 3\sqrt{6}\cis\left(\frac{\pi}{2}+\frac{\pi}{15}\right) = 3\sqrt{6}\cis\left(\frac{17\pi}{30}\right)\)
   \(\frac{z_1}{z_2} = \frac{\sqrt{6}}{9}\cis\left(\frac{\pi}{2}-\frac{\pi}{15}\right) = \frac{\sqrt{6}}{9}\cis\left(\frac{13\pi}{30}\right)\)

2. \(z_1 = -5+5i = 5\sqrt{2}\cis\left(\frac{3\pi}{4}\right),\; z_2 = -3i = 3\cis\left(-\frac{\pi}{2}\right)\)៖
   \(z_1 z_2 = 15\sqrt{2}\cis\left(\frac{3\pi}{4}-\frac{\pi}{2}\right) = 15\sqrt{2}\cis\left(\frac{\pi}{4}\right) = 15+15i\)
   \(\frac{z_1}{z_2} = \frac{5\sqrt{2}}{3}\cis\left(\frac{3\pi}{4}+\frac{\pi}{2}\right) = \frac{5\sqrt{2}}{3}\cis\left(\frac{5\pi}{4}\right) = -\frac{5}{3}-\frac{5}{3}i\)

3. \(z_1 = 2i = 2\cis\left(\frac{\pi}{2}\right),\; z_2 = -3i = 3\cis\left(-\frac{\pi}{2}\right)\)៖
   \(z_1 z_2 = 6\cis(0) = 6\)
   \(\frac{z_1}{z_2} = \frac{2}{3}\cis(\pi) = -\frac{2}{3}\)

លំហាត់ ១៣៖ បំប្លែងវិល
\(M'(z')\) ជារូបភាព \(M(z=-4+4i)\) តាមបំប្លែងវិល \(\mathcal{R}(O, \frac{\pi}{6})\)៖
\(z' = z \cdot e^{i\pi/6} = (-4+4i)\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\)
\(= -2\sqrt{3} - 2i + 2i\sqrt{3} - 2\)
\(= 2(-\sqrt{3}-1) + 2(\sqrt{3}-1)i\)

លំហាត់ ១៤៖ កំណត់មុំវិល \(\theta\)

1. \(z' = \frac{1}{2}(z+i\sqrt{3}z) = z\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = z e^{i\pi/3} \implies \theta = \frac{\pi}{3}\)

2. \(z' = \frac{1}{2}(\sqrt{3}z-iz) = z\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right) = z e^{-i\pi/6} \implies \theta = -\frac{\pi}{6}\)

3. \(z' = iz = z e^{i\pi/2} \implies \theta = \frac{\pi}{2}\)

លំហាត់ ១៥៖ គណនាម៉ូឌុល

1. \(z = -7+24i \implies |z| = \sqrt{(-7)^2+24^2} = \sqrt{49+576} = 25\)

2. \(z = (1+i)^4 \implies |z| = |1+i|^4 = (\sqrt{1^2+1^2})^4 = (\sqrt{2})^4 = 4\)

3. \(z = \sqrt{2}-i\sqrt{7} \implies |z| = \sqrt{(\sqrt{2})^2+(-\sqrt{7})^2} = \sqrt{2+7} = 3\)

លំហាត់ ១៦៖ ស្រាយបញ្ជាក់លក្ខណៈម៉ូឌុល
ចំពោះ \(z = a+bi \implies |z| = \sqrt{a^2+b^2}\)៖
\(\bar{z} = a-bi \implies |\bar{z}| = \sqrt{a^2+(-b)^2} = \sqrt{a^2+b^2}\)
\(-z = -a-bi \implies |-z| = \sqrt{(-a)^2+(-b)^2} = \sqrt{a^2+b^2}\)
ដូចនេះ \( |z| = |\bar{z}| = |-z|\)

លំហាត់ ១៧៖ គណនាអាគុយម៉ង់

1. \(z = 1+i\sqrt{3} = 2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) \implies \arg(z) = \frac{\pi}{3}\)

2. \(z = -\sqrt{2}-i\sqrt{2} = 2\left(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right) \implies \arg(z) = -\frac{3\pi}{4}\)

3. \(z = 5 = 5(1+0i) \implies \arg(z) = 0\)

4. \(z = -3i = 3(0-i) \implies \arg(z) = -\frac{\pi}{2}\)

លំហាត់ ១៨៖ អាគុយម៉ង់នៃផលគុណ និងផលចែក
\(z_1 = \sqrt{3}+i \implies \arg(z_1) = \frac{\pi}{6}\)
\(z_2 = 1+i \implies \arg(z_2) = \frac{\pi}{4}\)
\(\arg(z_1 z_2) = \frac{\pi}{6}+\frac{\pi}{4} = \frac{5\pi}{12}\)
\(\arg\left(\frac{z_1}{z_2}\right) = \frac{\pi}{6}-\frac{\pi}{4} = -\frac{\pi}{12}\)

លំហាត់ ១៩៖ ទម្រង់ត្រីកោណមាត្រ

1. \(z_1 = -4 = 4(\cos\pi+i\sin\pi)\)

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

3. \(z_3 = -1+i\sqrt{3} = 2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)

4. \(z_4 = \sqrt{6}-i\sqrt{2} = 2\sqrt{2}\left[\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right]\)

លំហាត់ ២០៖ គណនា \(z^2, z^4\)
\(z = 2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\)៖
\(z^2 = 2^2\left(\cos\frac{2\pi}{4}+i\sin\frac{2\pi}{4}\right) = 4\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) = 4i\)
\(z^4 = 2^4\left(\cos\frac{4\pi}{4}+i\sin\frac{4\pi}{4}\right) = 16(\cos\pi+i\sin\pi) = -16\)

លំហាត់ ២១៖ បំប្លែងត្រីកោណមាត្រ
\(z = \frac{(1+i\sqrt{3})(1+i)}{1-i} = \frac{2e^{i\pi/3} \cdot \sqrt{2}e^{i\pi/4}}{\sqrt{2}e^{-i\pi/4}} = 2e^{i(\frac{\pi}{3}+\frac{\pi}{4}+\frac{\pi}{4})} = 2e^{i\frac{5\pi}{6}}\)
\(= 2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)\)

លំហាត់ ២២៖ កំណត់ \(a, b\)
\(2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right) = 2\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = -1+i\sqrt{3}\)
\(\implies a+bi = -1+i\sqrt{3} \implies a = -1,\; b = \sqrt{3}\)

លំហាត់ ២៣៖ ស្វ័យគុណទី ១០
\(z = \sqrt{2}-i\sqrt{2}\)៖

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

2. \(z^{10} = 2^{10}\left[\cos\left(-\frac{10\pi}{4}\right)+i\sin\left(-\frac{10\pi}{4}\right)\right]\)
   \(= 1024\left[\cos\left(-\frac{5\pi}{2}\right)+i\sin\left(-\frac{5\pi}{2}\right)\right] = 1024(0-i) = -1024i\)

លំហាត់ ២៤៖ អាគុយម៉ង់នៃ \(\bar{z}\)
\(z = -1-i = \sqrt{2}\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)\)
\(\bar{z} = -1+i = \sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right) \implies \arg(\bar{z}) = \frac{3\pi}{4}\)

លំហាត់ ២៥៖ ផលគុណមុំដឺក្រេ
\(z_1 = 3\cis(20^\circ),\; z_2 = 2\cis(40^\circ)\)
\(z_1 z_2 = 6\cis(20^\circ+40^\circ) = 6\cis(60^\circ)\)
\(= 6\left(\cos 60^\circ+i\sin 60^\circ\right) = 6\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 3+3i\sqrt{3}\)

លំហាត់ ២៦៖ ទម្រង់ \(r\cis\varphi\)
\(z = \frac{2(\cos\pi+i\sin\pi)}{4(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})} = \frac{1}{2}\left[\cos\left(\pi-\frac{\pi}{3}\right)+i\sin\left(\pi-\frac{\pi}{3}\right)\right]\)
\(= \frac{1}{2}\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)

លំហាត់ ២៧៖ ទម្រង់ត្រីកោណមាត្រត្រឹមត្រូវ
\(z = \cos\alpha - i\sin\alpha\)
ដោយ \(\cos(-\alpha) = \cos\alpha\) និង \(\sin(-\alpha) = -\sin\alpha\) គេបាន៖
\(z = \cos(-\alpha) + i\sin(-\alpha)\) (ដែលមាន \(r = 1 > 0\))

លំហាត់ ២៨៖ ចំណុចបង្វិល \(M'\)
\(z = 1+i\sqrt{3} = 2e^{i\pi/3}\)
\(z' = z e^{i\pi/3} = 2e^{i 2\pi/3} = 2\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = -1+i\sqrt{3}\)
ដូចនេះ \( M'(-1, \sqrt{3})\)

លំហាត់ ២៩៖ ស្រាយបញ្ជាក់ \(\bar{z} = 1/z\)
ដោយ \(|z| = 1 \implies |z|^2 = 1\)
តាមរូបមន្ត \(|z|^2 = z \cdot \bar{z}\) គេបាន៖
\(z \cdot \bar{z} = 1 \implies \bar{z} = \frac{1}{z}\) (ពិត)

លំហាត់ ៣០៖ គណនា \((1+i)^{10}\)
\(1+i = \sqrt{2}e^{i\pi/4}\)
\(z = (1+i)^{10} = (\sqrt{2})^{10} e^{i 10\pi/4} = 32 e^{i 5\pi/2}\)
- ទម្រង់ត្រីកោណមាត្រ៖ \(z = 32\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\)
- ទម្រង់ពិជគណិត៖ \(z = 32(0+i) = 32i\)

លំហាត់ ៣១៖ ម៉ូឌុល \(1+\cos\theta+i\sin\theta\)
\(z = 2\cos^2\frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)\)
ដោយ \(-\pi < \theta < \pi \implies -\frac{\pi}{2} < \frac{\theta}{2} < \frac{\pi}{2} \implies \cos\frac{\theta}{2} > 0\)
ដូចនេះ \( |z| = 2\cos\frac{\theta}{2}\)

លំហាត់ ៣២៖ រក \(z\) ពីម៉ូឌុល និងអាគុយម៉ង់
\(z = 3\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right) = 3\left(-\frac{\sqrt{3}}{2}+i\frac{1}{2}\right) = -\frac{3\sqrt{3}}{2}+\frac{3}{2}i\)

លំហាត់ ៣៣៖ កំណត់មុំបង្វិល \(\theta\)
\(z' = z e^{i\theta} \implies e^{i\theta} = \frac{z'}{z} = \frac{-1+i}{1+i} = \frac{(-1+i)(1-i)}{2} = \frac{2i}{2} = i\)
\(\implies \cos\theta+i\sin\theta = i \implies \theta = \frac{\pi}{2}\)