رياضيات فصل أول

التوجيهي علمي

الشرح الملخص أوراق العمل حل اسئلة الدرس النتاجات الملفات

أجد ناتج كلاً مما يأتي:

$a) (7 + 8 i) + (- 9 + 14 i) S o l u t i o n : (7 + 8 i) + (- 9 + 14 i) = 7 + - 9 + (8 + 14) i = - 2 + 22 i$

$b) (11 + 9 i) - (4 - 6 i) S o l u t i o n : = 11 - 4 + (9 - - 6) i = 7 + 14 i$

أجد ناتج كل مما يأتي ثم أكتبه بالصورة القياسية:

$a) - 3 i (4 - 5 i) S o l u t i o n : - 3 i (4 - 5 i) = - 3 i \times 4 - 3 i \times - 5 i = - 12 i + 15 i^{2} i^{2} = - 1 = - 15 - 12 i$

$b) (5 + 4 i) (7 - 4 i) S o l u t i o n : = (5) (7 - 4 i) + (4 i) (7 - 4 i) = (5 \times 7) - (5 \times 4 i) + (4 i \times 7) - (4 i \times 4 i) = 35 - 20 i + 28 i - 16 i^{2} i^{2} = - 1 = 35 + 8 i + 16 = 51 + 8 i$

$c) {(3 + 6 i)}^{2} S o l u t i o n : = (3 + 6 i) \times (3 + 6 i) = (3 \times 3) + (3 \times 6 i) + (6 i \times 3) + (6 i \times 6 i) = 9 + 18 i + 18 i + 36 i^{2} i^{2} = - 1 = 9 + 36 i - 36 = - 27 + 36 i O r {(3 + 6 i)}^{2} = {(3)}^{2} + 2 (3 \times 6 i) + {(6 i)}^{2} = 9 + 36 i - 36 = - 27 + 36 i$

أجد ناتج كل مما يأتي ثم أكتبه بالصورة القياسية:

$a) \frac{- 4 + 3 i}{1 + i} S o l u t i o n : = \frac{- 4 + 3 i}{1 + i} \times \frac{1 - i}{1 - i} (a - b i) (a + b i) = a^{2} + b^{2} = \frac{(- 4 + 3 i) (1 - i)}{1 + 1} = \frac{(- 4 \times 1) + (- 4 \times - i) + (3 i \times 1) + (3 i \times - i)}{1 - - 1} = \frac{- 4 + 4 i + 3 i - 3 i^{2}}{2} i^{2} = - 1 = \frac{- 4 + 7 i + 3}{2} = \frac{- 1 + 7 i}{2} = - \frac{1}{2} + \frac{7}{2} i$

$b) \frac{2 - 6 i}{- 3 i} S o l u t i o n : = \frac{2 - 6 i}{- 3 i} \times \frac{+ 3 i}{+ 3 i} (a - b i) (a + b i) = a^{2} + b^{2} = \frac{(2 \times 3 i) - (6 i \times 3 i)}{9} = \frac{6 i - 18 i^{2}}{9} = = \frac{6 i + 18}{9} i^{2} = - 1 = 2 + \frac{2}{3} i O r \frac{2 - 6 i}{- 3 i} = - \frac{2}{3 i} + \frac{- 6 i}{- 3 i} = - \frac{2}{3} {(i)}^{- 1} + 2 {(i)}^{- 1} = - i = 2 + \frac{2}{3} i$

$c) \frac{7 i}{4 - 4 i} S o l u t i o n : = \frac{7 i}{4 - 4 i} \times \frac{4 + 4 i}{4 + 4 i} (a - b i) (a + b i) = a^{2} + b^{2} = \frac{28 i + 28 i^{2}}{16 + 16} i^{2} = - 1 = \frac{- 28 + 28 i}{32} = - \frac{7}{8} + \frac{7}{8} i$

أجد ناتج مما يأتي بالصورة المثلثية:

$a) 6 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})) \times 2 (c o s (\frac{π}{6}) + i s i n (\frac{π}{6})) S o l u t i o n : z_{1} z_{2} = r_{1} r_{2} (c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})) = 6 \times 2 (c o s (\frac{π}{3} + \frac{π}{6}) + i s i n (\frac{π}{3} + \frac{π}{6})) = 12 (c o s (\frac{π}{2}) + i s i n (\frac{π}{2})) = 12 (c o s (\frac{π}{2}) + i s i n (\frac{π}{2}))$

$b) 6 (c o s (- \frac{π}{3}) + i s i n (- \frac{π}{3})) \div 2 (c o s (\frac{5 π}{6}) + i s i n (\frac{5 π}{6})) S o l u t i o n : \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (c o s (θ_{1} - θ_{2}) + i s i n (θ_{1} - θ_{2})) = \frac{6}{2} (c o s (- \frac{π}{3} - \frac{5 π}{6}) + i s i n (- \frac{π}{3} - \frac{5 π}{6})) = 3 (c o s (- \frac{7 π}{6}) + i s i n (- \frac{7 π}{6})) = 3 (c o s (\frac{7 π}{6}) - i s i n (\frac{7 π}{6})) c o s (- θ) = c o s (θ) a n d \sin (- θ) = - \sin (θ)$

أجد الجذرين التربيعيين لكل من الأعداد المركَبة الآتية:

$a) - 5 - 12 i S o l u t i o n : z = - 5 - 12 i l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} - 5 - 12 i = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ - 5 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ - 12 & = & 2 x y \dots 2 \end{array} F r o m 1 a n d 2 5 = y^{2} - {(- \frac{6}{y})}^{2} 5 = y^{2} - \frac{36}{y^{2}} 5 y^{2} = y^{4} - 36 y^{4} - 5 y^{2} - 36 = 0 (y^{2} - 9) (y^{2} + 4) = 0 y^{2} - 9 = 0 \Rightarrow y = 3 \Rightarrow x = - \frac{6}{3} = - 2 z = - 2 + 3 i \Rightarrow y = - 3 \Rightarrow x = - \frac{6}{- 3} = 2 z = 2 - 3 i$

$b) - 9 i S o l u t i o n : z = - 9 i l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} - 9 i = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ 0 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ - 9 & = & 2 x y \dots 2 \end{array} x = - \frac{9}{2 y} F r o m 1 a n d 2 y^{2} = {(- \frac{9}{2 y})}^{2} y^{2} = \frac{81}{4 y^{2}} 4 y^{4} = 81 4 y^{4} - 81 = 0 (2 y^{2} - 9) (2 y^{2} + 9) = 0 2 y^{2} - 9 = 0 \Rightarrow y = \frac{3}{\sqrt{2}} \Rightarrow x = - \frac{9}{2 \times \frac{3}{\sqrt{2}}} = - \frac{3}{\sqrt{2}} z = - \frac{3}{\sqrt{2}} + \frac{3}{\sqrt{2}} i \Rightarrow y = - \frac{3}{\sqrt{2}} \Rightarrow x = \frac{9}{\sqrt{2} \times 3} = \frac{3}{\sqrt{2}} z = \frac{3}{\sqrt{2}} - \frac{3}{\sqrt{2}} i$

$c) - \frac{1}{2} + i \frac{\sqrt{3}}{2} S o l u t i o n : z = - \frac{1}{2} + i \frac{\sqrt{3}}{2} l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} - \frac{1}{2} + i \frac{\sqrt{3}}{2} = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ - \frac{1}{2} & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ \frac{\sqrt{3}}{2} & = & 2 x y \dots 2 \end{array} x = \frac{\sqrt{3}}{4 y} F r o m 1 a n d 2 \frac{1}{2} = y^{2} - {(\frac{\sqrt{3}}{4 y})}^{2} \frac{1}{2} = y^{2} - \frac{3}{16 y^{2}} 8 y^{2} = 16 y^{4} - 3 16 y^{4} - 8 y^{2} - 3 = 0 (4 y^{2} + 1) (4 y^{2} - 3) = 0 4 y^{2} - 3 = 0 \Rightarrow y = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{\sqrt{3}}{4 \times \frac{\sqrt{3}}{2}} = \frac{1}{2} z = \frac{1}{2} + \frac{\sqrt{3}}{2} i \Rightarrow y = - \frac{\sqrt{3}}{2} \Rightarrow x = \frac{\sqrt{3}}{4 \times - \frac{\sqrt{3}}{2}} = - \frac{1}{2} z = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

أجد جميع الجذور الحقيقية والجذور المُركَبة للمعادلة: $z^{3} - z^{2} - 7 z + 15 = 0$

الحل :

بأستخدام عوامل الحد المطلق(الثابت) : $\pm 1, \pm 3, \pm 5, \pm 15$

بتجريب العدد $- 3$

${(- 3)}^{3} - {(- 3)}^{2} - 7 (- 3) + 15 = 0 - 27 - 9 + 21 + 15 = 0 - 36 + 36 = 0$

بتنفيذ القسمة التركيبية أو المطولة على العامل $(z + 3)$ :

$z^{3} - z^{2} - 7 z + 15 = (z^{2} - 4 z + 5) (z + 3) = 0 \Rightarrow z^{2} - 4 z + 5 = 0 z = \frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \times 1 \times 5}}{2} z = \frac{4 \pm \sqrt{16 - 20}}{2} z = \frac{4 \pm \sqrt{- 4}}{2} = \frac{4 \pm 2 i}{2} z = 2 \pm i z + 3 = 0 z = - 3 z = {- 3, 2 + i, 2 - i}$

إذا كان: $2 - i$ هو أحد جذور المعادلة: $x^{2} + a x + b = 0$ فأجد قيمة كل من b و a.

الحل :

بما أن جذري المعادلة التربيعية مترافقان فأنّ الجذر الثاني هو $2 + i$ :

وأن ناتج تعويض كل منهما في المعادلة يساوي صفر :

$x^{2} + a x + b = 0 {(2 - i)}^{2} + a (2 - i) + b = 0 4 - 4 i + i^{2} + 2 a - a i + b = 0 : i^{2} = - 1 3 - (4 + a) i + 2 a + b = 0 \dots 1 {(2 + i)}^{2} + a (2 + i) + b = 0 4 + 4 i + i^{2} + 2 a + a i + b = 0 3 + (4 + a) i + 2 a + b = 0 \dots 2 2 - 1 0 + 2 (4 + a) i = 0 0 = 4 i + a i a = - 4 \frac{1}{i} = - i \Rightarrow a = - 4 \Rightarrow 3 + (4 + a) i + 2 a + b = 0 \Rightarrow 3 + (4 + - 4) i + 2 (- 4) + b = 0 \Rightarrow 3 - 8 + b = 0 \Rightarrow - 5 + b = 0 \Rightarrow b = 5 O r x = 2 \pm i x - 2 = \pm i {(x - 2)}^{2} = - 1 x^{2} - 4 x + 4 = - 1 x^{2} - 4 x + 5 = 0 \Rightarrow a = - 4, b = 5$

أجد ناتج كل مما يأتي ثم أكتبه بالصورة القياسية:

$1) (7 + 2 i) + (3 - 11 i) S o l u t i o n : = 7 + 3 + (2 - 11) i = 10 - 9 i$

$2) (5 - 9 i) - (- 4 + 7 i) S o l u t i o n : = 5 - - 4 + (- 9 - 7) i = 9 - 16 i$

$3) (4 - 3 i) \times (1 + 3 i) S o l u t i o n : = 4 (1 + 3 i) - 3 i (1 + 3 i) = 4 + 12 i - 3 i + 9 = 13 + 9 i$

$4) (4 - 6 i) \times (1 - 2 i) \times (2 - 3 i) S o l u t i o n : = (4 (1 - 2 i) - 6 i (1 - 2 i)) \times (2 - 3 i) = (4 - 8 i - 6 i - 12) \times (2 - 3 i) = (- 8 - 14 i) \times (2 - 3 i) = - 8 \times (2 - 3 i) - 14 i \times (2 - 3 i) = - 16 + 24 i - 28 i - 42 = - 58 - 4 i$

$5) {(9 - 2 i)}^{2} S o l u t i o n : = 81 - 36 i + 4 i^{2} i^{2} = - 1 = 81 - 36 i - 4 = 77 - 36 i$

$6) \frac{10}{3 - i} S o l u t i o n : = \frac{10}{3 - i} \times \frac{3 + i}{3 + i} = \frac{10 (3 + i)}{9 + 1} = 3 + i$

أجد ناتج كل ممّا يأتي بالصورة المثلثية:

$7) 6 (c o s (π) + i s i n (π)) \times 2 (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})) S o l u t i o n : z_{1} z_{2} = r_{1} r_{2} (c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})) = 6 \times 2 (c o s (π - \frac{π}{4}) + i s i n (π - \frac{π}{4})) = 12 (c o s (\frac{3 π}{4}) + i s i n (\frac{3 π}{4}))$

$8) (c o s (\frac{3 π}{10}) + i s i n (\frac{3 π}{10})) \div (c o s (\frac{2 π}{5}) + i s i n (\frac{2 π}{5})) S o l u t i o n : \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (c o s (θ_{1} - θ_{2}) + i s i n (θ_{1} - θ_{2})) = (c o s (\frac{3 π}{10} - \frac{2 π}{5}) + i s i n (\frac{3 π}{10} - \frac{2 π}{5})) = (c o s (- \frac{π}{10}) + i s i n (- \frac{π}{10})) = (c o s (\frac{π}{10}) - i s i n (\frac{π}{10})) c o s (- θ) = c o s (θ) a n d s i n (- θ) = - s i n (θ)$

$9) 12 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4})) \div 4 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})) S o l u t i o n : \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (c o s (θ_{1} - θ_{2}) + i s i n (θ_{1} - θ_{2})) = \frac{12}{4} (c o s (\frac{π}{4} - \frac{π}{3}) + i s i n (\frac{π}{4} - \frac{π}{3})) = 3 (c o s (- \frac{π}{12}) + i s i n (- \frac{π}{12})) = 3 (c o s (\frac{π}{12}) - i s i n (\frac{π}{12})) c o s (- θ) = c o s (θ) a n d s i n (- θ) = - s i n (θ)$

$10) 11 (c o s (- \frac{π}{6}) + i s i n (- \frac{π}{6})) \times 2 (c o s (\frac{3 π}{2}) + i s i n (\frac{3 π}{2})) S o l u t i o n : z_{1} z_{2} = r_{1} r_{2} (c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})) = 11 \times 2 (c o s (- \frac{π}{6} + \frac{3 π}{2}) + i s i n (- \frac{π}{6} + \frac{3 π}{2})) = 22 (c o s (\frac{4 π}{3}) + i s i n (\frac{4 π}{3}))$

أجد القِيّم الحقيقية للثابتين b و a في كلٍّ ممّا يأتي:

$11) (a + 6 i) + (7 - b i) = - 2 + 5 i S o l u t i o n : a + 7 + (6 - b) i = - 2 + 5 i R e = R e a + 7 = - 2 \Rightarrow a = - 9 I m = I m 6 - b = 5 \Rightarrow b = 1$

$12) (11 - i a) - (b - 9 i) = 7 - 6 i S o l u t i o n : 11 - b + (- a + 9) i = 7 - 6 i R e = R e 11 - b = 7 \Rightarrow b = 4 I m = I m - a + 9 = - 6 \Rightarrow a = 15$

$13) (a + i b) (2 - i) = 5 + 5 i S o l u t i o n : a (2 - i) + i b (2 - i) = 5 + 5 i 2 a - a i + 2 b i + b = 5 + 5 i 2 a + b + (- a + 2 b) i = 5 + 5 i R e = R e 2 a + b = 5 \dots 1 I m = I m - a + 2 b = 5 \dots 2 \Rightarrow 5 b = 15 \Rightarrow b = 3 \Rightarrow 2 a + 3 = 5 \Rightarrow a = 1$

$14) \frac{a - 6 i}{1 - 2 i} = b + 4 i S o l u t i o n : 1 (b + 4 i) - 2 i (b + 4 i) = a - 6 i b + 4 i - 2 b i + 8 = a - 6 i b + 8 + (4 - 2 b) i = a - 6 i R e = R e b + 8 = a \dots 1 I m = I m 4 - 2 b = - 6 \dots 2 \Rightarrow 2 - b = - 3 \Rightarrow b = 5 \Rightarrow 5 + 8 = a \Rightarrow a = 13$

15) أضرب العدد المركّب $8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4}))$ في مُرافقه.

$8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) S o l u t i o n : 8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) = 8 (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{8}{\sqrt{2}} - \frac{8}{\sqrt{2}} i (\frac{8}{\sqrt{2}} - \frac{8}{\sqrt{2}} i) \times (\frac{8}{\sqrt{2}} + \frac{8}{\sqrt{2}} i) = (\frac{64}{2} + \frac{64}{2}) = 64 O r 8 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) \times 8 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4})) = = 64 (c o s^{2} (\frac{π}{4}) + s i n^{2} (\frac{π}{4})) = 64 \times 1 = 64$

أجد الجذرين التربيعيين لكلٌّ من الأعداد المركّبة الآتية:

$16) 3 - 4 i S o l u t i o n : z = 3 - 4 i l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} 3 - 4 i = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ 3 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ - 4 & = & 2 x y \dots 2 \end{array} x = - \frac{2}{y} F r o m 1 a n d 2 - 3 = y^{2} - {(- \frac{2}{y})}^{2} - 3 = y^{2} - \frac{4}{y^{2}} - 3 y^{2} = y^{4} - 4 y^{4} + 3 y^{2} - 4 = 0 (y^{2} - 1) (y^{2} + 4) = 0 y^{2} - 1 = 0 \Rightarrow y = 1 \Rightarrow x = \frac{- 2}{1} = - 2 z = - 2 + i \Rightarrow y = - 1 \Rightarrow x = \frac{- 2}{- 1} = 2 z = 2 - i$

$17) - 15 + 8 i S o l u t i o n : l e t z = - 15 + 8 i \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} - 15 + 8 i = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ - 15 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ 8 & = & 2 x y \dots 2 \end{array} x = \frac{4}{y} F r o m 1 a n d 2 15 = y^{2} - {(\frac{4}{y})}^{2} 15 = y^{2} - \frac{16}{y^{2}} 15 y^{2} = y^{4} - 16 y^{4} - 15 y^{2} - 16 = 0 (y^{2} + 1) (y^{2} - 16) = 0 y^{2} - 16 = 0 \Rightarrow y = 4 \Rightarrow x = \frac{4}{4} = 1 z = 1 + 4 i \Rightarrow y = - 4 \Rightarrow x = \frac{4}{- 4} = - 1 z = - 1 - 4 i$

$18) 5 - 12 i S o l u t i o n : z = 5 - 12 i l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} 5 - 12 i = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ 5 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ - 12 & = & 2 x y \dots 2 \end{array} x = - \frac{6}{y} F r o m 1 a n d 2 - 5 = y^{2} - {(- \frac{6}{y})}^{2} - 5 = y^{2} - \frac{36}{y^{2}} - 5 y^{2} = y^{4} - 36 y^{4} + 5 y^{2} - 36 = 0 (y^{2} - 4) (y^{2} + 9) = 0 y^{2} - 4 = 0 \Rightarrow y = 2 \Rightarrow x = - \frac{6}{2} = - 3 z = - 3 + 2 i \Rightarrow y = - 2 \Rightarrow x = - \frac{6}{- 2} = 3 z = 3 - 2 i$

$19) - 7 - 24 i S o l u t i o n : z = - 7 - 24 i l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} - 7 - 24 i = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ - 7 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ - 24 & = & 2 x y \dots 2 \end{array} x = - \frac{12}{y} F r o m 1 a n d 2 7 = y^{2} - {(- \frac{12}{y})}^{2} 7 = y^{2} - \frac{144}{y^{2}} 7 y^{2} = y^{4} - 144 y^{4} - 7 y^{2} - 144 = 0 (y^{2} + 9) (y^{2} - 16) = 0 y^{2} - 16 = 0 \Rightarrow y = 4 \Rightarrow x = - \frac{12}{4} = - 3 z = - 3 + 4 i \Rightarrow y = - 4 \Rightarrow x = - \frac{12}{- 4} = 3 z = 3 - 4 i$

إذا كان: $z = 2 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})),$ ، فأجد كلا مما يأتي بالصورة المثلثية:

$20) z ω S o l u t i o n : z = 2 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})), z = 2 (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})), z_{1} z_{2} = r_{1} r_{2} (c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})) = 2 \times 2 (c o s (- \frac{π}{4} + \frac{π}{3}) + i s i n (- \frac{π}{4} + \frac{π}{3})) = 4 (c o s (\frac{π}{12}) + i s i n (\frac{π}{12}))$

$21) \frac{z}{ω} S o l u t i o n : z = 2 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})), z = 2 (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})), \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (c o s (θ_{1} - θ_{2}) + i s i n (θ_{1} - θ_{2})) = \frac{2}{2} (c o s (- \frac{π}{4} - \frac{π}{3}) + i s i n (- \frac{π}{4} - \frac{π}{3})) = 1 (c o s (- \frac{7 π}{12}) + i s i n (- \frac{7 π}{12})) = (c o s (\frac{7 π}{12}) - i s i n (\frac{7 π}{12}))$

$22) \frac{ω}{z} S o l u t i o n : z = 2 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})), z = 2 (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})), ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})), \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (c o s (θ_{1} - θ_{2}) + i s i n (θ_{1} - θ_{2})) = \frac{2}{2} (c o s (\frac{π}{3} - - \frac{π}{4}) + i s i n (\frac{π}{3} - - \frac{π}{4})) = (c o s (\frac{7 π}{12}) + i s i n (\frac{7 π}{12}))$

$23) \frac{1}{z} S o l u t i o n : z = 2 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})), 1 = 1 (c o s (0) + i s i n (0)), \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} (c o s (θ_{1} - θ_{2}) + i s i n (θ_{1} - θ_{2})) = \frac{1}{2} (c o s (0 - \frac{π}{4}) + i s i n (0 - \frac{π}{4})) = \frac{1}{2} (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})) = \frac{1}{2} (c o s (\frac{π}{4}) - i s i n (\frac{π}{4}))$

$24) ω^{2} S o l u t i o n : ω = 2 (c o s (\frac{π}{3}) + i s i n (\frac{π}{3})) ω^{2} = ω \times ω z_{1} z_{2} = r_{1} r_{2} (c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})) = 2 \times 2 (c o s (\frac{π}{3} + \frac{π}{3}) + i s i n (\frac{π}{3} + \frac{π}{3})) = 4 (c o s (\frac{2 π}{3}) + i s i n (\frac{2 π}{3}))$

$25) 5 i z S o l u t i o n : z = 2 (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})), 5 i = 5 (c o s (\frac{π}{2}) + i s i n (\frac{π}{2})), z = 2 (c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4})), 5 i = 5 (c o s (\frac{π}{2}) + i s i n (\frac{π}{2})), z_{1} z_{2} = r_{1} r_{2} (c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})) = 2 \times 5 (c o s (- \frac{π}{4} + \frac{π}{2}) + i s i n (- \frac{π}{4} + \frac{π}{2})) = 10 (c o s (\frac{π}{4}) + i s i n (\frac{π}{4}))$

أجد جميع الجذور الحقيقية والمركبة لكل من المعادلات الآتية:

$26) z^{2} + 104 = 20 z S o l u t i o n : z^{2} - 20 z - 104 = 0 \Rightarrow z = \frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \times 1 \times 104}}{2} z = \frac{20 \pm \sqrt{400 - 416}}{2} z = \frac{20 \pm \sqrt{- 16}}{2} = \frac{20 \pm 4 i}{2} z = 10 \pm 2 i$

$27) z^{2} + 18 z + 202 = 0 S o l u t i o n : z^{2} + 18 z + 202 = 0 \Rightarrow z = \frac{- 18 \pm \sqrt{{(18)}^{2} - 4 \times 1 \times 202}}{2} z = \frac{- 18 \pm \sqrt{324 - 808}}{2} z = \frac{- 18 \pm \sqrt{- 484}}{2} = \frac{- 18 \pm 22 i}{2} z = - 9 \pm 11 i$

$28) 9 z^{2} + 68 = 0 S o l u t i o n : 9 z^{2} + 68 = 0 \Rightarrow z^{2} = - \frac{68}{9} z = \frac{\sqrt{- 68}}{3} = \pm \frac{2 \sqrt{17}}{3} i$

$29) 3 z^{3} - 2 z^{2} + 2 z + 1 = 0$

الحل :

بأستخدام الاصفار النسبية المحتملة : ${\pm 1 ، \pm \frac{1}{3}}$

بالتجريب :

$3 {(- \frac{1}{3})}^{3} - 2 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 1 = 0 - \frac{1}{9} - \frac{2}{9} - \frac{6}{9} + \frac{9}{9} = 0$

ومنه فإنّ المقدار $(3 z + 1)$ هو أحد عوامل كثير الحدود وسنبدأ القسمة التركيبية على النحو :

$3 z^{3} - 2 z^{2} + 2 z + 1 = (3 z + 1) (z^{2} - z + 1) z^{2} - z + 1 = 0 \Rightarrow z = \frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \times 1 \times 1}}{2} z = \frac{1 \pm \sqrt{1 - 4}}{2} z = \frac{1 \pm \sqrt{- 3}}{2} = \frac{1 \pm \sqrt{3} i}{2} z = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i, - \frac{1}{3}$

$30) z^{3} + 4 z + 10 = 5 z^{2}$

الحل :

بأستخدام عوامل الحد المطلق وتجريب العدد 1 - الذي يحقق المعادلة :

$z^{3} - 5 z^{2} + 4 z + 10 = (z + 1) (z^{2} - 6 z + 10) z^{2} - 6 z + 10 = 0 \Rightarrow z = \frac{6 \pm \sqrt{{(6)}^{2} - 4 \times 1 \times 10}}{2} z = \frac{6 \pm \sqrt{36 - 10}}{2} z = \frac{6 \pm \sqrt{- 4}}{2} = \frac{6 \pm 2 i}{2} z = 3 \pm i, - 1$

$31) 2 z^{3} = 8 z^{2} + 13 z - 87$

الحل :

بأستخدام عوامل الحد المطلق وتجريب العدد 3 - الذي يحقق المعادلة :

$2 z^{3} - 8 z^{2} - 13 z + 87 = (z + 3) (2 z^{2} - 14 z + 29) 2 z^{2} - 14 z + 29 = 0 \Rightarrow z = \frac{14 \pm \sqrt{{(14)}^{2} - 4 \times 2 \times 29}}{4} z = \frac{14 \pm \sqrt{196 - 232}}{4} z = \frac{14 \pm \sqrt{- 36}}{4} = \frac{14 \pm 6 i}{4} z = \frac{7 \pm 3 i}{2}, - 3$

أجد معادلة تربيعية لها الجذران المُركَبَان المعطيان في كلٍّ ممّا يأتي:

$32) 2 \pm 5 i S o l u t i o n : z = 2 \pm 5 i z - 2 = \pm 5 i {(z - 2)}^{2} = {(\pm 5 i)}^{2} z^{2} - 4 z + 4 = 25 i^{2} z^{2} - 4 z + 4 = - 25 i^{2} = - 1 z^{2} - 4 z + + 29 = 0$

$33) 7 \pm 4 i S o l u t i o n : z = 7 \pm 4 i z - 7 = \pm 4 i {(z - 7)}^{2} = {(\pm 4 i)}^{2} z^{2} - 14 z + 49 = 16 i^{2} z^{2} - 14 z + 49 = - 16 i^{2} = - 1 z^{2} - 14 z + + 65 = 0$

$34) - 8 \pm 20 i S o l u t i o n : z = - 8 \pm 20 i z + 8 = \pm 20 i {(z + 8)}^{2} = {(\pm 20 i)}^{2} z^{2} + 16 z + 64 = 400 i^{2} z^{2} + 16 z + 64 = - 400 i^{2} = - 1 z^{2} + 16 z + + 464 = 0$

$35) - 3 \pm 2 i S o l u t i o n : z = - 3 \pm 2 i z + 3 = \pm 2 i {(z + 3)}^{2} = {(\pm 2 i)}^{2} z^{2} + 6 z + 9 = 4 i^{2} z^{2} + 6 z + 9 = - 4 i^{2} = - 1 z^{2} + 6 z + + 13 = 0$

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 مما يأتي:

$36) \frac{z_{2}}{z_{1}} S o l u t i o n : \frac{z_{2}}{z_{1}} = \frac{\sqrt{5} - i \sqrt{15}}{\sqrt{12} - 2 i} \times \frac{\sqrt{12} + 2 i}{\sqrt{12} + 2 i} = \frac{(\sqrt{5} - i \sqrt{15}) (\sqrt{12} + 2 i)}{12 + 4} = \frac{\sqrt{5} (\sqrt{12} + 2 i) - i \sqrt{15} (\sqrt{12} + 2 i)}{16} = \frac{\sqrt{60} + 2 i \sqrt{5} - i \sqrt{180} + 2 \sqrt{15}}{16} = \frac{2 \sqrt{15} + 2 i \sqrt{5} - i 6 \sqrt{5} + 2 \sqrt{15}}{16} = \frac{4 \sqrt{15} - i 4 \sqrt{5}}{16} = \frac{\sqrt{15}}{4} - \frac{i \sqrt{5}}{4} | \frac{z_{2}}{z_{1}} | = \sqrt{{(\frac{\sqrt{15}}{4})}^{2} + {(\frac{\sqrt{5}}{4})}^{2}} = \sqrt{\frac{15}{16} + \frac{5}{16}} = \sqrt{\frac{20}{16}} = \frac{\sqrt{5}}{2}$

لكن من خصائص المقياس : $| \frac{z_{1}}{z_{2}} | = \frac{| z_{1} |}{| z_{2} |}$ ليصبح الحل بشكل افضل :

$| \frac{z_{2}}{z_{1}} | = \frac{| z_{2} |}{| z_{1} |} | z_{2} | = \sqrt{{(\sqrt{5})}^{2} + {(\sqrt{15})}^{2}} = \sqrt{20} = 2 \sqrt{5} | z_{1} | = \sqrt{{(\sqrt{12})}^{2} + {(2)}^{2}} = \sqrt{16} = 4 \Rightarrow | \frac{z_{2}}{z_{1}} | = \frac{| z_{2} |}{| z_{1} |} = \frac{2 \sqrt{5}}{4} = \frac{\sqrt{5}}{2} \frac{z_{2}}{z_{1}} = \frac{\sqrt{15}}{4} - \frac{i \sqrt{5}}{4} \Rightarrow A r g (\frac{z_{2}}{z_{1}}) = - t a n^{- 1} (\frac{b}{a}) \Rightarrow A r g (\frac{z_{2}}{z_{1}}) = - t a n^{- 1} (\frac{\frac{\sqrt{5}}{4}}{\frac{\sqrt{15}}{4}}) = - t a n^{- 1} (\frac{1}{\sqrt{3}}) = - \frac{π}{6} T h e b i t t e r o n e : A r g (\frac{z_{2}}{z_{1}}) = A r g (z_{2}) - A r g (z_{1}) = - t a n^{- 1} (\frac{\sqrt{15}}{\sqrt{5}}) - (- t a n^{- 1} (\frac{2}{\sqrt{12}})) = - t a n^{- 1} (\sqrt{3}) + t a n^{- 1} (\frac{1}{\sqrt{3}}) = - \frac{π}{3} + \frac{π}{6} = - \frac{π}{6}$

$37) \frac{1}{z_{3}} S o l u t i o n : \frac{1}{z_{3}} = \frac{1}{2 - 2 i} \times \frac{2 + 2 i}{2 + 2 i} = \frac{2 + 2 i}{4 + 4} = \frac{2 + 2 i}{8} = \frac{1}{4} + \frac{1}{4} i | \frac{1}{z_{3}} | = \sqrt{{(\frac{1}{4})}^{2} {+ (\frac{1}{4})}^{2}} = \sqrt{\frac{2}{16}} = \frac{1}{2 \sqrt{2}} | \frac{1}{z_{3}} | = \frac{| 1 |}{| z_{3} |} = \frac{1}{\sqrt{{(2)}^{2} {+ (2)}^{2}}} = \frac{1}{2 \sqrt{2}} A r g (\frac{1}{z_{3}}) = A r g (\frac{1}{4} + \frac{1}{4} i) = t a n^{- 1} (\frac{\frac{1}{4}}{\frac{1}{4}}) = t a n^{- 1} (1) = \frac{π}{4} T h e b i t t e r o n e A r g (\frac{1}{z_{3}}) = A r g (1) - A r g (z_{3}) = 0 - (- t a n^{- 1} (\frac{2}{2})) = \frac{π}{4}$

$38) \frac{z_{3}}{z_{2}} S o l u t i o n : | z_{3} | = \sqrt{{(2)}^{2} + {(2)}^{2}} = 2 \sqrt{2} | z_{2} | = \sqrt{{(\sqrt{5})}^{2} + {(\sqrt{15})}^{2}} = 2 \sqrt{5} | \frac{z_{3}}{z_{2}} | = \frac{| z_{3} |}{| z_{2} |} = \frac{2 \sqrt{2}}{2 \sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}} A r g (\frac{z_{3}}{z_{2}}) = A r g (z_{3}) - A r g (z_{2}) = - t a n^{- 1} (\frac{2}{2}) - t a n^{- 1} (\frac{\sqrt{15}}{\sqrt{5}}) = - t a n^{- 1} (1) - t a n^{- 1} (\sqrt{3}) = - \frac{π}{4} - \frac{π}{3} = - \frac{7 π}{12}$

إذا كان : $8 (c o s (\frac{2 π}{3}) - i s i n (\frac{2 π}{3}))$ ، فأجيب عن السؤالين الآتيين تباعًا:

39) أَمثل العدد بيانيًا في المستوى المُركب.

$S o l u t i o n : 8 (c o s (- \frac{2 π}{3}) + i s i n (- \frac{2 π}{3})) r = 8 a n d θ = - \frac{2 π}{3} O r 8 (c o s (- \frac{2 π}{3}) + i s i n (- \frac{2 π}{3})) = 8 ((- \frac{1}{2}), (- \frac{\sqrt{3}}{2})) = (- 4, - 4 \sqrt{3})$

40) أجد الجذرين التربيعيين للعدد z .

$S o l u t i o n : 8 (c o s (- \frac{2 π}{3}) + i s i n (- \frac{2 π}{3})) = 8 (- \frac{1}{2} - i \frac{\sqrt{3}}{2}) z = - 4 - 4 i \sqrt{3} l e t \sqrt{z} = x + y i \Rightarrow z = {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} {(x + y i)}^{2} = x^{2} + 2 x y i - y^{2} - 4 - 4 i \sqrt{3} = x^{2} - y^{2} + 2 x y i \begin{array}{rcl} R e_{1} & = & R e_{2} \\ - 4 & = & x^{2} - y^{2} \dots 1 \end{array} \begin{array}{rcl} I m_{1} & = & I m_{2} \\ - 4 \sqrt{3} & = & 2 x y \dots 2 \end{array} x = - \frac{2 \sqrt{3}}{y} F o r m 1 a n d 2 4 = y^{2} - {(- \frac{2 \sqrt{3}}{y})}^{2} 4 = y^{2} - \frac{12}{y^{2}} 4 y^{2} = y^{4} - 12 y^{4} - 4 y^{2} - 12 = 0 (y^{2} - 6) (y^{2} + 2) = 0 y^{2} - 6 = 0 \Rightarrow y = \sqrt{6} \Rightarrow x = - \frac{2 \sqrt{3}}{\sqrt{6}} = - \sqrt{2} z = - \sqrt{2} + i \sqrt{6} \Rightarrow y = - \sqrt{6} \Rightarrow x = - \frac{2 \sqrt{3}}{- \sqrt{6}} = \sqrt{2} z = \sqrt{2} - i \sqrt{6}$

41) إذا كان: $(b + i c), (a - 3 i)$ هما الجذرين التربيعيين للعدد المُركب: $55 - 48 i$ ، فأجد قيمة كلّ من الثوابت الحقيقية: a ،b ، c .

$S o l u t i o n : {(a - 3 i)}^{2} = 55 - 48 i a^{2} - 6 a i - 9 = 55 - 48 i a^{2} - 9 - 6 a i = 55 - 48 i R e = R e a^{2} - 9 = 55 \Rightarrow a^{2} = 64 \Rightarrow a = \pm 8 I m = I m - 6 a i = - 48 i \Rightarrow a = 8 {(b + i c)}^{2} = 55 - 48 i b^{2} + 2 b c i - c = 55 - 48 i b^{2} - c^{2} + 2 b c i = 55 - 48 i R e = R e b^{2} - c^{2} = 55 \dots 1 I m = I m 2 b c i = - 48 i b c = - 24 \dots 2 b^{2} + {(\frac{24}{b})}^{2} = 55 \Rightarrow b^{2} + \frac{576}{b^{2}} = 55 b^{4} - 55 b^{2} + 576 = 0 (b^{2} - 64) (b^{2} + 9) = 0 b^{2} = 64 \Rightarrow b = \pm 8 w h e n b = 8 \Rightarrow c = - 3 w h e n b = - 8 \Rightarrow c = 3$

أحلٌّ المعادلة المعطى أحد جذورها في كلّ مما يأتي:

$42) x^{3} + x^{2} + 15 x = 225, x = 5$

بما أن $(x - 5)$ أحد عوامل كثير الحدود وبعد القسمة التركيبية سنجد :

$S o l u t i o n : x^{3} + x^{2} + 15 x - 255 = (x - 5) (x^{2} + 6 x + 45) x = \frac{- 6 \pm \sqrt{{(6)}^{2} - 4 \times 1 \times 45}}{2} x = \frac{- 6 \pm \sqrt{36 - 180}}{2} = \frac{- 6 \pm \sqrt{- 144}}{2} x = - 3 \pm 6 i, x = 5$

$43) x^{3} + 7 x^{2} - 13 x - 45 = 0, x = - 9$

بما أن $(x + 9)$ أحد عوامل كثير الحدود وبعد القسمة التركيبية سنجد :

$S o l u t i o n : x^{3} + 7 x^{2} - 13 x - 45 = (x + 9) (x^{2} - 2 x + 5) x = \frac{2 \pm \sqrt{{(2)}^{2} - 4 \times 1 \times 5}}{2} x = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{- 16}}{2} x = 1 \pm 2 i, x = - 9$

$44) 3 x (x^{2} + 45) = 2 (19 x^{2} + 37), x = 6 - i$

بما أنّ $(x = 6 - i)$ أحد عوامل كثير الحدود فإن مرافقه $(x = 6 + i)$ كذلك :

$S o l u t i o n : x = 6 \pm i x - 6 = \pm i {(x - 6)}^{2} = {(\pm i)}^{2} x^{2} - 12 x + 36 = i^{2} x^{2} - 12 x + 36 = - 1 x^{2} - 12 x + 37 = 0$

وبعد قسمة كثيرات الحدود : $3 x^{3} + 135 x - 38 x^{2} - 74 = 0$ ، $x^{2} - 12 x + 37$

$3 x^{3} + 135 x - 38 x^{2} - 74 = (x^{2} - 12 x + 37) (3 x - 2) \Rightarrow x = \frac{2}{3}, 6 \pm i$

$45) x^{3} + 10 x^{2} + 29 x + 30 = 0, x = - 2 + i$

بما أنّ $(x = - 2 + i)$ أحد عوامل كثير الحدود فإن مرافقه $(x = - 2 - i)$ كذلك :

$S o l u t i o n : x = - 2 \pm i x + 2 = \pm i {(x + 2)}^{2} = {(\pm i)}^{2} x^{2} + 4 x + 4 = i^{2} x^{2} + 4 x + 4 = - 1 x^{2} + 4 x + 5 = 0$

وبعد قسمة كثيرات الحدود :

$x^{3} + 10 x^{2} + 29 x + 30, x^{2} + 4 x + 5 x^{3} + 10 x^{2} + 29 x + 30 = (x^{2} + 4 x + 5) (x + 6) x = - 2 \pm i, x = - 6$

إذا كان: $4 + 11 i$ هو أحد جذري المعادلة: $z^{2} - 8 z + k = 0$ حيث k عدد حقيقي ، فأجيب عن السؤالين الآتيين تباعًا:

46) أجد الجذر الآخر للمعادلة.

الحل :

بما أنّ $4 + 11 i$ أحد عوامل كثير الحدود فإن مرافقه $4 - 11 i$ هو العامل الآخر.

47) أجد قيمة الثابت k .

الحل :

الثابت $k$ هو حاصل ضرب الجذرين :

$k = (4 + 11 i) (4 - 11 i) = 16 - 121 i^{2} = 16 + 121 = 137$

تبرير: أجيب عن الأسئلة الثلاثة الآتية تباعَاً مُبررًا إجابتي:

48) أجد ناتج: ${(p + i q)}^{2}$ حيث p و q عددان حقيقيان.

$S o l u t i o n : {(p + i q)}^{2} = p^{2} + 2 p q i + i^{2} q^{2} = p^{2} + 2 p q i - q^{2}$

49) إذا كان: ${(p + i q)}^{2} = 45 + i m$ ، حيث p و q عددان صحيحان موجبان ، حيث p > q ، فأجد ثلاث قِيّم مُمكنة للعدد لحقيقي m .

$S o l u t i o n : {(p + i q)}^{2} = p^{2} + 2 p q i - q^{2} 45 + i m = p^{2} + 2 p q i - q^{2} R e_{1} = R e_{2} 45 = p^{2} - q^{2} . . . . 1 I m_{1} = I m_{2} m = 2 p q . . . . 2 45 = (p + q) (p - q) = 45 \times 1 \Rightarrow p = 23, q = 22 \Rightarrow m = 2 \times 23 \times 22 = 1012 45 = (p + q) (p - q) = 15 \times 3 \Rightarrow p = 9, q = 6 \Rightarrow m = 2 \times 9 \times 6 = 108 45 = (p + q) (p - q) = 9 \times 5 \Rightarrow p = 7, q = 2 \Rightarrow m = 2 \times 7 \times 2 = 28$

50) أستعمل إجابة السؤال السابق لإيجاد الجذرين التربيعيين للعدد المُركَّب: $45 - 108 i$ .

$S o l u t i o n : z = 45 - 108 i - 108 = 2 p q - 54 = p q \Rightarrow p = 9, q = - 6 \Rightarrow m = - 6 \times 9 = - 54 p + i q = 9 - 6 i O r \Rightarrow p = - 9, q = 6 \Rightarrow m = - 9 \times 6 = - 54 p + i q = - 9 + 6 i$

51) برهان: أثبت أن $z \bar{z} = {(| z |)}^{2}$ لأيّ عدد مركب z .

$S o l u t i o n : z = x + y i \Rightarrow \bar{z} = x - y i z \bar{z} = (x + y i) (x - y i) = x^{2} + y^{2} . . . . 1 | z | = \sqrt{x^{2} + y^{2}} {(| z |)}^{2} = {(\sqrt{x^{2} + y^{2}})}^{2} = x^{2} + y^{2} . . . . 2 \Rightarrow z \bar{z} = {(| z |)}^{2}$

52) برهان : إذا كان z عددًا مُركَبًا $A r g (z) = t a n^{- 1} (\frac{1}{2})$ حيث: $| z | = 5 \sqrt{5}$ وكان: $\frac{z}{3 + 4 i} = p + i q$ ، فأثبت أن $p + q = 1$ .

$S o l u t i o n : z = a + b i A r g (z) = t a n^{- 1} (\frac{1}{2}) \Rightarrow \frac{b}{a} = \frac{1}{2} \Rightarrow a = 2 b . . . 1 | z | = 5 \sqrt{5} | z | = \sqrt{a^{2} + b^{2}} \Rightarrow 5 \sqrt{5} = \sqrt{a^{2} + b^{2}} {(5 \sqrt{5})}^{2} = {(\sqrt{a^{2} + b^{2}})}^{2} 125 = a^{2} + b^{2} . . . 2 125 = {(2 b)}^{2} + b^{2} 125 = 5 b^{2} \Rightarrow b = 5 > 0 \Rightarrow a = 10 > 0 \Rightarrow z = 10 + 5 i \frac{z}{3 + 4 i} = \frac{10 + 5 i}{3 + 4 i} \times \frac{3 - 4 i}{3 - 4 i} = \frac{50 - 25 i}{9 + 16} = \frac{50 - 25 i}{25} = 2 - i p + i q = 2 - i p = 2, q = - 1 \Rightarrow p + q = 2 + - 1 = 1$

53) تحدّ: العدد المركّب: $z = (10 - i) - (2 - 7 i)$ هو أحد جذور المعادلة: $z^{3} - 20 z^{2} + 164 z - 400 = 0$

أجد بقية جذور هذه المعادلة ثم أحل المعادلة الآتية: $x^{6} + 164 x^{2} = 20 (x^{4} + 20)$

$S o l u t i o n : z = (10 - i) - (2 - 7 i) z_{1} = 8 + 6 i z_{2} = 8 - 6 i z = 8 \pm i z - 8 = \pm 6 i {(z - 8)}^{2} = {(\pm 6 i)}^{2} z^{2} - 16 z + 64 = 36 i^{2} z^{2} - 16 z + 64 = - 36 z^{2} - 16 z + 100 = 0$

وبعد قسمة كثيرات الحدود : $z^{3} - 20 z^{2} + 164 z - 400, z^{2} - 16 z + 100$

$z^{3} - 20 z^{2} + 164 z - 400 = (z^{2} - 16 z + 100) (z - 4) \Rightarrow z = 4, z = 8 \pm 6 i$

ولحل المعادلة : $x^{6} + 164 x^{2} = 20 (x^{4} + 20)$ ستكتب على الصورة :

$x^{6} + 164 x^{2} - 20 x^{4} - 400 = 0 l e t u = x^{2} {(x^{2})}^{3} + 164 x^{2} - 20 {(x^{2})}^{2} - 400 = 0 u^{3} + 164 u - 20 u^{2} - 400 = (u^{2} - 16 u + 100) (u - 4) \Rightarrow u = 4 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2 \Rightarrow u = 8 \pm 6 i \Rightarrow x^{2} = 8 \pm 6 i \Rightarrow x = \pm \sqrt{8 + 6 i} x = \pm \sqrt{8 + 6 i} l e t x = a + i b a + i b = \pm \sqrt{8 + 6 i} {(a + i b)}^{2} = {(\pm \sqrt{8 + 6 i})}^{2} a^{2} + 2 a b i - b^{2} = 8 + 6 i R e_{1} = R e_{2} a^{2} - b^{2} = 8 . . . . 1 I m_{1} = I m_{2} a b = 3 . . . . 2 a^{2} - {(\frac{3}{a})}^{2} = 8 a^{4} - 8 a^{2} - 9 = 0 (a^{2} - 9) (a^{2} + 1) = 0 \Rightarrow a^{2} - 9 = 0 \Rightarrow a = \pm 3 \Rightarrow b = \pm 1 x = 2, x = - 2, x = 3 + i x = 3 - i x = - 3 + i x = - 3 - i$