أتدرب وأحل المسائل

العمليات على الأعداد المركبة

أجد ناتج كلّ ممّا يأتي، ثم أكتبه بالصورة القياسية:

1) (7 + 2i) + (3 – 11i)

$(7+2i)+(3-11i)=10-9i$

(2)  (5 – 9i) – (-4 + 7i)

 $(5-9i)-(-4+7i)=9-16i$

(3) (4 – 3i) (1 + 3i)

$(4-3i)(1+3i)=4+12i-3i+9=13+9i$

(4) (4 – 6i) (1 – 2i) (2 – 3i)

$\begin{array}{rl}(4-6i)(1-2i)(2-3i)& =(4-6i)(2-3i-4i-6)\\ & =(4-6i)(-4-7i)\\ & =-16-28i+24i-42\\ & =-58-4i\end{array}$

(5) (x2 + y2)2 = 50(x2 – y2)

$(9-2i{)}^2=81-36i-4=77-36i$

(6) $\frac{48+19i}{5-4i}$

$\begin{array}{rl}\frac{48+19i}{5-4i}& =\frac{48+19i}{5-4i}\times \frac{5+4i}{5+4i}\\ & =\frac{240+192i+95i-76}{25+16}\\ & =\frac{164+287i}{41}\\ & =4+7i\end{array}$

أجد ناتج كلّ ممّا يأتي بالصورة المثلثية:

(7) $6(\cos \pi +i\sin \pi )\times 2(\cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4}))$

$\begin{array}{rl}& 6(cos \pi +isin \pi )\times 2(cos (-\frac{\pi }{4})+isin (-\frac{\pi }{4}))\\ & =12(cos (\pi -\frac{\pi }{4})+isin (\pi -\frac{\pi }{4}))=12(cos\frac{ 3\pi }{4}+isin \frac{3\pi }{4})\end{array}$

(8) $(\cos \frac{3\pi }{10}+i\sin \frac{3\pi }{10})÷(\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5})$

 $\begin{array}{rl}& (cos \left(\frac{3\pi }{10}\right)+isin \left(\frac{3\pi }{10}\right))÷(cos \frac{2\pi }{5}+isin \frac{2\pi }{5})\\ & =cos (\frac{3\pi }{10}-\frac{2\pi }{5})+isin (\frac{3\pi }{10}-\frac{2\pi }{5})\\ & =cos (-\frac{\pi }{10})+isin (-\frac{\pi }{10})\end{array}$

(9) $12(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})÷4(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$

$\begin{array}{rl}& 12(cos \left(\frac{\pi }{4}\right)+isin \left(\frac{\pi }{4}\right))÷4(cos \frac{\pi }{3}+isin \frac{\pi }{3})\\ & =\frac{12}{4}(cos (\frac{\pi }{4}-\frac{\pi }{3})+isin (\frac{\pi }{4}-\frac{\pi }{3}))\\ & =3(cos (-\frac{\pi }{12})+isin (-\frac{\pi }{12}))\end{array}$

(10) $11(\cos (-\frac{\pi }{6})+i\sin (-\frac{\pi }{6}))\times 2(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2})$

$\begin{array}{rl}& 11(cos(-\frac{\pi }{6})+isin(-\frac{\pi }{6}))\times 2(cos\left(\frac{3\pi }{2}\right)+isin\left(\frac{3\pi }{2}\right))\\ & =22(cos(-\frac{\pi }{6}+\frac{3\pi }{2})+isin(-\frac{\pi }{6}+\frac{3\pi }{2}))\\ & =22(cos \left(\frac{4\pi }{3}\right)+isin \left(\frac{4\pi }{3}\right))\\ & =22(cos (\frac{4\pi }{3}-2\pi )+isin (\frac{4\pi }{3}-2\pi ))\\ & =22(cos (-\frac{2\pi }{3})+isin (-\frac{2\pi }{3}))\end{array}$

أجد القيم الحقيقية للثابتين a و b في كل ممّا يأتي:

(11) $(a+6i)+(7-ib)=-2+5i$

$\begin{array}{rl}(a+6i)+(7-bi)=-2+5i& \\ a+7+(6-b)i=-2+5i& \to a+7=-2,6-b=5\\ & \to a=-9,b=1\end{array}$

(12) $(11-ia)-(b-9i)=7-6i$

$\begin{array}{rl}& (11-ia)-(b-9i)=7-6i\\ & 11-b+(9-a)i=7-6i\to 11-b=7,9-a=-6\end{array}$

(13) $(a+ib)(2-i)=5+5i$

$\begin{array}{rl}(a+ib)(2-i)=5+5i& \\ 2a+b+(2b-a)i=5+5i& \to 2a+b=5,2b-a=5\\ & \to b=3,a=1\end{array}$

طريقة ثانية للحل:

$\begin{array}{rl}a+ib& =\frac{5+5i}{2-i}\\ & =\frac{5+5i}{2-i}\times \frac{2+i}{2+i}\\ & =\frac{10+5i+10i-5}{4+1}\\ & =\frac{5+15i}{5}\\ \to a& =1+3i\\ \to a& 1,b=3\end{array}$

(14) $\frac{a-6i}{1-2i}=b+4i$

$\begin{array}{rl}\frac{a-6i}{1-2i}=b+4i& \to \frac{a-6i}{1-2i}\times \frac{1+2i}{1+2i}=b+4i\\ & \to \frac{a+2ai-6i+12}{1+4}=b+4i\\ & \to \frac{a+12}{5}+\frac{2a-6}{5}i=b+4i\\ & \to \frac{a+12}{5}=b,\frac{2a-6}{5}=4\to a=13\end{array}$

بتعويض قيمة a في المعادلة الأولى ينتج أن: b = 5

طريقة ثانية للحل:

$\begin{array}{rl}a-6i=(b+4i)(1-2i)& \to a-6i=b+8+(-2b+4)i\\ & \to a=b+8,-6=-2b+4\\ & \to b=5,a=13\end{array}$

(15) أضرب العدد المركب $8(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4})$ في مرافقه.

$\begin{array}{rl}z& =8(cos \left(\frac{\pi }{4}\right)-isin \left(\frac{\pi }{4}\right))=8(cos \left(\frac{-\pi }{4}\right)+isin \left(\frac{-\pi }{4}\right))\\ \to \overline{z}& =8(cos \left(\frac{\pi }{4}\right)+isin \left(\frac{\pi }{4}\right))\\ \to z\overline{z}& =8(cos \left(\frac{\pi }{4}\right)-isin \left(\frac{\pi }{4}\right))\times 8(cos \left(\frac{\pi }{4}\right)+isin \left(\frac{\pi }{4}\right))\\ & =64(co{s}^2 \left(\frac{\pi }{4}\right)+si{n}^2 \left(\frac{\pi }{4}\right))=64\end{array}$

الحل الثاني: نكتب كلاً من العددين بالصورة المثلثية أولاً ثم نطبق القاعدة:

$\begin{array}{rl}& z=8(cos \left(\frac{\pi }{4}\right)-isin \left(\frac{\pi }{4}\right))=8(cos \left(\frac{-\pi }{4}\right)+isin \left(\frac{-\pi }{4}\right))\\ & \to \overline{z}=8(cos \left(\frac{\pi }{4}\right)+isin \left(\frac{\pi }{4}\right))\\ & \to z\overline{z}=64(cos (\frac{-\pi }{4}+\frac{\pi }{4})+isin (\frac{-\pi }{4}+\frac{\pi }{4}))=64\end{array}$

الحل الثالث: نكتب كلاً من العددين بالصورة القياسية أولاً ثم إجراء عملية الضرب:

$\begin{array}{rl}& z=8(cos \left(\frac{\pi }{4}\right)-isin \left(\frac{\pi }{4}\right))=8(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)=4\sqrt{2}-4\sqrt{2}i\\ & \to \overline{z}=4\sqrt{2}+4\sqrt{2}i\\ & \to z\overline{z}=(4\sqrt{2}-4\sqrt{2}i)(4\sqrt{2}+4\sqrt{2}i)=32+32=64\end{array}$

إذا كان: ${\mathit{z}}_{\mathbf{1}}\mathbf{=}\sqrt{\mathbf{12}}\mathbf{-}\mathbf{2}\mathit{i}\mathbf{,}{\mathit{z}}_{\mathbf{2}}\mathbf{=}\sqrt{\mathbf{5}}\mathbf{-}\mathit{i}\sqrt{\mathbf{15}}\mathbf{,}{\mathit{z}}_{\mathbf{3}}\mathbf{=}\mathbf{2}\mathbf{-}\mathbf{2}\mathit{i}$ ، فأجد المقياس والسعة الرئيسة لكل ممّا يأتي:

(16) $\frac{z_2}{z_1}$

$\begin{array}{rl}& z_1=2\sqrt{3}-2i,z_2=\sqrt{5}-i\sqrt{15},z_3=2-2i\\ & \left|z_1\right|=\sqrt{12+4}=4\\ & \left|z_2\right|=\sqrt{5+15}=2\sqrt{5}\\ & \left|z_3\right|=\sqrt{4+4}=2\sqrt{2}\\ & Arg\left(z_1\right)=-ta{n}^{-1}\left(\frac{2}{2\sqrt{3}}\right)=-ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)=-\frac{\pi }{6}\\ & Arg\left(z_2\right)=-ta{n}^{-1}\left(\frac{\sqrt{15}}{\sqrt{5}}\right)=-ta{n}^{-1}\left(\sqrt{3}\right)=-\frac{\pi }{3}\\ & Arg\left(z_3\right)=-ta{n}^{-1}\left(\frac{2}{2}\right)=-ta{n}^{-1}\left(1\right)=-\frac{\pi }{4}\\ & \left|\frac{z_2}{z_1}\right|=\frac{\left|z_2\right|}{\left|z_1\right|}=\frac{2\sqrt{5}}{4}=\frac{\sqrt{5}}{2}\\ & Arg\left(\frac{z_2}{z_1}\right)=Arg\left(z_2\right)-Arg\left(z_1\right)=-\frac{\pi }{3}-(-\frac{\pi }{6})=-\frac{\pi }{6}\end{array}$

(17) $\frac{1}{z_3}$

$\begin{array}{rl}& \left|\frac{1}{z_3}\right|=\frac{\left|1\right|}{\left|z_3\right|}=\frac{1}{2\sqrt{2}}\\ & Arg\left(\frac{1}{z_3}\right)=Arg\left(1\right)-Arg\left(z_3\right)=0-(-\frac{\pi }{4})=\frac{\pi }{4}\end{array}$

(18) $\frac{z_3}{z_2}$

$\begin{array}{rl}& \overline{z_2}=\sqrt{5}+i\sqrt{15}\to \left|\overline{z_2}\right|=\left|z_2\right|=2\sqrt{5},Arg\left(\overline{z_2}\right)=-Arg\left(z_2\right)=\frac{\pi }{3}\\ & \left|\frac{z_3}{\overline{z_2}}\right|=\frac{\left|z_3\right|}{\left|\overline{z_2}\right|}=\frac{2\sqrt{2}}{2\sqrt{5}}=\frac{\sqrt{2}}{\sqrt{5}}\\ & Arg\left(\frac{z_3}{\overline{z_2}}\right)=Arg\left(z_3\right)-Arg\left(\overline{z_2}\right)=-\frac{\pi }{4}-\frac{\pi }{3}=-\frac{7\pi }{12}\end{array}$