أتدرب وأحل المسائل

أتدرب وأحل المسائل

العمليات على الأعداد المركبة

أجد ناتج كلّ ممّا يأتي، ثم أكتبه بالصورة القياسية:

1) (7 + 2i) + (3 – 11i)

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

(2) (5 – 9i) – (-4 + 7i)

$(5 - 9 i) - (- 4 + 7 i) = 9 - 16 i$

(3) (4 – 3i) (1 + 3i)

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

(4) (4 – 6i) (1 – 2i) (2 – 3i)

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

(5) (x2 + y2)2 = 50(x2 – y2)

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

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

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

أجد ناتج كلّ ممّا يأتي بالصورة المثلثية:

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

$\begin{aligned} 6 (c o s π + i s i n π) \times 2 (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})) \\ = 12 (c o s (π - \frac{π}{4}) + i s i n (π - \frac{π}{4})) = 12 (c o s \frac{3 π}{4} + i s i n \frac{3 π}{4}) \end{aligned}$

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

$\begin{aligned} (c o s (\frac{3 π}{10}) + i s i n (\frac{3 π}{10})) \div (c o s \frac{2 π}{5} + i s i n \frac{2 π}{5}) \\ = c o s (\frac{3 π}{10} - \frac{2 π}{5}) + i s i n (\frac{3 π}{10} - \frac{2 π}{5}) \\ = c o s (- \frac{π}{10}) + i s i n (- \frac{π}{10}) \end{aligned}$

(9) $12 (\cos \frac{π}{4} + i \sin \frac{π}{4}) \div 4 (\cos \frac{π}{3} + i \sin \frac{π}{3})$

$\begin{aligned} 12 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4})) \div 4 (c o s \frac{π}{3} + i s i n \frac{π}{3}) \\ = \frac{12}{4} (c o s (\frac{π}{4} - \frac{π}{3}) + i s i n (\frac{π}{4} - \frac{π}{3})) \\ = 3 (c o s (- \frac{π}{12}) + i s i n (- \frac{π}{12})) \end{aligned}$

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

$\begin{aligned} 11 (c o s (- \frac{π}{6}) + i s i n (- \frac{π}{6})) \times 2 (c o s (\frac{3 π}{2}) + i s i n (\frac{3 π}{2})) \\ = 22 (c o s (- \frac{π}{6} + \frac{3 π}{2}) + i s i n (- \frac{π}{6} + \frac{3 π}{2})) \\ = 22 (c o s (\frac{4 π}{3}) + i s i n (\frac{4 π}{3})) \\ = 22 (c o s (\frac{4 π}{3} - 2 π) + i s i n (\frac{4 π}{3} - 2 π)) \\ = 22 (c o s (- \frac{2 π}{3}) + i s i n (- \frac{2 π}{3})) \end{aligned}$

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

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

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

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

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

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

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

طريقة ثانية للحل:

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

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

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

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

طريقة ثانية للحل:

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

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

$\begin{aligned} z & = 8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) = 8 (c o s (\frac{- π}{4}) + i s i n (\frac{- π}{4})) \\ \to \bar{z} & = 8 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4})) \\ \to z \bar{z} & = 8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) \times 8 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4})) \\ = 64 (c o s^{2} (\frac{π}{4}) + s i n^{2} (\frac{π}{4})) = 64 \end{aligned}$

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

$\begin{aligned} z = 8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) = 8 (c o s (\frac{- π}{4}) + i s i n (\frac{- π}{4})) \\ \to \bar{z} = 8 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4})) \\ \to z \bar{z} = 64 (c o s (\frac{- π}{4} + \frac{π}{4}) + i s i n (\frac{- π}{4} + \frac{π}{4})) = 64 \end{aligned}$

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

$\begin{aligned} z = 8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) = 8 (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i) = 4 \sqrt{2} - 4 \sqrt{2} i \\ \to \bar{z} = 4 \sqrt{2} + 4 \sqrt{2} i \\ \to z \bar{z} = (4 \sqrt{2} - 4 \sqrt{2} i) (4 \sqrt{2} + 4 \sqrt{2} i) = 32 + 32 = 64 \end{aligned}$

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

(16) $\frac{z_{2}}{z_{1}}$

$\begin{aligned} z_{1} = 2 \sqrt{3} - 2 i, z_{2} = \sqrt{5} - i \sqrt{15}, z_{3} = 2 - 2 i \\ | z_{1} | = \sqrt{12 + 4} = 4 \\ | z_{2} | = \sqrt{5 + 15} = 2 \sqrt{5} \\ | z_{3} | = \sqrt{4 + 4} = 2 \sqrt{2} \\ A r g (z_{1}) = - t a n^{- 1} (\frac{2}{2 \sqrt{3}}) = - t a n^{- 1} (\frac{1}{\sqrt{3}}) = - \frac{π}{6} \\ A r g (z_{2}) = - t a n^{- 1} (\frac{\sqrt{15}}{\sqrt{5}}) = - t a n^{- 1} (\sqrt{3}) = - \frac{π}{3} \\ A r g (z_{3}) = - t a n^{- 1} (\frac{2}{2}) = - t a n^{- 1} (1) = - \frac{π}{4} \\ | \frac{z_{2}}{z_{1}} | = \frac{| z_{2} |}{| z_{1} |} = \frac{2 \sqrt{5}}{4} = \frac{\sqrt{5}}{2} \\ A r g (\frac{z_{2}}{z_{1}}) = A r g (z_{2}) - A r g (z_{1}) = - \frac{π}{3} - (- \frac{π}{6}) = - \frac{π}{6} \end{aligned}$

(17) $\frac{1}{z_{3}}$

$\begin{aligned} | \frac{1}{z_{3}} | = \frac{| 1 |}{| z_{3} |} = \frac{1}{2 \sqrt{2}} \\ A r g (\frac{1}{z_{3}}) = A r g (1) - A r g (z_{3}) = 0 - (- \frac{π}{4}) = \frac{π}{4} \end{aligned}$

(18) $\frac{z_{3}}{z_{2}}$

$\begin{aligned} \bar{z_{2}} = \sqrt{5} + i \sqrt{15} \to | \bar{z_{2}} | = | z_{2} | = 2 \sqrt{5}, A r g (\bar{z_{2}}) = - A r g (z_{2}) = \frac{π}{3} \\ | \frac{z_{3}}{\bar{z_{2}}} | = \frac{| z_{3} |}{| \bar{z_{2}} |} = \frac{2 \sqrt{2}}{2 \sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}} \\ A r g (\frac{z_{3}}{\bar{z_{2}}}) = A r g (z_{3}) - A r g (\bar{z_{2}}) = - \frac{π}{4} - \frac{π}{3} = - \frac{7 π}{12} \end{aligned}$

إعداد : شبكة منهاجي التعليمية

10 / 07 / 2023

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