تمارين (4-1)

تمارين (4-1)

(1)- احسب ما يلي:

${[\cos \frac{5}{24}\pi +i\sin \frac{5}{24}\pi ]}^4$

${[\cos \frac{5}{24}\pi +i\sin \frac{5}{24}\pi ]}^4=[\cos 4\frac{5}{24}\pi +i\sin 4\frac{5}{24}\pi ]=[\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}]$

تقع في الربع الثاني $\frac{\pi }{6}=30 , 5\times 30=150$

$=-\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}=[-\frac{\sqrt{3}}{2}+\frac{1}{2}i]$

${[\cos 3\frac{7}{12}\pi +i\sin \frac{7}{12}\pi ]}^{-3}$

$[\cos (-3\frac{7}{12}\pi )+i\sin (-3\frac{7}{12}\pi )]=[\cos \frac{-7\pi }{4}+i\sin \frac{-7\pi }{4}]=[\cos \frac{7\pi }{4}-i\sin \frac{7\pi }{4}]$

تقع في الربع الرابع $7\times 45=315 , \frac{\pi }{4}=45$

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

(2)- احسب باستخدام مبرهنة ديموافر (أو التعميم) ما يأتي:

$(1-i{)}^7$

$\begin{array}{rl}& z=1-i\Rightarrow r=\sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt{2}\\ & \cos \theta =\frac{x}{r}=\frac{1}{\sqrt{2}} , \sin \theta =\frac{y}{r}=\frac{-1}{\sqrt{2}}\end{array}$

زاوية الإسناد $\frac{\pi }{4}$ تقع في الربع الرابع $\theta =2\pi -\frac{\pi }{4}=\frac{8\pi -\pi }{4}=\frac{7\pi }{4}$

الصيغة القطبية:

$Z=\sqrt{2}(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4})$

نبدأ بتطبيق مبرهنة ديموافر:

$\begin{array}{rl}& Z^7=\left(\sqrt{2}{)}^7\right(\cos \theta +i\sin \theta {)}^7=(\sqrt{2}{)}^7{(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4})}^7\\ & Z^7=(\sqrt{2}{)}^7(\cos 7\frac{7\pi }{4}+i\sin 7\frac{7\pi }{4})=(\sqrt{2}{)}^6\sqrt{2}(\cos \frac{49\pi }{4}+i\sin \frac{49\pi }{4})\end{array}$

$\frac{49}{4}=12$ زوجي دائماً والباقي 1 لذلك تكون الزاوية الجديدة باقي $\frac{\pi }{4}$ وهي $\frac{\pi }{4}$ تقع في الربع الرابع.

$\begin{array}{rl}& =\left(2^3\right)\sqrt{2}(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})\\ & =8\sqrt{2}(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})=\frac{8\sqrt{2}}{\sqrt{2}}+i\frac{8\sqrt{2}}{\sqrt{2}}=8+8i\end{array}$

$(\sqrt{3}+i{)}^{-9}$

$\begin{array}{rl}& Z=\sqrt{3}+i\Rightarrow r=\sqrt{x^2+y^2}=\sqrt{3+1}=\sqrt{4}=2\\ & \cos \theta =\frac{x}{r}=\frac{\sqrt{3}}{2} , \sin \theta =\frac{y}{r}=\frac{1}{2}\end{array}$

زاوية الإسناد $\theta =\frac{\pi }{6}$ تقع في الربع الأول.

الصيغة القطبية:

$Z=2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$

نبدأ بتطبيق مبرهنة ديموافر:

$$\begin{gathered} \begin{array}{rl}& Z^{-9}=(\sqrt{3}+1{)}^{-9}=r^{-9}(\cos \theta +i\sin \theta {)}^{-9}=(2{)}^{-9}{(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^{-9}\\ & Z^{-9}=\frac{1}{(2{)}^9}{(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^{-9}=\frac{1}{512}(\cos \left(\frac{-9\pi }{6}\right)+i\sin \left(\frac{-9\pi }{6}\right))\end{array} \\ \begin{array}{rl}& Z^{-9}=\frac{1}{512}(\cos \left(\frac{-3\pi }{2}\right)+i\sin \left(\frac{-3\pi }{2}\right))=\frac{1}{512}(\cos \frac{3\pi }{2}-i\sin \frac{3\pi }{2})\\ & Z^{-9}=\frac{1}{512}(0-i(-1\left)\right)=\frac{i}{512}\end{array} \end{gathered}$$

(3)- بسط ما يأتي:

$\frac{[\cos 2\theta +i\sin 2\theta {]}^5}{[\cos 3\theta +i\sin 3\theta {]}^3}$

$\frac{[\cos 2\theta +i\sin 2\theta {]}^5}{[\cos 3\theta +i\sin 3\theta {]}^3}=\frac{{\left[\right(\cos \theta +i\sin \theta {)}^2]}^5}{{\left[\right(\cos \theta +i\sin \theta {)}^3]}^3}=\frac{(\cos \theta +i\sin \theta {)}^{10}}{(\cos \theta +i\sin \theta {)}^9}=\cos \theta +i\sin \theta$

$(\cos \theta +i\sin \theta {)}^8(\cos \theta -i\sin \theta {)}^4$

$\begin{array}{rl}& (\cos \theta +i\sin \theta {)}^8(\cos \theta -i\sin \theta {)}^4\\ & =(\cos \theta +i\sin \theta {)}^8(\cos (-\theta )+i\sin (-\theta ){)}^4\\ & =(\cos \theta +i\sin \theta {)}^8(\cos \theta +i\sin \theta {)}^{-4}=(\cos \theta +i\sin \theta {)}^4\\ & =\cos 4\theta +i\sin 4\theta \end{array}$

(4)- جد الجذور التربيعية للعدد المركب $-1+\sqrt{3}i$ باستخدام نتيجة مبرهنة ديموافر.

$\begin{array}{rl}& Z=-1+\sqrt{3}i\Rightarrow r=\sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=\sqrt{1+3}=\sqrt{4}=2\\ & \cos \theta =\frac{x}{r}=\frac{-1}{2} , \sin \theta =\frac{y}{r}=\frac{\sqrt{3}}{2}\end{array}$

زاوية الإسناد $\theta =\frac{\pi }{3}$ تقع في الربع الثاني $\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

$$\begin{gathered} \begin{array}{rl}& \sqrt{Z}=\sqrt{-1+\sqrt{3i}}\Rightarrow (-1+\sqrt{3i}{)}^{\frac{1}{2}}=(r{)}^{\frac{1}{2}}(\cos \theta +i\sin \theta {)}^{\frac{1}{2}}\\ & Z^{\frac{1}{2}}=2^{\frac{1}{2}}[\cos \frac{\frac{2\pi }{3}+2k\pi }{2}+i\sin \frac{\frac{2\pi }{3}+2k\pi }{2}]\\ & Z^{\frac{1}{2}}=\sqrt{2}(\cos \frac{\frac{2\pi +6k\pi }{3}}{2}+i\sin \frac{\frac{2\pi +6k\pi }{3}}{2})=\sqrt{2}(\cos \frac{2\pi +6k\pi }{6}+i\sin \frac{2\pi +6k\pi }{6})\end{array} \\ \begin{array}{ll}{Z}_1=\sqrt{2}(\cos \frac{2\pi }{6}+i\sin \frac{2\pi }{6})=\sqrt{2}(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})& \left(k=0\right)\\ Z_1=\sqrt{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2})=\frac{1}{\sqrt{2}}+i\frac{\sqrt{3}}{\sqrt{2}}& \\ Z_2=\sqrt{2}(\cos \frac{8\pi }{6}+i\sin \frac{8\pi }{6})=\sqrt{2}(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3})& \left(k=1\right)\end{array} \end{gathered}$$

زاوية الإسناد $\frac{4\pi }{3}$ تقع في الربع الثالث $\theta =\pi -\frac{4\pi }{3}=\frac{4\pi -3\pi }{3}=\frac{\pi }{3}$

$Z_2=\sqrt{2}(-\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})=\sqrt{2}(-\frac{1}{2}-i\frac{\sqrt{3}}{2})=\frac{-1}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i$

(5)- باستخدام نتيجة مبرهنة ديموافر جد الجذور التكعيبية للعدد $27i$

$\begin{array}{rl}& Z=27i\Rightarrow r=\sqrt{x^2+y^2}=\sqrt{(27{)}^2}=27\\ & \cos \theta =\frac{x}{r}=\frac{0}{27}=0 , \sin \theta =\frac{y}{r}=\frac{27}{27}=1\end{array}$

زاوية الإسناد $\theta =\frac{\pi }{2}$ تقع في الربع الأول.

$\begin{array}{rl}& \sqrt[3]{Z}=\sqrt[3]{27i}\Rightarrow Z^{\frac{1}{3}}=r^{\frac{1}{3}}(\cos \theta +i\sin \theta {)}^{\frac{1}{3}}=(27{)}^{\frac{1}{3}}{(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})}^{\frac{1}{3}}\\ & Z^{\frac{1}{3}}=\sqrt[3]{27}(\cos \left(\frac{\frac{\pi }{2}+2k\pi }{3}\right)+i\sin \left(\frac{\frac{\pi }{2}+2k\pi }{3}\right))k=0,1,2\\ & Z^{\frac{1}{3}}=3(\cos \left(\frac{\pi +4k\pi }{6}\right)+i\sin \left(\frac{\pi +4k\pi }{6}\right))\\ & k=0\Rightarrow Z_1=3(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right))=3[\frac{\sqrt{3}}{2}+\frac{1}{2}i]=[\frac{3\sqrt{3}}{2}+\frac{3}{2}i]\\ & k=1\Rightarrow Z_2=3\left(\cos \left(\frac{5\pi }{6}\right)+i\sin \left(\frac{5\pi }{6}\right)\right)\end{array}$

زاوية الإسناد $\frac{5\pi }{6}$ تقع في الربع الثاني $\theta =\pi -\frac{5\pi }{6}=\frac{\pi }{6}$

$$\begin{gathered} Z_2=3(-\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right))=3[\frac{-\sqrt{3}}{2}+\frac{1}{2}i]=[\frac{-3\sqrt{3}}{2}+\frac{3}{2}i] \\ k=2\Rightarrow Z_3=3(\cos \left(\frac{9\pi }{6}\right)+i\sin \left(\frac{9\pi }{6}\right))=3(\cos \left(\frac{3\pi }{2}\right)+i\sin \left(\frac{3\pi }{2}\right)) \\ {Z}_3=3(0+(-1\left)i\right)=0-3i=-3i \end{gathered}$$

(6)- جد الجذور الأربع للعدد $(-16)$ باستخدام نتيجة مبرهنة ديموافر.

الصيغة الوزارية:

جد حل المعادلة حيث $x\in \mathrm{ℂ}$ وباستخدام نتيجة مبرهنة ديموافر $x^4+16=0$

$\begin{array}{rl}& Z=-16\Rightarrow r=\sqrt{x^2+y^2}=\sqrt{(16{)}^2}=16\\ & \cos \theta =\frac{x}{r}=\frac{-16}{16}=-1 , \sin \theta =\frac{y}{r}=\frac{0}{16}=0\end{array}$

زاوية الإسناد $\theta =\pi$

$$\begin{gathered} \begin{array}{rl}\sqrt[4]{Z}& =\sqrt[4]{-16}\Rightarrow Z^{\frac{1}{4}}=r^{\frac{1}{4}}(\cos \theta +i\sin \theta {)}^{\frac{1}{4}}=(16{)}^{\frac{1}{4}}(\cos \pi +i\sin \pi {)}^{\frac{1}{4}}\\ \sqrt[n]{Z}& =r^{\frac{1}{n}}[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)]\\ Z^{\frac{1}{4}}& =\sqrt[4]{16}(\cos \left(\frac{\pi +2k\pi }{n}\right)+i\sin \left(\frac{\pi +2k\pi }{n}\right))n=4 , k=0,1,2,3\end{array} \\ \begin{array}{rl}k=0\Rightarrow Z_1& =2(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})=2[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i]=\sqrt{2}+\sqrt{2}i\\ k=1\Rightarrow Z_2& =2(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})=2(-\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})\\ Z_2& =2(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=-\sqrt{2}+\sqrt{2}i\\ k=2\Rightarrow Z_3& =2(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4})=2(-\cos \frac{\pi }{4}-i\sin \frac{5\pi }{4})\\ Z_3& =2(\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)=-\sqrt{2}-\sqrt{2}i\\ k=3\Rightarrow & Z_4=2(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4})=2(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4})\\ Z_4& =2(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)=\sqrt{2}-\sqrt{2}i\end{array} \end{gathered}$$

(7)- جد الجذور الستة للعدد $(-64i)$ باستخدام نتيجة مبرهنة ديموافر.

$\begin{array}{rl}& Z=-64i\Rightarrow r=\sqrt{x^2+y^2}=\sqrt{(-64{)}^2}=\sqrt{(64{)}^2}=64\\ & \cos \theta =\frac{x}{r}=\frac{0}{64}=0 , \sin \theta =\frac{y}{r}=\frac{-64}{64}=-1\end{array}$

زاوية الإسناد $\theta =\frac{3\pi }{2}$

$$\begin{gathered} \begin{array}{rl}& \sqrt[6]{Z}=\sqrt[6]{-64i}\Rightarrow Z^{\frac{1}{6}}=r^{\frac{1}{6}}(\cos \theta +i\sin \theta {)}^{\frac{1}{6}}=(64{)}^{\frac{1}{6}}{(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2})}^{\frac{1}{6}}\\ & \sqrt[n]{Z}=r^{\frac{1}{n}}[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)]\\ & Z^{\frac{1}{6}}=\sqrt[6]{64}=[\cos \left(\frac{\frac{3\pi }{2}+2k\pi }{6}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2k\pi }{6}\right)]\\ & Z^{\frac{1}{6}}=2[\cos \left(\frac{3\pi +4k\pi }{12}\right)+i\sin \left(\frac{3\pi +4k\pi }{12}\right)] , k=0,1,2,\dots 5\\ & k=0\Rightarrow Z_1=2(\cos \frac{3\pi }{12}+i\sin \frac{3\pi }{12})=2(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})=2(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=\sqrt{2}+\sqrt{2}i\end{array} \\ \begin{array}{rl}& k=1\Rightarrow Z_2=2(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12})\\ & k=2\Rightarrow Z_3=2(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12})\\ & k=3\Rightarrow Z_4=2(\cos \frac{15\pi }{12}+i\sin \frac{15\pi }{12})=2(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4})=2(-\cos \frac{\pi }{4}-i\sin \frac{\pi }{4})\\ & k=4\Rightarrow Z_4=2(\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)=-\sqrt{2}-\sqrt{2}\mathbf{1}\\ & k=5\Rightarrow Z_6=2(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12})\end{array} \end{gathered}$$