ចំនួនកុំផ្លិច៖ ភាពខុសគ្នារវាងកំណែនានា

<math>cos\alpha = \frac{a}{r} ; sin\alpha = \frac{b}{r}\!</math>

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ក)​ <math>z_1z_2 = r_1r_2[cos(\alpha + \alpha) + isin(\alpha_1 + \alpha_2)]\!</math><br><br>

ខ) <math>\frac{z_1}{z_2} = \frac{r_1}{r_2}[cos(\alpha_1 - \alpha_2) + isin(\alpha_1 - \alpha_2)]\!</math><br>

<math>(i+i)^{50} = \sqrt{2}^{50}[cos(50 \cdot \frac{\pi}{4}) + isin(50 \cdot \frac{\pi}{4})] = 2^{25}(cos\frac{25\pi}{2} + isin\frac{25\pi}{2}) = 2^{25}[cos(12\pi+\frac{\pi}{2}) + isin(12\pi+\frac{\pi}{2})] = 2^{25}(cos\frac{\pi}{2} + isin\frac{\pi}{2}) \!</math><br><br>

ដូចនេះ <math>(1+i)^{50}= 2^{25}i = 33554432i\!</math>

គេបាន <math>W^n = s^n(cosn\alpha+isinn\alpha)\!</math><br><br>

ដោយ <math>W^n = Z\!</math> គេបាន <math> s^n(cosn\alpha+isinn\alpha) = r(cos\theta+isin\theta)\!</math><br><br>

ចំនួនកុំផ្លិចពីរស្មើគ្នា ម៉ូឌុលរបស់វាក៏ស្មើគ្នាដែរ។

 

 

<math>cosn\alpha + isinn\alpha = cos\theta + isin\theta\!</math><br><br>

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តាង Z = -1 + 0i គេបាន <math>r = \sqrt{1} = 1</math> ។<br><br>

<math>cos\theta = \frac{a}{r} = -1 </math> និង <math>sin\theta = \frac{b}{r} = 0</math> នាំអោយ <math>\theta = \pi</math>។ <br><br>

<math>Z = -1 + 0i = (cos\pi + isin\pi)\!</math><br><br>

<math>w_k = cos(\frac{\pi + 2k\pi}{6}) + isin(\frac{\pi + 2k\pi}{6})\!</math><br><br>

<math>w_k = cos(\frac{\pi}{6} + \frac{k\pi}{3}) + isin(\frac{\pi}{6} + \frac{k\pi}{3})\!</math> បើ k=0;1;2;3;4;5 គេបាន

 

 

== សូមមើលផងដែរ ==