دروس مادة الرياضيات للفصل الاول للشعب العلمية سنة ثالثة ثانوي

‫ﺇﺫﻥ ‪ x2 − x − 6 = 0 :‬ﻭﻤﻨﻪ ‪x2 = 3 ; x1 = −2 ; ∆ = 25 :‬‬ ‫ﻭ ﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻭﺤﻴﺩ }‪s = {3‬‬ ‫‪ (2‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪2 ( logx )2 + 5logx − 3 = 0 :‬‬‫ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ‪ x > 0‬ﺒﻭﻀﻊ ‪ logx = t‬ﻨﺠﺩ ‪ 2t 2 + 5t − 3 = 0‬ﻭﻤﻨﻪ ‪:‬‬ ‫‪t2 = −3‬‬ ‫;‬ ‫‪t1‬‬ ‫=‬ ‫‪1‬‬ ‫‪; ∆ = 49‬‬ ‫‪2‬‬‫= ‪lnx‬‬ ‫‪1‬‬ ‫‪ln10‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪lnx‬‬ ‫=‬ ‫‪1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪logx‬‬ ‫=‬ ‫‪1‬‬ ‫‪:‬‬ ‫‪t‬‬ ‫=‬ ‫‪1‬‬ ‫ﻟﻤﺎ‬ ‫‪2‬‬ ‫‪ln10‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪ lnx = ln 10 :‬ﻭﻋﻠﻴﻪ ‪x = 10 :‬‬ ‫‪lnx‬‬ ‫=‬ ‫‪−3‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﻟﻤﺎ ‪logx = −3 : τ = −3‬‬ ‫‪ln10‬‬ ‫ﻭﻋﻠﻴﻪ ‪ lnx = ln10 −3 :‬ﻭ ﻤﻨﻪ ‪x = 10−3 :‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ ‪{ }s = 10;10−3 :‬‬ ‫‪log x > 3‬‬ ‫‪ (3‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪:‬‬ ‫ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺘﻜﻭﻥ ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ‪x > 0‬‬‫ﻭﻋﻠﻴﻪ ‪x > 103 :‬‬ ‫ﺃﻱ ‪lnx > ln103 :‬‬ ‫‪lnx‬‬ ‫‪>3‬‬ ‫‪:‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ‬ ‫‪ln10‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ‪s = 103;+∞ :‬‬ ‫‪ (3 (4‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪:‬‬ ‫‪log ( x − 6) > 2logx‬‬‫ﺘﻜﻭﻥ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ‪ x > 0‬ﻭ ‪ x − 6 > 0‬ﻭﻋﻠﻴﻪ ‪x > 6 :‬‬ ‫ﻭﻤﻨﻪ ﻓﻬﻲ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ∞‪] [. 6;+‬‬‫ﻭ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺘﻜﺎﻓﺊ ‪ log ( x − 6) > logx2 :‬ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪x − 6 > x2 :‬‬ ‫ﺇﺫﻥ ‪ − x2 + x − 6 < 0 :‬ﻭﻋﻠﻴﻪ ﻟﻴﺱ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ﺤﻠﻭل ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ‪18‬‬ ‫ﺤﺴﺎﺏ ‪: S‬‬

S = log  1 × 2 × 3 × ..... × 88 × 99   2 3 4 99 1000  S = log  1  = −log1000  1000  S = −log103 = −3 19 ‫ﺍﻟﺘﻤﺭﻴﻥ‬1) f ( x) = x + log x ]−∞;0[ ∪ ]0;+∞[ ‫ ﻤﻌﺭﻓﺔ ﻋﻠﻰ‬f ‫ﺍﻟﺩﺍﻟﺔ‬ lim f ( x) = lim x + ln x ln10 x→−∞ x→−∞ = lim x 1 + ln(− x) × 1 = −∞   x → −∞ −x ln10 lim f (x) = lim 1 + lnx × 1 = +∞ x ln10 x→+∞ x → +∞ lim f (x) = lim x + lnx = −∞ > > ln10 x→0 x→0 lim f ( x) = lim x + ln(− x) = −∞ < < ln10 x→0 x→6( )2) f ( x) = x2 − 1 − log x2 − 1 x2 − 1 > 0 ‫ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻤﻥ ﺃﺠل‬f ‫ﺍﻟﺩﺍﻟﺔ‬ x ∈ ]−∞;1[ ∪ ]1;+∞[ : ‫ﻭ ﻋﻠﻴﻪ‬ ]−∞;1[ ∪ ]1;+∞[ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻑ ﻫﻲ‬( )lim f ( x) = lim x2 − 1 − log x2 − 1x→−∞ x → −∞ ( ) ( )= lim x → −∞ x2 −1  log x2 − 1  1 −  = +∞  x2 −1 

( ) ( )lim f ( x) = lim x2 −1  log x2 − 1  1 −  = +∞ x→+∞ x→+∞  x2 −1  ( )lim f ( x) = lim x2 − 1 − log x2 − 1 = +∞ x→−1 x→−1 x>−1 x>−1 ( )lim f ( x) = lim x2 − 1 − log x2 − 1 = −∞ x→1 x→1 x>1 x>13) f ( x) = 1 −1 Logx x > 0 ‫ ﻭ‬Logx − 1 ≠ 0 ‫ ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل‬f ‫ﺍﻟﺩﺍﻟﺔ‬ x ≠ 10 : ‫ ﺃﻱ‬lnx ≠ ln10 : ‫ﻭﻋﻠﻴﻪ‬ lnx ≠1 : ‫ﻤﻌﻨﺎﻩ‬ logx ≠ 1 ln10 ]0,10[ ∪ ]10;+∞[ : ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ﻫﻲ‬ lim f (x) = lim 1 = 0 > > logx − 1 x→0 x→0 ( )lim f x = lim 1 −1 = −∞ logx x →10 x →10 x>10 x>10 ( )lim f x = lim 1 − 1 = +∞ > logx > x →10 x →10 lim f (x) = lim 1 = 0 logx − 1 x→+∞ x → +∞ 4) f ( x) = (log x)2 ] [0;+∞ ‫ ﺃﻱ ﻤﻌﺭﻓﺔ ﻋﻠﻰ‬x > 0 ‫ﺍﻟﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل‬ lim f ( x) = lim (logx)2 = +∞ x→0 x→0 x>0 x>0 lim f ( x) = lim (logx)2 = +∞ x→+∞ x → +∞

‫ﺍﻟﺘﻤﺭﻴﻥ ‪20‬‬ ‫‪ – 1‬ﺩﺭﺍﺴﺔ ﺍﻟﺘﻐﻴﺭﺍﺕ ‪:‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ‪ x − 1 ≠ 0‬ﺃﻱ ‪x ≠ 1‬‬ ‫ﻭ ﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ﻫﻲ ‪]−∞;1[ ∪ ]1;+∞[ :‬‬ ‫∞‪lim f ( x) = lim log x − 1 = +‬‬ ‫∞‪x→−‬‬ ‫∞‪x→−‬‬ ‫∞‪lim f ( x) = lim log x − 1 = −‬‬ ‫‪x→1‬‬ ‫‪x→1‬‬ ‫‪x<1 x<1‬‬ ‫∞‪lim f ( x) = lim log x − 1 = −‬‬ ‫‪x→1‬‬ ‫‪x→1‬‬ ‫‪x>1 x>1‬‬ ‫∞‪lim f ( x) = lim log x − 1 = +‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫‪f‬‬ ‫‪′‬‬ ‫(‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫‪1‬‬ ‫×‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪ln10‬‬ ‫‪−‬‬‫ﻟﻤﺎ ‪ f ′ ( x ) > 0 : x > 1‬ﻭ ﻤﻨﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞‪ 1;+‬ﻭ ﻋﻠﻴﻪ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ[ ]‬ ‫[‪. ]−∞;1‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪1‬‬ ‫∞‪+‬‬ ‫‪+‬‬‫)‪f ′(x‬‬ ‫‪-‬‬ ‫∞‪+‬‬ ‫∞‪+‬‬‫)‪f (x‬‬ ‫∞‪−∞ −‬‬ ‫‪ -2‬ﺩﺭﺍﺴﺔ ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ﻭ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﻤﻘﺎﺭﺒﺔ ﻫﻨﺎﻙ ‪4‬‬ ‫ﻓﺭﻭﻉ ﻻﻨﻬﺎﺌﻴﺔ ﻭ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ ﻤﻌﺎﺩﻟﺘﻪ ‪x = 1 :‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim‬‬ ‫)‪log ( x − 1‬‬ ‫‪=0‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬ ‫∞‪x → +‬‬ ‫‪x‬‬

‫ﺇﺫﻥ ﻴﻭﺠﺩ ﻓﺭﻉ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﺒﺎﺘﺠﺎﻩ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻋﻨﺩ ∞‪+‬‬‫‪lim‬‬ ‫‪f‬‬ ‫)‪( x‬‬ ‫=‬ ‫‪lim‬‬ ‫)‪log (− x + 1‬‬ ‫‪=0‬‬‫∞‪x→−‬‬ ‫‪x‬‬ ‫∞‪x → −‬‬ ‫‪x‬‬ ‫ﺇﺫﻥ ﻴﻭﺠﺩ ﻓﺭﻉ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﺒﺎﺘﺠﺎﻩ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻋﻨﺩ ∞‪−‬‬ ‫‪ -3‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ‪y = f ′ ( 2) .( x − 2) + f ( 2) :‬‬‫‪f‬‬ ‫‪′‬‬ ‫(‬ ‫)‪2‬‬ ‫=‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪f (2) = 0‬‬ ‫‪ln10‬‬ ‫‪y‬‬ ‫=‬ ‫‪1‬‬ ‫(‬ ‫‪x‬‬ ‫‪−‬‬ ‫)‪2‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪ln10‬‬ ‫‪ – 4‬ﺇﻴﺠﺎﺩ ﺍﻟﻤﻤﺎﺴﺎﺕ ﺫﺍﺕ ﻤﻌﺎﻤل ﺍﻟﺘﻭﺠﻴﻪ ‪: 10‬‬ ‫‪1‬‬ ‫×‬ ‫‪1‬‬ ‫=‬ ‫‪10‬‬ ‫‪:‬‬ ‫ﻤﻨﻪ‬ ‫ﻭ‬ ‫‪f ′ ( x) = 10‬‬ ‫‪ln10‬‬ ‫‪x−1‬‬‫=‪x−1‬‬ ‫‪1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﺇﺫﻥ ‪10(ln10) .( x − 1) = 1 :‬‬ ‫‪10.ln10‬‬ ‫ﺇﺫﻥ ﻫﻨﺎﻙ ﻤﻤﺎﺱ ﻭﺍﺤﺩ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ‬ ‫‪x‬‬ ‫=‬ ‫‪1+‬‬ ‫‪1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪ln10‬‬ ‫ﻤﻌﺎﻤل ﺘﻭﺠﻴﻬﻪ ‪. 10‬‬ ‫‪1+‬‬ ‫‪1‬‬ ‫ﺍﻟﻔﺎﺼﻠﺔ‬ ‫‪ln10‬‬ ‫‪ -5‬ﺇﻨﺸﺎﺀ ‪( ): C‬‬ ‫‪y‬‬ ‫‪1‬‬ ‫‪0,5‬‬‫‪-1,5 -1 -0,5 0‬‬ ‫‪0,5 1 1,5 2 2,5 3‬‬ ‫‪x‬‬ ‫‪-0,5‬‬ ‫‪-1‬‬

‫* ‪x−1 =1: y = 0‬‬ ‫‪ -‬ﻨﻘﻁ ﺍﻟﺘﻘﺎﻁﻊ ‪y = 0 : x = 0 :‬‬ ‫* }‪(C ) ∩ ( y′y) = {0‬‬‫ﻭ ﻤﻨﻪ ‪ x − 1 = 1 :‬ﺃﻭ ‪ x = 1 = −1‬ﻭﻋﻠﻴﻪ ‪ x = 2 :‬ﺃﻭ ‪x = 0‬‬ ‫}‪ (C ) ∩ ( x′x) = {O, A‬ﺤﻴﺙ ‪A(2;0) :‬‬ ‫‪ -6‬ﺍﻟﻤﻨﺎﻗﺸﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ‪:‬‬ ‫ﻴﺘﻘﺎﻁﻊ ) ‪ (C‬ﻭ ) ∆ ( ﻓﻲ ﻨﻘﻁﺘﻴﻥ ﻤﺘﻤﺎﻴﺯﺘﻴﻥ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 21‬‬ ‫[∞‪Df = ]0;+‬‬ ‫‪-1‬‬ ‫∞‪lim f ( x) = lim− 4 + 4 logx = −‬‬ ‫>>‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫∞‪lim f ( x) = lim − 4 + 4logx = +‬‬‫∞‪x→+‬‬ ‫∞‪x → +‬‬ ‫‪f‬‬ ‫(‪′‬‬ ‫)‪x‬‬ ‫=‬ ‫‪4‬‬ ‫×‬ ‫‪1‬‬ ‫‪Ln10‬‬ ‫‪x‬‬‫ﻭ ﻤﻨﻪ ‪ f ′ ( x ) > 0 :‬ﻭ ﻋﻠﻴﻪ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ [∞‪]0;+‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫∞‪+‬‬‫)‪f ′(x‬‬ ‫‪+‬‬‫)‪f (x‬‬ ‫∞‪+‬‬ ‫∞‪−‬‬ ‫‪ – 2‬ﺩﺭﺍﺴﺔ ﺍﻟﻔﺭﻭﻉ ﺍﻟﻼﻨﻬﺎﺌﻴﺔ ﻭ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﻤﻘﺎﺭﺒﺔ ‪:‬‬ ‫ﻫﻨﺎﻙ ﻓﺭﻋﻴﻥ ﻻﻨﻬﺎﺌﻴﻴﻥ ﻭ ﻤﺴﺘﻘﻴﻡ ﻤﻘﺎﺭﺏ ﻤﻌﺎﺩﻟﺘﻪ ‪x = 0 :‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪−4‬‬ ‫‪+‬‬ ‫‪logx‬‬ ‫‪=0‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫∞‪x→+‬‬ ‫‪x‬‬ ‫∞‪x → +‬‬

‫ﻭ ﻋﻠﻴﻪ ﻴﻭﺠﺩ ﻓﺭﻉ ﻗﻁﻊ ﻤﻜﺎﻓﺊ ﺒﺎﺘﺠﺎﻩ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ‪.‬‬ ‫‪ – 3‬ﺇﻨﺸﺎﺀ ) ‪: (C‬‬ ‫‪y‬‬‫‪1,5‬‬ ‫‪0‬‬ ‫‪1,5 3 4,5 6 7,5 9 10,5 12 13,5 15 16,5 x‬‬‫‪-1,5‬‬‫‪-3‬‬‫‪-4,5‬‬‫‪-6‬‬‫‪-7,5‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 22‬‬ ‫}‪Df = { x ∈ : x > 0 , logx ≠ 0‬‬ ‫[∞‪Df = ]0;+1[;]1;+‬‬ ‫‪ -‬ﺩﺭﺍﺴﺔ ﺍﻟﺘﻐﻴﺭﺍﺕ ‪:‬‬ ‫‪( )lim f‬‬ ‫‪x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫>‬ ‫>‬ ‫‪logx‬‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫=)‪(x‬‬ ‫‪lim 1‬‬ ‫=‬ ‫∞‪−‬‬ ‫<‬ ‫<‬ ‫‪logx‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪x→1‬‬ ‫→‬ ‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim 1‬‬ ‫=‬ ‫∞‪+‬‬ ‫>‬ ‫>‬ ‫‪logx‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪x→1‬‬ ‫→‬

‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫‪logx‬‬ ‫∞‪x → +‬‬ ‫∞‪x→ +‬‬ ‫‪11‬‬ ‫‪− ln10 × x‬‬ ‫= )‪f ′(x‬‬ ‫=‬ ‫‪−1‬‬ ‫‪x )2‬‬ ‫‪(log x )2‬‬ ‫‪( xln10)(log‬‬‫[‪ ]0;1‬و [∞‪]1;+‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f ′ x < 0‬ﻭ ﻤﻨﻪ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ) (‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪1 ∞+‬‬ ‫‪0‬‬ ‫)‪f ′(x‬‬ ‫‪--‬‬ ‫)‪f (x‬‬ ‫∞‪+‬‬ ‫∞‪−‬‬ ‫‪0‬‬ ‫ﺇﻨﺸﺎﺀ ) ‪: (C‬‬ ‫‪ y = 0 , x = 1‬ﻤﻌﺎﺩﻟﺘﻲ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﻥ ﺍﻟﻤﻘﺎﺭﺒﻴﻥ ‪.‬‬

y 5 4 3 2 1-1 0 1 2 3 4 5 6 7x -1 -2 -3 23 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ ] [Df = 0;+∞ : ‫ﺩﺭﺍﺴﺔ ﺍﻟﺘﻐﻴﺭﺍﺕ‬lim f (x) = lim logx − 1 = lim 1 ( log x − 1) = −∞ > > x > xx→0 x→0 x→0 lim f ( x) = lim logx − 1 = 0 x x x→+∞ x → +∞ 1 . 1 .x − 1(logx − 1) ln10 x f ′( x) = x2 f ′( x) = 1 − log + 1 ln10 x2

‫)‪′( x‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪lnx‬‬ ‫‪+1‬‬ ‫‪ln10‬‬ ‫‪ln10‬‬ ‫‪f‬‬ ‫=‬ ‫‪x2‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪lnx + ln10‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪ln‬‬ ‫‪10‬‬ ‫‪x 2 ln10‬‬ ‫‪x‬‬‫‪f‬‬ ‫(‪′‬‬ ‫)‪x‬‬ ‫=‬ ‫=‬ ‫‪x 2 ln10‬‬ ‫‪10‬‬ ‫=‬ ‫‪e −1‬‬ ‫‪:‬‬ ‫ﻤﻨﻪ‬ ‫ﻭ‬ ‫‪ln‬‬ ‫‪10‬‬ ‫=‬ ‫‪−1‬‬ ‫‪:‬‬ ‫ﺘﻜﺎﻓﺊ‬ ‫‪f ′(x) = 0‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪x = 10e‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪x‬‬ ‫=‬ ‫‪10‬‬ ‫‪ 10 = xe−1‬ﺃﻱ ‪:‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪e −1‬‬ ‫‪10‬‬ ‫>‬ ‫‪e −1‬‬ ‫‪:‬‬ ‫ﻤﻨﻪ‬ ‫ﻭ‬ ‫‪ln‬‬ ‫‪10‬‬ ‫>‬ ‫‪−1‬‬ ‫‪:‬‬ ‫ﺘﻜﺎﻓﺊ‬ ‫‪f ′(x) > 0‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫ﺃﻱ ‪ 10 > xe−1 :‬ﻭﻋﻠﻴﻪ ‪x < 10e :‬‬ ‫ﻭ ﻤﻨﻪ ‪ f :‬ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪ 0 ; 10e‬ﻭ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ] ]‬ ‫ﻋﻠﻰ [∞ ‪[10e ; +‬‬‫‪x0‬‬ ‫∞‪10e +‬‬‫)‪f ′(x‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫)‪f (10e‬‬‫)‪f (x‬‬ ‫‪0‬‬ ‫∞‪−‬‬ ‫‪f‬‬ ‫) ‪(10e‬‬ ‫=‬ ‫‪log2e‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫ﺇﺫﻥ ‪0, 02 :‬‬ ‫‪2e‬‬ ‫ﺇﻨﺸﺎﺀ ﺍﻟﻤﻨﺤﻨﻲ ‪:‬‬

1y 1 2 3 4 5 6 7x-1 0 -1 -2 -3 -4 -5 -6

‫‪ - 6‬ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‬ ‫ﺍﻟﻜﻔﺎﺀﺓ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‬ ‫‪ -1‬ﺇﺠﺭﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﺤﺴﺎﺒﻴﺔ ﻋﻠﻰ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‪.‬‬ ‫‪ - 2‬ﺍﺴﺘﻌﻤﺎل ﺨﻭﺍﺹ ﻤﺭﺍﻓﻕ ﻋﺩﺩ ﻤﺭﻜﺏ ‪.‬‬ ‫‪ -3‬ﺤﺴﺎﺏ ﺍﻟﻁﻭﻴﻠﺔ ﻭﻋﻤﺩﺓ ﻟﻌﺩﺩ ﻤﺭﻜﺏ‬ ‫‪ -4‬ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﺇﻟﻰ ﺍﻟﻤﺜﻠﺜﻲ ﻭﺍﻟﻌﻜﺱ‪.‬‬ ‫‪ -5‬ﺍﻟﺘﻌﺒﻴﺭ ﻋﻥ ﺨﻭﺍﺹ ﺍﻷﺸﻜﺎل ﺍﻟﻬﻨﺩﺴﻴﺔ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‪.‬‬‫‪ -6‬ﺘﻭﻅﻴﻑ ﺨﻭﺍﺹ ﺍﻟﻁﻭﻴﻠﺔ ﻭﻋﻤﺩﺓ ﻟﺤل ﺍﻟﻤﺴﺎﺌل ﻓﻲ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﻓﻲ ﺍﻟﻬﻨﺩﺴﺔ‪.‬‬ ‫‪ -7‬ﺘﻭﻅﻴﻑ ﺩﺴﺘﻭﺭ ﻤﻭﺍﻓﺭ ﻟﺤل ﻤﺴﺎﺌل ﻓﻲ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﻓﻲ ﺍﻟﻬﻨﺩﺴﺔ ‪.‬‬ ‫‪ -8‬ﺤل ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ‪.‬‬ ‫‪ – 9‬ﺤل ﻤﻌﺎﺩﻻﺕ ﻴﺅﻭل ﺤﻠﻬﺎ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ‬ ‫‪ -10‬ﺘﻌﻴﻴﻥ ﺍﻟﻜﺘﺎﺒﺔ ﺍﻟﻤﺭﻜﺒﺔ ﻟﻠﺘﺤﻭﻴﻼﺕ ﺍﻟﻤﺄﻟﻭﻓﺔ ‪.‬‬ ‫‪ -11‬ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺘﺤﻭﻴل ﺍﻨﻁﻼﻗﺎ ﻤﻥ ﻜﺘﺎﺒﺘﻪ ﺍﻟﻤﺭﻜﺒﺔ ‪.‬‬ ‫‪ -12‬ﺘﻭﻅﻴﻑ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﻓﻲ ﺍﻟﺘﺤﻭﻴﻼﺕ ‪.‬‬ ‫ﺗﺼﻤﻴﻢ اﻟﺪرس‬‫‪ – 1‬ﺘﻌﺭﻴﻑ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‬ ‫ﺃﻨﺸﻁﺔ‬‫‪ -3‬ﺘﻌﺎﺭ ﻴﻑ ﻭ ﻤﺼﻁﻠﺤﺎﺕ‬ ‫‪ -2‬ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﻟﻌﺩﺩ ﻤﺭﻜﺏ‬ ‫‪ -5‬ﻤﺭﺍﻓﻕ ﻋﺩﺩ ﻤﺭﻜﺏ‬ ‫‪ -4‬ﺍﻟﺤﺴﺎﺏ ﻓﻲ ‪C‬‬‫‪ -6‬ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﻋﺩﺩ ﻤﺭﻜﺏ ‪ -7‬ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻌﺩﺩ ﻤﺭﻜﺏ )ﺘﺭﻤﻴﺯ ﺃﻭﻟﻴﺭ(‬‫ﺘﻜﻨﻭﻟﻭﺠﻴﺎ ﺍﻹﻋﻼﻡ ﻭ ﺍﻻﺘﺼﺎل‬ ‫‪ – 8‬ﺍﻟﻤﻌﺎﺩﻻﺕ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻓﻲ ‪C‬‬‫ﺍﻟﺤـﻠــــــﻭل‬ ‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬

‫ﺃﻨﺸﻁﺔ‬ ‫ﺍﻟﻨﺸﺎﻁ ‪:‬‬‫ﻨﻌﺘﺒﺭ ﻤﺠﻤﻭﻋﺔ ‪ E‬ﻋﻨﺎﺼﺭﻫﺎ ﻤﻥ ﺍﻟﺸﻜل ‪ x + iy‬ﺤﻴﺙ ‪ x‬ﻭ ‪ y‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﻭ ‪ i‬ﻋﺩﺩ ﺤﻴﺙ‬ ‫‪ i2 = -1‬ﻟﻴﻜﻥ ﺍﻟﻌﺩﺩﺍﻥ ‪ Z′ , Z‬ﺤﻴﺙ ‪:‬‬ ‫‪ Z = 5 + 2i‬ﻭ ‪Z′ = 3 - i‬‬ ‫‪ -1‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻋﻤﻠﻴﺘﻲ ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻀﺭﺏ ﻓﻲ ‪ E‬ﻟﻬﺎ ﻨﻔﺱ ﺨﻭﺍﺹ‬ ‫ﻋﻤﻠﻴﺘﻲ ﺍﻟﺠﻤﻊ ﻭﺍﻟﻀﺭﺏ ﻓﻲ \ ‪ ،‬ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪:‬‬‫‪Z × Z′ ; Z2; 2Z -3Z′ ; 8Z ; Z +Z′ ; (Z- Z′)2 ; (2Z +Z′)2‬‬ ‫‪1‬‬ ‫=‬ ‫‪5 - 2i‬‬ ‫ﻋﻠﻰ ﺍﻟﺸﻜل ‪:‬‬ ‫‪1‬‬ ‫‪ (2‬ﺒﻜﺘﺎﺒﺔ‬ ‫‪Z‬‬ ‫)‪(5 + 2i) (5 - 2i‬‬ ‫‪Z‬‬ ‫‪1‬‬ ‫‪=α‬‬ ‫‪+ iβ‬‬ ‫ﻋﻴﻥ ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ α‬و ‪ β‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪Z‬‬ ‫‪1‬‬‫‪.‬‬ ‫‪Z′‬‬ ‫‪= a + ib‬‬ ‫ﻭﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ ﻋﻴﻥ ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ a‬ﻭ ‪ b‬ﺒﺤﻴﺙ‪:‬‬ ‫‪ (3‬ﺍﺤﺴﺏ ‪ i 2008‬ﺜﻡ ‪ in‬ﺒﺩﻻﻟﺔ ‪( ). n‬‬ ‫‪ (4‬ﺍﺴﺘﻨﺘﺞ ﻁﺭﻴﻘﺔ ﻟﺤﺴﺎﺏ ‪( ). 1 + i 2008‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪ (1‬ﺍﻟﺤﺴﺎﺏ ‪:‬‬‫‪• Z + Z′ = (5 + 2i) + (3 - i) = (5 + 3) + (2 - 1) i = 8 + i‬‬‫‪• 8Z = 8 (5 +2i) = 40 + 16i‬‬‫‪• 2Z - 3Z′ = 2 (5 + 2i) - 3 (3 - i) = 10 + 4i - 9 + 3i = 1 + 7i‬‬‫‪• Z2 = (5 + 2i)2 = (5)2 + 2 × 5 × 2i + (2i)2 = 21 + 20i‬‬‫‪• Z . Z′ = (5 +2i) (3 - i)= 15 - 5i + 6i - 2i2 = 17 + i‬‬‫‪• (2Z + Z′)2 = [2 (5 + 2i) + 3 - i]2 = (10 + 4i + 3 - i)2‬‬

= (13 + 3i)2 = (13)2 + 2 × 13 × 3i + (3i)2 = 169 + 78i -9 = 160 + 78i• (Z - Z′)2 = (5 + 2i - 3 + i)2 = (2 + 3i)2 = (2)2 + 2.2.3i + (3i)2 = 4 + 12i - 9 = -5 + 12i 1 = 5 - 2i = 5 - 2i (2 Z (5 + 2i) (5 - 2i) (5)2 - (2i)2 1 = 5 - 2i = 5 - 2 i : ‫ﻭ ﻤﻨﻪ‬ Z 25 + 4 29 29 -2 5 β = 29 ‫و‬ α = 29 : ‫ﺇﺫﻥ‬1 = 1 = 3+i = 3 + i = 3+i = 3+iZ′ 3-i (3 - i) (3 + i) (3)2 - (i)2 9+1 10 . b = 1 ‫و‬ a = 3 : ‫ﺇﺫﻥ‬ 1 = 3 + 1 i : ‫ﻭ ﻤﻨﻪ‬ 10 10 Z′ 10 10 ( ): i 2008 ‫ ﺤﺴﺎﺏ‬- (3( )( ) ( )i 2008 = i2 1004 = -1 1004 = 1 (i)n )2 n n = ( i  2 = ( -1) 2 : in ‫ ﺤﺴﺎﺏ‬-  ( )1 + i 2008 ‫( ﺍﺴﺘﻨﺘﺎﺝ ﻁﺭﻴﻘﺔ ﻟﺤﺴﺎﺏ‬4 2 1004 ( )=1004 1 +2i -1 1004( )(1 + i)2008 = (1 + i )  = 1 + 2i + i2  = 21004 i2 502 = 21004 . -1 502 = 21004 ( ) ( ) ( )=2i 1004 .

‫‪ – 1‬ﺘﻌﺭﻴﻑ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ‪:‬‬ ‫‪( )G G‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﺯﻭﺩ ﺒﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ‪O ; i , j‬‬ ‫‪ -‬ﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺘﻤﺜل ﻋﺩﺩ ﻤﺭﻜﺏ ﻭﻋﺩﺩ ﻤﺭﻜﺏ ﻭﺤﻴﺩ‪.‬‬ ‫ﻭﻜل ﻋﺩﺩ ﻤﺭﻜﺏ ﻴﻤﺜل ﺒﻨﻘﻁﺔ ﻭ ﺒﻨﻘﻁﺔ ﻭﺤﻴﺩﺓ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ‪.‬‬ ‫‪ -‬ﺍﻟﻨﻘﻁﺔ )‪ J(o ; 1‬ﺘﻤﺜل ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ﺍﻟﺫﻱ ﻨﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ‪. i‬‬ ‫‪ -‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ x‬ﻭ ‪ ، y‬ﻴﺭﻤﺯ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ﺍﻟﻤﻤﺜل‬ ‫ﺒﺎﻟﻨﻘﻁﺔ ‪ M x ; y‬ﺒﺎﻟﺭﻤﺯ ‪( ). x + iy‬‬ ‫‪ -‬ﻴﺭﻤﺯ ﻟﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﺒﺎﻟﺭﻤﺯ ^ ‪.‬‬ ‫‪ -2‬ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ‪:‬‬‫ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ x‬ﻭ ‪ : y‬ﺍﻟﺸﻜل ‪ x + iy‬ﻴﺴﻤﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ‪. Z‬‬ ‫‪ -3‬ﺘﻌﺎﺭ ﻴﻑ ﻭ ﻤﺼﻁﻠﺤﺎﺕ ‪:‬‬ ‫ﻟﻴﻜﻥ ‪ Z = x + iy‬ﻋﺩﺩ ﻤﺭﻜﺏ ‪ x ،‬ﻭ ‪ y‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ‬ ‫‪ -‬ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ‪ x‬ﻴﺴﻤﻰ ﺍﻟﺠﺯﺀ ﺃﻭ ﺍﻟﻘﺴﻡ ﺍﻟﺤﻘﻴﻘﻲ ﻟﻠﻌﺩﺩ‬ ‫ﺍﻟﻤﺭﻜﺏ ‪ Z‬ﻭ ﻨﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ‪ Re Z‬ﺃﻱ ‪( ) ( )Re Z = x :‬‬ ‫‪ -‬ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ‪ y‬ﻴﺴﻤﻰ ﺍﻟﺠﺯﺀ ﺃﻭ ﺍﻟﻘﺴﻡ ﺍﻟﺘﺨﻴﻠﻲ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪Z‬‬ ‫ﻭﻴﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ‪ Im Z‬ﺃﻱ ‪. Im ( Z) = y ( ):‬‬ ‫‪ -‬ﺍﻟﻨﻘﻁﺔ )‪ M ( x ; y‬ﺘﺴﻤﻰ ﺼﻭﺭﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪Z‬‬ ‫ﻭ ﺍﻟﻌﺩﺩ ‪ Z‬ﻴﺴﻤﻰ ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ )‪M ( x ; y‬‬

‫‪ -‬ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ y′ , x′ , y , x‬ﻓﺈﻥ ﺍﻟﻌﺩﺩ ‪ x + iy‬ﻴﺴﺎﻭﻱ‬ ‫ﺍﻟﻌﺩﺩ ‪ x′ + iy′‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪ x = x′ :‬ﻭ ‪. y = y′‬‬ ‫‪ -‬ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻫﻭ ﻋﺩﺩ ﻤﺭﻜﺏ ﻭ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫\ ∈ ‪ Z‬ﻴﻜﺎﻓﺊ‪. Im ( Z) = 0 :‬‬ ‫‪ -‬ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z‬ﺘﺨﻴﻠﻲ ﺼﺭﻑ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ‪( )Re Z = 0 :‬‬ ‫‪ -‬ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻴﺩﻋﻰ ﺍﻟﻤﺤﻭﺭ ﺍﻟﺤﻘﻴﻘﻲ ﻭ ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ ﻴﺩﻋﻰ‬ ‫ﺍﻟﻤﺤﻭﺭ ﺍﻟﺘﺨﻴﻠﻲ ‪.‬‬ ‫‪ -‬ﺇﺫﺍ ﻜﺎﻥ ‪ Z = 0‬ﻓﺈﻥ ‪ Z‬ﺤﻘﻴﻘﻲ ﻭ ﺘﺨﻴﻠﻲ ﺼﺭﻑ ﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﻭ ﻴﻤﺜل‬ ‫ﺒﺎﻟﻨﻘﻁﺔ )‪. O (0 ; 0‬‬ ‫‪ -4‬ﺍﻟﺤﺴﺎﺏ ﻓﻲ ‪:C‬‬ ‫‪ -‬ﺍﻟﻤﺠﻤﻭﻉ ﻭ ﺍﻟﺠﺩﺍﺀ ﻓﻲ ^ ‪:‬‬ ‫ﺍﻟﻤﺠﻤﻭﻋﺔ ^ ﻤﺯﻭﺩﺓ ﺒﻌﻤﻠﻴﺘﻴﻥ ﻫﻤﺎ ﺍﻟﺠﻤﻊ ‪ +‬ﻭ ﺍﻟﻀﺭﺏ × ﻤﻌﺭﻓﺘﺎﻥ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ‬ ‫‪ Z‬و ‪ Z′‬ﺤﻴﺙ‪:‬‬ ‫‪ Z = x + iy‬ﻭ ‪ Z′ = x′ + iy′‬ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬‫)‪Z + Z′ = (x + x′) + i (y + y′‬‬‫)‪Z × Z′ = (xx′ - yy′) + i (xy′ + x′y‬‬ ‫ﻫﺎﺘﻴﻥ ﺍﻟﻌﻤﻠﻴﺘﻴﻥ ﻟﻬﻤﺎ ﻨﻔﺱ ﺨﻭﺍﺹ ﺍﻟﺠﻤﻊ ‪ +‬ﻭ ﺍﻟﻀﺭﺏ × ﻓﻲ \‬ ‫‪ -‬ﻗﻭﻯ ﻋﺩﺩ ﻤﺭﻜﺏ ‪:‬‬ ‫ﺍﻟﻘﻭﻯ ﺍﻟﺼﺤﻴﺤﺔ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ﻟﻬﺎ ﻨﻔﺱ ﺨﻭﺍﺹ ﺍﻟﻘﻭﻯ ﺍﻟﺼﺤﻴﺤﺔ ﻟﻌﺩﺩ ﺤﻘﻴﻘﻲ ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫)‪i2 = (0 + 1 . i) × (0 + 1 . i‬‬ ‫‪= (0 - 1) + i (0 . 1 + 0 . 1) = -1‬‬ ‫ﻭﻋﻠﻴﻪ ‪i2 = -1 :‬‬

‫ﺃﻤﺜﻠﺔ ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﻤﺭﻜﺒﺎﻥ‪. Z2 = -4 + 5i ; Z1 = 3 + 2i :‬‬ ‫‪ (1‬ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪. Z1 × Z2 , Z1 + Z2‬‬ ‫‪.‬‬ ‫‪Z‬‬ ‫‪3‬‬ ‫;‬ ‫‪Z12‬‬ ‫‪ (2‬ﺍﺤﺴﺏ‬ ‫‪2‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪Z1 + Z2 = (3 - 4) + i (2 + 5) • (1‬‬ ‫ﻭﻤﻨﻪ ‪Z1 + Z2 = - 1 + 7i :‬‬ ‫)‪Z1 × Z2 = (3 + 2i) (-4 + 5i‬‬ ‫‪x‬‬ ‫‪= -12 + 15i - 8i + 10i2 = -12 + 7i - 10‬‬ ‫ﻭ ﻤﻨﻪ ‪Z1 × Z2 = -22 + 7i :‬‬ ‫‪Z12 = (3 + 2i)2 = (3)2 + 2(3) × 2i + (2i)2‬‬ ‫‪(2‬‬ ‫ﻭ ﻤﻨﻪ ‪ Z12 = 9 + 12i - 4 :‬ﺃﻱ‪Z12 = 5 + 12i :‬‬‫‪Z‬‬ ‫‪3‬‬ ‫‪= (-4 + 5i)3‬‬ ‫‪= (-4)3‬‬ ‫‪+ 3(-4)2‬‬ ‫×‬ ‫‪5i + 3(-4) (5i)2‬‬ ‫‪+ (5i)3‬‬ ‫‪2‬‬ ‫‪= -64 + 240i + 300 - 125i = 236 + 115i‬‬ ‫‪( JJJJG JJJJG‬‬ ‫ﺨﻭﺍﺹ ‪:‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ ‪ Z‬ﻭ ‪ Z′‬ﻻﺤﻘﺘﻲ ﺍﻟﻨﻘﻁﺘﻴﻥ ‪ M‬ﻭ ‪) M′‬ﺃﻭ ﺍﻟﺸﻌﺎﻋﻴﻥ ‪ OM‬و ‪ OM′‬ﻋﻠﻰ‬ ‫‪M′ S‬‬ ‫ﺍﻟﺘﺭﺘﻴﺏ ﻓﺈﻥ ‪:‬‬ ‫‪ Z + Z′ x‬ﻫ‪JG‬ﻭ‪J‬ﻻ‪J‬ﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪S‬‬ ‫)ﺃ‪G‬ﻭ‪ J‬ﺍ‪J‬ﻟ‪J‬ﺸﻌ‪J‬ﺎﻉ ‪JJ( OJJGS‬ﺤﻴﺙ‪M JJJG :‬‬ ‫‪O JJJG OS = OM + OM′‬‬ ‫‪JGZ - Z′ x‬ﻫ‪J‬ﻭ‪JJ‬ﻻﺤ‪J‬ﻘﺔ ﺍﻟﻨﻘ‪JG‬ﻁ‪J‬ﺔ‪) DJJ‬ﺃﻭ ﺍﻟ‪JG‬ﺸﻌ‪J‬ﺎ‪J‬ﻉ‪JJ(JOG DJ‬ﺤﻴ‪J‬ﺙ‪M′′ D JJJG :‬‬‫‪JJJG OD = OM - OM′ = OM + OM′′‬‬‫ﻭﻋﻠﻴﻪ ﺇﺫﺍ ﻜﺎﻨﺕ ‪ A‬ﻭ ‪ B‬ﻨﻘﻁﺘﺎﻥ ﻻﺤﻘﺘﺎﻫﻤﺎ ‪ ZA‬ﻭ ‪ ZB‬ﻋﻠﻲ ﺍﻟﺘﺭﺘﻴﺏ ﻓﺈﻥ ﻻﺤﻘﺔ ﺍﻟﺸﻌﺎﻉ ‪AB‬‬ ‫‪ZJJJG‬‬ ‫ﺍﻟﻤﺭﻜﺏ‬ ‫ﺍﻟﻌﺩﺩ‬ ‫ﻫﻭ‬ ‫‪AB‬‬

‫‪Z JJJG‬‬ ‫‪= ZB‬‬ ‫‪- ZA‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪AB‬‬ ‫‪ZI‬‬ ‫=‬ ‫‪ZA‬‬ ‫‪+ ZB‬‬ ‫ﻭﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪ I‬ﻤﻨﺘﺼﻑ ‪ AB‬ﻫﻭ ‪ ZI‬ﺤﻴﺙ] [‬ ‫‪2‬‬ ‫‪ -‬ﻤﻘﻠﻭﺏ ﻋﺩﺩ ﻤﺭﻜﺏ ‪:‬‬ ‫‪ Z‬ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ‪ .‬ﺤﻴﺙ ‪. Z = x + iy‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫)‪( x - iy‬‬ ‫=‬ ‫‪x - iy‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪Z‬‬ ‫‪x + iy‬‬ ‫)‪(x + iy) (x - iy‬‬ ‫‪x2 + y2‬‬ ‫‪1‬‬ ‫=‬ ‫‪x2‬‬ ‫‪x‬‬ ‫‪-i‬‬ ‫‪y‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪Z‬‬ ‫‪+ y2‬‬ ‫‪x2 + y2‬‬ ‫ﻭﻫﻭ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﻟﻤﻘﻠﻭﺏ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z‬ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ‪.‬‬ ‫ﺃﻱ ‪ x‬ﻭ‪ y‬ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ﻤﻌﺎ‪.‬‬ ‫‪ -‬ﺤﺎﺼل ﻗﺴﻤﺔ ﻋﺩﺩﻴﻥ ﻤﺭﻜﺒﻴﻥ ‪:‬‬ ‫‪ Z‬ﻭ ‪ Z′‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﺤﻴﺙ‪Z′ ≠ 0 :‬‬ ‫ﻤﻊ ‪ Z = x + iy‬ﻭ ‪Z′ = x′ + iy′‬‬‫‪Z‬‬ ‫‪=Z‬‬ ‫×‬ ‫‪1‬‬ ‫=‬ ‫‪(x‬‬ ‫)‪+iy‬‬ ‫×‬ ‫‪ x′‬‬ ‫‪-i‬‬ ‫‪y′ ‬‬‫‪Z′‬‬ ‫‪Z′‬‬ ‫‪ x′2 + y′2‬‬ ‫‪x′2 + y′2 ‬‬ ‫=‬ ‫‪xx′‬‬ ‫‪-i‬‬ ‫‪xy′‬‬ ‫‪+i‬‬ ‫‪x′y‬‬ ‫‪+‬‬ ‫‪yy′‬‬ ‫‪x′2 + y′2‬‬ ‫‪x′2 + y′2‬‬ ‫‪x′2 + y′2‬‬ ‫‪x′2 + y′2‬‬ ‫=‬ ‫‪xx′ + yy′‬‬ ‫‪+i‬‬ ‫‪x′y - xy′‬‬ ‫‪x′2 + y′2‬‬ ‫‪x′2 + y′2‬‬ ‫‪Z‬‬ ‫ﻭﻫﻭ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‬‫‪ Z′‬ﺃﻱ ﺤﺎﺼل ﻗﺴﻤﺔ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z‬ﻋﻠﻰ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ﻏﻴﺭ‬ ‫ﺍﻟﻤﻌﺩﻭﻡ ‪. Z′‬‬

‫‪ -5‬ﻤﺭﺍﻓﻕ ﻋﺩﺩ ﻤﺭﻜﺏ ‪:‬‬ ‫ﺘﻌﺭﻴﻑ ‪:‬‬‫ﻟﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z = x + iy‬ﺤﻴﺙ ‪ x‬ﻭ‪ y‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﻨﻅﻴﺭﺓ‬‫ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻫﻲ ﺍﻟﻨﻘﻁﺔ ‪ M′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ . x - i y‬ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪x - iy‬‬ ‫ﻴﺴﻤﻰ ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ x + i y‬ﻭﻨﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ‪ Z‬ﺃﻱ ‪Z = x - i y :‬‬ ‫ﺃﻤﺜﻠﺔ ‪:‬‬ ‫ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪ Z1 = 1 + i :‬ﻫﻭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪Z1 = 1 - i :‬‬ ‫ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪ Z2 = i :‬ﻫﻭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪Z2 = -i :‬‬ ‫ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪ Z3 = 8 + 3 i :‬ﻫﻭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪:‬‬ ‫‪Z3 = 8 - 3 i‬‬ ‫ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪ Z4 = 10 :‬ﻫﻭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‪Z4 = 10 :‬‬ ‫ﺃﻱ ﺃﻥ ‪Z4 = Z4 :‬‬ ‫ﺨﻭﺍﺹ ‪:‬‬ ‫‪ x (a‬ﻭ ‪y‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ Z = x + iy .‬ﻋﺩﺩ ﻤﺭﻜﺏ ‪.‬‬ ‫‪ Z = x - iy‬ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪. Z‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪ Z = x + iy :‬ﻭﻤﻨﻪ ‪Z = x + iy :‬‬ ‫ﻭﻋﻠﻴﻪ ‪Z = Z :‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪ Z + Z = 2 x :‬ﻭﻤﻨﻪ ‪Z + Z = 2 Re( Z) :‬‬ ‫‪ Z - Z = 2 i y (3‬ﻭﻤﻨﻪ ‪Z - Z = 2 Im ( Z) :‬‬ ‫‪Z . Z = x2 + y2 (4‬‬ ‫‪ Z ∈ \ (5‬ﺘﻜﺎﻓﺊ ‪Z = Z‬‬ ‫‪ Z (6‬ﺘﺨﻴﻠﻲ ﺼﺭﻑ ﻴﻜﺎﻓﺊ ‪Z = -Z :‬‬ ‫‪ y′, y , x′, x (b‬ﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ‪ Z2 , Z1 .‬ﻋﺩﺍﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﺤﻴﺙ ‪:‬‬

Z2 = x′ + i y′ ‫ ؛‬Z1 = x + i y (1 Z1+ Z2 = ( x + x′ + i (y + y′)) = x + x′ − i (y + y′) = x - i y + x′ - i y′ Z1 + Z2 = Z1 + Z2 : ‫ﻭﻤﻨﻪ‬ (2 Z1 . Z2 = ( xx′ - yy′) + i ( xy′ + x′y) = ( xx′ - yy′) - i ( xy′ + x′y )Z1 . Z2 = ( x - iy) .( x′ - iy) = ( xx′ - yy′) - i ( xy′ + x′y ) Z1 . Z2 = Z1 . Z2 : ‫ﻭﻋﻠﻴﻪ‬  1  =  x2 x -i y (3  Z1   + y2 x2 + y2    = x2 x +i y + y2 x2 + y21 = 1 = x + iy = x2 x +i yZ1 x -iy x2 + y2 + y2 x2 + y2 1 = 1 : ‫ﻭﻋﻠﻴﻪ‬   Z1  Z1   Z1  = Z1 (4  Z2  Z2   ( )Z1n = Z1 n : n ∈ `* ‫( ﻤﻥ ﺃﺠل‬5 ( )Z1n = Z1 n : n ∈ ` ‫ ﻭ‬Z1 ≠ 0 :‫ﻭﺇﺫﺍ ﻜﺎﻥ‬

‫ﺃﻤﺜﻠﺔ ‪:‬‬‫)‪1) (1 + 2i) (3 - i) = (1 + 2i) (3 - i‬‬ ‫)‪= (1 - 2i) (3 + i‬‬‫)‪2‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪3+i‬‬ ‫‪3-i‬‬‫)‪3‬‬ ‫‪‬‬ ‫‪2 + 3i‬‬ ‫‪‬‬ ‫=‬ ‫)‪(2 + 3i‬‬ ‫=‬ ‫‪2 - 3i‬‬ ‫‪‬‬ ‫‪5+i‬‬ ‫‪‬‬ ‫)‪(5 + i‬‬ ‫‪5-i‬‬‫)‪4‬‬ ‫‪‬‬ ‫‪a+b‬‬ ‫‪‬‬ ‫=‬ ‫‪a‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪‬‬ ‫‪1 - ab‬‬ ‫‪‬‬ ‫‪1-‬‬ ‫‪a‬‬ ‫‪.b‬‬ ‫ﺤﻴﺙ‪ a :‬ﻭ‪ b‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻤﻊ ‪. ab ≠ 1‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬ ‫‪ M‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﻻﺤﻘﺘﻬﺎ ‪ M′ ، Z = x + iy‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﻻﺤﻘﺘﻬﺎ‬ ‫= ‪Z′‬‬ ‫‪Z+1‬‬ ‫‪Z-1‬‬ ‫‪ (1‬ﺍﻜﺘﺏ ‪ Z′‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ‪.‬‬ ‫‪ (2‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ Z′‬ﺤﻘﻴﻘﻲ ‪.‬‬ ‫‪ (3‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ Z′‬ﺘﺨﻴﻠﻲ ﺼﺭﻑ ‪.‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪ (1‬ﻜﺘﺎﺒﺔ ‪ Z′‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ‪:‬‬‫= ‪Z′‬‬ ‫‪x + iy + 1‬‬ ‫=‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1 + iy‬‬ ‫=‬ ‫)‪( x + iy + 1) (x - 1 - iy‬‬ ‫‪x + iy - 1‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1+ iy‬‬ ‫)‪( x + iy - 1) (x - 1 - iy‬‬‫= ‪Z′‬‬ ‫‪x2 + y2 - 1‬‬ ‫‪+i‬‬ ‫‪-2y‬‬ ‫‪(x - 1)2 + y2‬‬ ‫‪(x - 1)2 + y2‬‬ ‫‪ (2‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ Z′‬ﺤﻘﻴﻘﻲ ‪:‬‬

‫‪-2y‬‬ ‫‪=0‬‬ ‫‪ Z′‬ﺤﻘﻴﻘﻲ ﻴﻜﺎﻓﺊ ‪:‬‬‫‪(x - 1)2 + y2‬‬ ‫‪y = 0‬‬‫ﻭﻴﻜﺎﻓﺊ ‪. (x ; y) ≠ (1 , 0) :‬‬‫ﻭﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻫﻲ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﻨﻘﻁﺔ‬ ‫)‪. A (1 ; 0‬‬‫‪ (3‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ Z′‬ﺘﺨﻴﻠﻲ ﺼﺭﻑ ‪:‬‬ ‫‪x2 + y2 - 1‬‬ ‫‪ Z′‬ﺘﺨﻴﻠﻲ ﺼﺭﻑ ﻴﻜﺎﻓﺊ ‪= 0 :‬‬‫‪(x - 1)2 + y2‬‬‫ﻭﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ O‬ﻭﻨﺼﻑ‬ ‫‪x2 + y2 = 1‬‬ ‫ﻭﻴﻜﺎﻓﺊ ‪(x ; y) ≠ (1 , 0) :‬‬ ‫ﻗﻁﺭﻫﺎ ‪ 1‬ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﻨﻘﻁﺔ )‪. A (1 ; 0‬‬ ‫‪ -6‬ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﻋﺩﺩ ﻤﺭﻜﺏ ‪:‬‬ ‫‪( )G G‬‬‫ﺍﻟﻤﺴﺘﻭﻯ ﻤﻨﺴﻭﺏ ﻓﻲ ﻤﺎ ﻴﻠﻲ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭ ﻤﺘﺠﺎﻨﺱ ﻭ ﻤﺒﺎﺸﺭ ‪ M . O ; i , j‬ﻨﻘﻁﺔ ﻤﻥ‬‫ﺍﻟﻤﺴﺘﻭﻱ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ ﺍﻟﻘﻁﺒﻴﺎﻥ ‪ ρ ; θ‬ﺤﻴﺙ ‪ ρ‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﻭ ‪ θ‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻭ ﻋﻠﻴﻪ ‪[ ]Z‬‬‫)‪ρ (cosθ + i sinθ‬‬ ‫ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪ M‬ﻴﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ‪:‬‬‫‪ ρ (cosθ + i sinθ) x‬ﻴﺴﻤﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻠﻌﺩﺩ ‪Z‬‬‫‪ x‬ﻨﺼﻑ ﺍﻟﻘﻁﺭ ﺍﻟﻘﻁﺒ‪JG‬ﻲ‪ OJMJJ‬ﻴ‪G‬ﺤﻘﻕ ‪ OM = ρ‬ﻭﻴﺴﻤﻰ ﻁﻭﻴﻠ‪JG‬ﺔ‪JZJJ‬ﻭﻨﺭﻤ‪G‬ﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ‪. Z‬‬‫‪ x‬ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻘﻁﺒﻴﺔ ‪ i ; OM‬ﺘﺤﻘﻕ ‪ i ; OM = θ + 2kπ‬ﺤﻴﺙ ] ∈ ‪( ) ( )k‬‬ ‫ﻭ ﺘﺴﻤﻰ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ . Z‬ﻭ ﻨﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ‬‫)‪ arg (Z‬ﻭﻨﻜﺘﺏ ‪ arg(Z) = θ 2π :‬ﻭ ﺘﻘﺭﺃ ‪ θ‬ﺒﺘﺭﺩﻴﺩ ‪[ ]2π‬‬

‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬ ‫ﻟﺘﻜﻥ ‪ M‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ‬ ‫‪y‬‬ ‫‪M‬‬ ‫ﺍﻟﻘﻁﺒﻴﺎﻥ ‪ ρ ; θ‬ﻓﻴﻜﻭﻥ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ] [‬ ‫‪ρ‬‬‫‪G‬‬ ‫)‪ (x ; y‬ﻤﻌﺭﻓﺎﻥ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬‫‪j‬‬ ‫‪G‬‬ ‫‪θ‬‬ ‫‪ x = ρ cosθ‬‬ ‫‪i‬‬ ‫‪y = ρ sinθ‬‬‫‪O‬‬ ‫‪x‬‬ ‫ﻭﻋﻠﻴﻪ ﺇﺫﺍ ﻜﺎﻥ ‪ Z‬ﻻﺤﻘﺔ ‪ M‬ﻓﺈﻥ ‪:‬‬ ‫‪Z = x + iy = ρ cosθ + iρ sinθ‬‬ ‫ﻭﻤﻨﻪ ‪Z = ρ (cosθ + i sinθ) :‬‬ ‫‪JJJJG‬‬ ‫‪JJJJG‬‬ ‫ﻤﻼﺤﻅﺎﺕ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ OM = x2 + y2 :‬ﻭﻋﻠﻴﻪ ‪Z = OM = ρ :‬‬ ‫= ‪cosθ‬‬ ‫‪x‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪x2 + y2‬‬ ‫‪‬‬ ‫= ‪sinθ‬‬ ‫‪y‬‬ ‫‪x2 + y2‬‬ ‫ﻭﺇﺫﺍ ﻜﺎﻥ ‪ Z = 0‬ﻓﺈﻥ ‪ ρ = 0 :‬ﻟﻜﻥ ‪ Z‬ﻟﻴﺱ ﻟﻪ ﻋﻤﺩﺓ‪.‬‬ ‫ﺃﻤﺜﻠﺔ ‪:‬‬ ‫ﻋﻴﻥ ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻵﺘﻴﺔ ‪:‬‬ ‫‪Z3 = 3 - i ; Z2 = i ; Z1 = 1 + i‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪Z1 = (1)2 + (1)2 = 2 ; Z1 = 1 + i x‬‬‫= ‪sinθ1‬‬ ‫‪y‬‬ ‫و‬ ‫= ‪cosθ1‬‬ ‫‪x‬‬ ‫ﻨﻔﺭﺽ ‪ θ1‬ﻋﻤﺩﺓ ‪ Z1‬ﻓﻴﻜﻭﻥ ‪:‬‬ ‫‪Z1‬‬ ‫‪Z1‬‬

sinθ1 = 1= 2 ‫ و‬cosθ1 = 1= 2 2 2 2 2 : ‫ﻭﻤﻨﻪ‬ θ1 = π + 2kπ ; k∈] : ‫ﺇﺫﻥ‬ 4 Z2 = 02 + (1)2 = 1 ; Z2 = i xsinθ2 = y ‫و‬ cosθ2 = x : ‫ ﻓﻴﻜﻭﻥ‬Z2 ‫ ﻋﻤﺩﺓ‬θ2 ‫ﻨﻔﺭﺽ‬ Z2 Z2 sinθ2 = 1 , cosθ2 = 0 =0 1 1 θ2 = π + 2kπ ; k∈] : ‫ﺇﺫﻥ‬ 2 ( )Z3 = 3 2 + (-1)2 = 2 ; Z3 = 3 - i xsinθ3 = y , cosθ3 = x : ‫ ﻓﻴﻜﻭﻥ‬Z3 ‫ ﻋﻤﺩﺓ‬θ3 ‫ ﻨﻔﺭﺽ‬x Z3 Z3 sinθ3 = -1 ‫و‬ cosθ3 = 3 : ‫ﺇﺫﻥ‬ 2 2 θ3 = −π + 2kπ ; k∈ ] : ‫ﻭﻋﻠﻴﻪ‬ 6 : ‫ﺨﻭﺍﺹ‬ .‫ ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ‬Z (Aarg(Z) = 0 + 2kπ ; k ∈ ] : ‫ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﻴﻜﺎﻓﺊ‬Z (1arg(Z) = π + 2kπ ; k ∈ ] : ‫ ﺤﻘﻴﻘﻲ ﺴﺎﻟﺏ ﻴﻜﺎﻓﺊ‬Z (2 : ‫ ﻴﻜﺎﻓﺊ‬Re(Z) = 0 ‫ ﻭ‬Im(Z) > 0 (3 arg(Z) = π + 2kπ ; k ∈ ] 2

‫‪ Im(Z) < 0 (4‬ﻭ ‪ Re(Z) = 0‬ﻴﻜﺎﻓﺊ ‪:‬‬‫)‪arg(Z‬‬ ‫=‬ ‫‪-‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫;‬ ‫‪k‬‬ ‫∈‬ ‫]‬ ‫‪2‬‬ ‫ﺃﻤﺜﻠﺔ ‪:‬‬ ‫ﻋﻴﻥ ﻋﻤﺩﺓ ﻜﻼ ﻤﻥ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻵﺘﻴﺔ ﺩﻭﻥ ﺤﺴﺎﺏ‬‫‪Z4 = -2i ; Z3 = 5i ; Z2 = -4 ; Z1 = 3‬‬ ‫ﺍﻟﺤل ‪:‬‬‫‪ Z1 = 3 x‬ﻭﻤﻨﻪ ‪arg(Z1 ) = 0 + 2kπ ; k ∈ ] :‬‬‫‪ Z2 = -4 x‬ﻭﻤﻨﻪ‪arg(Z2 ) = π + 2kπ ; k ∈ ] :‬‬‫) ‪arg(Z3‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫‪;k‬‬ ‫∈‬ ‫]‬ ‫ﻭﻤﻨﻪ‪:‬‬ ‫‪Z3 = 5i‬‬ ‫‪x‬‬ ‫‪2‬‬‫= ) ‪arg(Z4‬‬ ‫‪-‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫∈ ‪;k‬‬ ‫]‬ ‫‪ Z4 = -2i‬ﻭﻤﻨﻪ‪:‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪ (B‬ﻤﺭﺍﻓﻕ ﻋﺩﺩ ﻤﺭﻜﺏ ‪:‬‬ ‫ﻟﺘﻜﻥ ‪ M′ , M‬ﺼﻭﺭﺘﻲ ‪ Z‬و ‪ Z‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‬ ‫ﻟﺩﻴﻨﺎ ‪ M‬و ‪ M′‬ﻤﺘﻨﺎﻅﺭﺘﺎﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻭ ﻋﻠﻴﻪ ‪:‬‬‫‪ Z = Z‬ﻭ ] ∈ ‪arg(Z) = -arg(Z) + 2kπ ; k‬‬ ‫‪ (C‬ﺠﺩﺍﺀ ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ‪:‬‬ ‫‪ Z , Z‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ﺤﻴﺙ ‪:‬‬ ‫)‪ Z = ρ (cosθ + i sinθ‬ﻭ )‪Z′ = ρ′ (cosθ′ + i sinθ′‬‬‫])‪Z . Z′ =ρρ′[cosθ.cosθ′- sinθ.sinθ′+i (sinθ.cosθ′+cosθ.sinθ′‬‬‫ﻭﻤﻨﻪ ‪ZZ′ = ρρ′ [cos (θ + θ′) + i sin (θ + θ′)] :‬‬ ‫‪Z . Z′ = Z . Z′‬‬ ‫ﺇﺫﻥ ‪:‬‬‫] ∈ ‪arg(Z . Z′) = arg(Z) + arg(Z′) + 2kπ ; k‬‬ ‫‪ (D‬ﻤﻘﻠﻭﺏ ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ ‪Z = ρ (cosθ + i sinθ) :‬‬

1 = 1 = cosθ - i sinθZ ρ(cosθ + i sinθ) ρ(cosθ+i sinθ) (cosθ- i sinθ)1 = cosθ - i sinθZ ρ(cos2θ + i sin2θ) 1 = 1 [cos(-θ) + i sin(-θ)] : ‫ﺇﺫﻥ‬ Z ρ 1 = 1 ‫ﻭ‬ arg  1  = -arg(Z) + 2kπ : ‫ﻭﻋﻠﻴﻪ‬ Z Z  Z  : ‫( ﺤﺎﺼل ﻗﺴﻤﺔ ﻋﺩﺩﻴﻥ ﻤﺭﻜﺒﻴﻥ‬E Z′ ≠ 0 ‫ ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﺤﻴﺙ‬Z′ ‫ ﻭ‬ZZ = Z. 1 = Z . 1 = Z . 1Z′ Z′ Z′ Z′ Z = Z : ‫ﻭﻤﻨﻪ‬ Z′ Z′ arg  Z  = arg  Z × 1  = arg(Z) + arg  1  Z′   Z′   Z′  arg  Z  = arg(Z) - arg(Z′) : ‫ﻭﻤﻨﻪ‬  Z′  : ‫( ﺘﺴﺎﻭﻱ ﻋﺩﺩﻴﻥ ﻤﺭﻜﺒﻴﻥ‬F : ‫ ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ﺤﻴﺙ‬Z′ ‫ ﻭ‬Z Z′ = ρ′ (cosθ′ + i sinθ′) ‫ ﻭ‬Z = ρ (cosθ + i sinθ) θ = θ′ + 2kπ ; k ∈ ] ‫ ﻭ‬ρ = ρ′ : ‫ ﻴﻜﺎﻓﺊ‬Z = Z′ : Zn ‫( ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ‬G .‫ ﻋﺩﺩ ﺼﺤﻴﺢ‬n ، ‫ ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ‬Z

arg(Zn ) = n . arg(Z) ‫ ﻭ‬Zn = Z n ‫ ﻟﺩﻴﻨﺎ‬x : ‫ﺍﻟﺒﺭﻫﺎﻥ‬ n ≥ 2 ‫ ﻨﺒﺭﻫﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻤﻥ ﺃﺠل‬n ∈ ] + ‫ﻤﻥ ﺃﺠل‬ Z2 = Z 2 : n = 2 ‫ ﻟﻤﺎ‬x .‫ ﻜﻤﺎ ﺴﺒﻕ‬arg(Z2 ) = arg(Z) + arg(Z) = 2arg(Z) : ‫ ﺃﻱ‬k ‫ ﻨﻔﺭﺽ ﺼﺤﺔ ﺍﻟﺨﺎﺼﻴﺔ ﺇﻟﻰ ﺭﺘﺒﺔ‬x arg(Zk ) = k arg(Z) ‫ ﻭ‬Zk = Z k : k + 1 ‫ﻭﻨﺒﺭﻫﻥ ﺼﺤﺘﻬﺎ ﻓﻲ ﺍﻟﺭﺘﺒﺔ‬Zk+1 = Zk . Z = Zk . Z = Z k . Z = Z k+1arg(Zk+1 ) = arg(Zk . Z) = arg(Zk ) + arg(Z) = k arg(Z) + arg(Z) = (k + 1) arg(Z) n = -p ‫ ﺒﻭﻀﻊ‬: n ∈ ] − ‫ﻤﻥ ﺃﺠل‬Zn = Z-p = 1 = 1 = 1 = 1 = Zn Zp Zp Zp Z -n( ) ( )arg Zn  1  = arg Z-p = arg  Zp  = -arg( Zp ) = -p arg(Z) = n arg(Z) : ‫ﻨﺘﻴﺠﺔ‬ (cosθ + i sinθ)n = cosθ + i sinθ . θ ∈ \ ; n ∈ ] ‫ﻤﻥ ﺃﺠل‬ . ‫ﻭﻫﻭ ﻤﺎ ﻴﻌﺭﻑ ﺒﺩﺴﺘﻭﺭ ﻤﻭﺍﻓﺭ‬ZB ‫ ﻭ‬ZB ‫ ﻭ‬ZA ‫ ﺜﻼﺙ ﻨﻘﻁ ﻤﺘﻤﺎﻴﺯﺓ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻟﻭﺍﺤﻘﻬﺎ‬C ‫ ﻭ‬B ‫ ﻭ‬A ‫( ﺇﺫﺍ ﻜﺎﻨﺕ‬H : ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﺈﻥ‬

• ZC - ZA = AC ZB - ZA AB  ZC - ZA  JJJG JJJG  ZB - ZA  AB , AC( ) [ ]•arg  = Z′ ‫ﻭ‬ 2π  vG uG ‫ﻻﺤﻘﺘﻴﻬﻤﺎ‬ : ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﺈﻥ‬ Z ‫ﺸﻌﺎﻋﺎﻥ‬ ‫ﻭ‬ ‫ﻜﺎﻥ‬ ‫( ﺇﺫﺍ‬I (uG , vG) = arg  Z′  [2π]  Z  : ‫ﺘﻁﺒﻴﻘﺎﺕ‬ Z2 = -1 + i 3 ; Z1 = 3i : ‫ ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﺤﻴﺙ‬Z2 , Z1 (1 Z12 , Z1 × Z2 , Z2 , Z1 ‫ﺍﺤﺴﺏ ﻁﻭﻴﻠﺔ ﻭﻋﻤﺩﺓ ﻜل ﻤﻥ‬ Z1 , 1 , Z100 Z2 Z1 2 : ‫ﺍﻟﺤل‬ Z2 = 2 , Z1 = 3 x π cosθ1 = 0 =0 2 3 θ1 ≡ [2π] : ‫ﻭﻤﻨﻪ‬  : Z1 ‫ ﻋﻤﺩﺓ‬θ1 ‫ﻟﺘﻜﻥ‬ sinθ1 3 = 3 =1 2π  cosθ2 = - 1 3  2 [ ]θ2 ≡ 2π : ‫ﻭﻤﻨﻪ‬  : Z2 ‫ﻋﻤﺩﺓ‬ θ2 ‫ﻭﻟﺘﻜﻥ‬ sinθ2 = 3 2 Z1 . Z2 = Z1 . Z2 = 6 xarg ( Z1 . Z2 ) = arg ( Z1 ) + arg ( Z2 )

= π + 2π = 7π 2 3 6 Z12 = Z1 2 = 32 = 9 x( )arg Z12 = 2arg( Z1 ) = 2. π [2π] 2 ( )arg Z12 = π [2π] : ‫ﻭ ﻤﻨﻪ‬ ( ) ( )arg Z100 Z 100 = 3100 Z100 = 100arg Z1 ‫؛‬ 1 = 1 x 1 ( )arg π Z100 = 100 2 [2π] = 0 [2π] : ‫ﻭﻤﻨﻪ‬ 1• 1 = 1 = 1 Z1 Z1 3 arg  1  = -arg( Z1 ) [2π] = - π [2π]  Z1  2  • Z1 = Z1 = 3 Z2 Z2 2 arg  Z1  = arg( Z1 ) - arg( Z2 ) = π - 2π [2π]  Z2  2 3   arg  Z1  = -π [2π] : ‫ﻭﻤﻨﻪ‬  Z2  6   : ‫ ﻫﻲ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‬A , B , C ‫ ﻤﺜﻠﺙ ﺤﻴﺙ ﻟﻭﺍﺤﻕ ﺍﻟﻨﻘﻁ‬ABC (2 . 2+i , 4+i , 2+3i . ‫ ﻭ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ‬A ‫ ﻗﺎﺌﻡ ﻓﻲ‬ABC ‫ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻤﺜﻠﺙ‬ : ‫ﺍﻟﺤل‬

‫‪ZC - ZA‬‬ ‫=‬ ‫‪2 + 3i - 2 - i‬‬ ‫=‬ ‫‪i‬‬ ‫‪=1‬‬‫‪ZB - ZA‬‬ ‫‪4+i-2-i‬‬ ‫ﻭﻋﻠﻴﻪ ‪AC = AB :‬‬ ‫‪AC‬‬ ‫‪=1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪AB‬‬‫‪arg‬‬ ‫‪‬‬ ‫‪ZC‬‬ ‫‪-‬‬ ‫‪ZA‬‬ ‫‪‬‬ ‫)‪= arg (i‬‬ ‫=‬ ‫‪π‬‬ ‫]‪[2π‬‬ ‫‪‬‬ ‫‪ZB‬‬ ‫‪-‬‬ ‫‪ZA‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫ﺇﺫﻥ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻗﺎﺌﻡ ﻓﻲ ‪ A‬ﻭ ﻤﺘﻘﺎﻴﺱ ﺍﻟﺴﺎﻗﻴﻥ ‪:‬‬ ‫‪ -7‬ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻌﺩﺩ ﻤﺭﻜﺏ )ﺘﺭﻤﻴﺯ ﺃﻭﻟﻴﺭ(‬ ‫‪ -‬ﺍﻟﺘﻌﺭﻴﻑ ‪:‬‬ ‫ﻨﻀﻊ ﺍﺼﻁﻼﺤﺎ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪cosθ + i sinθ = eiθ : θ‬‬ ‫ﻓﺈﺫﺍ ﻜﺎﻥ ‪ Z‬ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ﺤﻴﺙ ‪ Z = ρ cosθ + i sinθ :‬ﻓﺈﻥ ‪( ):‬‬ ‫‪. Z = ρ . eiθ‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﺍﻜﺘﺏ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z = -2 2 + 2 6 i‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ‪:‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪Z=4‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪e 2π‬‬ ‫‪i‬‬ ‫‪Z =4‬‬ ‫‪2‬‬ ‫‪,‬‬ ‫= )‪arg(Z‬‬ ‫‪2π‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪ -‬ﺨﻭﺍﺹ ‪:‬‬ ‫ﻟﻴﻜﻥ ‪ Z2 , Z1‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﺤﻴﺙ ‪:‬‬ ‫‪Z2‬‬ ‫=‬ ‫‪ρ‬‬ ‫‪eiθ2‬‬ ‫‪, Z2 = ρ1eiθ1‬‬ ‫‪2‬‬‫)‪1‬‬ ‫‪Z1‬‬ ‫‪.‬‬ ‫‪Z2‬‬ ‫=‬ ‫‪ρ1‬‬ ‫‪.‬‬ ‫) ‪ρ ei(θ1+θ2‬‬ ‫‪2‬‬

‫‪2) 1 = 1 . e-iθ1‬‬ ‫‪Z1 ρ1‬‬‫) ‪3) Z1 = ρ1 . ei(θ1−θ2‬‬ ‫‪Z2 ρ2‬‬‫‪4) Z1n = ρ1n . einθ‬‬‫‪5) Z1 = ρ1 . e-iθ1‬‬ ‫ﻤﻼﺤﻅﺔ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪ eiθ′ = cosθ′ + i sinθ′ ; eiθ = cosθ + i sinθ :‬ﻭﻋﻠﻴﻪ ‪:‬‬‫)‪eiθ . eiθ′ = ei(θ+θ′) = cos (θ + θ′) + i sin(θ + θ′) . . . (1‬‬‫ﻭﻟﺩﻴﻨﺎ ‪eiθ . e2θ′ = (cosθ + i sinθ) (cosθ′ + i sinθ′) :‬‬‫)‪= cosθ.cosθ′- sinθ. sinθ′+i(cosθ. sinθ′+ sinθ. cosθ′) . . .(2‬‬‫ﻤﻥ )‪cos(θ + θ′) = cosθ . cosθ′ - sinθ . sinθ′ : (2) ; (1‬‬‫‪sin(θ + θ′) = cosθ . sinθ′ + sinθ . cosθ′‬‬ ‫‪ -‬ﺍﻟﺘﻌﺒﻴﺭ ﻋﻥ ﺩﺍﺌﺭﺓ ﺒﺎﻟﻌﻼﻗﺔ ‪Z = Z0 + k . eiθ‬‬ ‫ﻟﺘﻜﻥ)‪ (C‬ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ ω‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪.k‬‬ ‫ﻨﻔﺭﺽ ‪ Z0‬ﻻﺤﻘﺔ ‪ k ، ω‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ‪.‬‬ ‫ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﻤﻥ )‪(C‬‬ ‫ﻟﺩﻴﻨﺎ ‪ M ∈ (C) :‬ﺘﻜﺎﻓﺊ ‪Z - Z0 = k :‬‬‫ﻭ ﺘﻜﺎﻓﺊ ‪ Z - Z0‬ﻫﻭ ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ ﻁﻭﻴﻠﺘﻪ ‪ k‬ﻭ ﺘﻜﺎﻓﺊ ‪ :‬ﻴﻭﺠﺩ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪) θ‬ﻴﻤﻜﻥ‬ ‫ﺍﻟﻘﻭل ﺃﻥ ‪[ [( θ ∈ 0 ; 2π‬‬ ‫ﺒﺤﻴﺙ ‪. Z = Z0 + k . eiθ :‬‬ ‫=‪Z‬‬ ‫‪Z0 + k . eiθ‬‬ ‫ﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ ﺒﺎﻟﻌﻼﻗﺔ‬ ‫‪ -‬ﺍﻟﺘﻌﺒﻴﺭ ﻋﻥ‬‫‪ vG‬ﻤﻌﻁﻰ‬ ‫‪ ω‬ﻭ ﺸﻌﺎﻉ ﺘﻭﺠﻴﻬﻪ‬ ‫ﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ ﻤﺒﺩﺃﻩ‬ ‫ﻟﻴﻜﻥ ‪wx‬‬‫‪[ ).‬‬

‫ﻨﻔﺭﺽ ‪ Z0‬ﻻﺤﻘﺔ ‪ u ، w‬ﻻﺤﻘﺔ ‪( )u ∈ ^* , vG‬‬ ‫ﺤﻴﺙ ‪u = 1 ; arg (u) = θ [2π] :‬‬‫ﻭﺘﻜﺎﻓﺊ‬ ‫(‬ ‫‪JJJJG‬‬ ‫=‬ ‫‪k‬‬ ‫‪.‬‬ ‫‪vG‬‬ ‫ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪wM‬‬ ‫‪ M ∈ wx‬ﺘﻜﺎﻓﺊ )ﻴﻭﺠﺩ ﻋﺩﺩ ‪ k‬ﻤﻥ ‪ \ +‬ﺒﺤﻴﺙ ‪[ ):‬‬ ‫)ﻴﻭﺠﺩ ﻋﺩﺩ ‪ k‬ﻤﻥ ‪ \ +‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪( Z = Z0 + k . eiθ‬‬ ‫‪ – 8‬ﺍﻟﻤﻌﺎﺩﻻﺕ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻓﻲ ^ ‪.‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ Z′ , Z‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻓﺈﻥ ‪:‬‬ ‫‪ Z . Z′ = 0‬ﺘﻜﺎﻓﺊ ‪ Z = 0 :‬ﺃﻭ ‪Z′ = 0‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬ ‫ﻨﻔﺭﺽ ‪ Z = x + iy‬ﻭ ‪Z′ = x′ + iy′‬‬ ‫ﻟﺩﻴﻨﺎ ‪Z . Z′ = ( xx′ + yy′) + i ( xy - x′y) :‬‬ ‫‪ xx′ - yy′ = 0‬‬ ‫‪‬‬ ‫‪‬‬ ‫و‬ ‫ﻭﻤﻨﻪ ‪ Z . Z′ = 0 :‬ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪ xy′ + x′y = 0‬‬ ‫ﻨﻔﺭﺽ ‪ Z′ ≠ 0‬ﺤل ﺍﻟﺠﻤﻠﺔ ﺫﺍﺕ ﺍﻟﻤﺠﻬﻭل ‪ x ; y‬ﻫﻭ ‪( ):‬‬ ‫‪-y′ 0‬‬ ‫=‪x‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫=‬ ‫‪0‬‬ ‫‪=0‬‬ ‫‪x′‬‬ ‫‪-y′‬‬ ‫‪x′2 + y′2‬‬ ‫‪y′ x′‬‬

‫‪0 x′‬‬ ‫=‪y‬‬ ‫‪0‬‬ ‫‪y′‬‬ ‫=‬ ‫‪0‬‬ ‫‪=0‬‬ ‫‪x′‬‬ ‫‪-y′‬‬ ‫‪x′2 + y′2‬‬ ‫‪y′ x′‬‬ ‫ﻓﻨﺠﺩ ‪. Z = 0‬‬ ‫ﻭﺇﺫﺍ ﻓﺭﻀﻨﺎ ‪ Z ≠ 0‬ﻨﺠﺩ ‪ Z′ = 0 :‬ﻭﻤﻨﻪ ‪Z . Z′ = 0 :‬‬ ‫ﺘﻜﺎﻓﺊ ‪ Z = 0 :‬ﺃﻭ ‪Z′ = 0‬‬ ‫* ﺤل ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺸﻜل ‪:‬‬ ‫)‪aZ2 + bZ + C = 0 . . . (1‬‬ ‫ﻤﻊ ‪ c , b , a‬ﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ﺤﻴﺙ ‪ Z ; a ≠ 0‬ﻤﺘﻐﻴﺭ ﻓﻲ ^‬ ‫ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ ﻟﻠﻌﺒﺎﺭﺓ ‪ aZ2 + bZ + C‬ﻫﻭ ‪:‬‬ ‫‪aZ2‬‬ ‫‪+ bZ + C = a‬‬ ‫‪ Z +‬‬ ‫‪b 2‬‬ ‫‪-‬‬ ‫‪b2 - 4ac ‬‬ ‫‪2a ‬‬ ‫‪‬‬ ‫‪4a2‬‬ ‫‪‬‬ ‫‪aZ2‬‬ ‫‪+‬‬ ‫‪bZ‬‬ ‫‪+‬‬ ‫‪C‬‬ ‫=‬ ‫‪a‬‬ ‫‪‬‬ ‫‪Z‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫∆‬ ‫‪‬‬ ‫‪:‬‬ ‫ﻨﺠﺩ‬ ‫∆‬ ‫=‬ ‫‪b2‬‬ ‫‪-‬‬ ‫‪4ac‬‬ ‫ﺒﻭﻀﻊ‬ ‫‪2a‬‬ ‫‪‬‬ ‫‪4a2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ‪ : ∆ ≥ 0‬ﻟﻠﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺤﻠﻴﻥ ﻜﻤﺎ ﺴﺒﻕ ﻓﻲ \ ‪.‬‬ ‫‪Z2‬‬ ‫=‬ ‫‪-b +‬‬ ‫∆‬ ‫ﻭ‬ ‫‪Z1‬‬ ‫=‬ ‫‪-b +‬‬ ‫∆‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫‪( )2‬‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ‪ : ∆ < 0‬ﻨﻀﻊ ∆‪∆ = i -‬‬ ‫ﻓﻴﻜﻭﻥ ‪:‬‬‫‪aZ2‬‬ ‫‪+ bZ + C = a‬‬ ‫‪‬‬ ‫‪Z‬‬ ‫‪-‬‬ ‫‪-b - i −∆ ‬‬ ‫‪‬‬ ‫‪Z‬‬ ‫‪-‬‬ ‫‪-b + i‬‬ ‫‪−∆ ‬‬ ‫‪‬‬ ‫‪2a ‬‬ ‫‪‬‬ ‫‪2a‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺘﻜﺎﻓﺊ ‪:‬‬

‫‪Z‬‬ ‫=‬ ‫‪-b‬‬ ‫‪+i‬‬ ‫∆‪−‬‬ ‫ﺃﻭ‬ ‫=‪Z‬‬ ‫∆‪-b - i −‬‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫ﺃﻱ ﻟﻠﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪.‬‬ ‫* ﺍﻟﺠﺫﻭﺭ ﺍﻟﺭﺒﻴﻌﻴﺔ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ‪:‬‬‫ﻟﻴﻜﻥ ‪ Z‬ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﻤﻌﺩﻭﻡ‪ .‬ﻟﻨﺒﺤﺙ ﻋﻥ ﻭﺠﻭﺩ ﻋﺩﺩ ﻤﺭﻜﺏ ‪ k‬ﺒﺤﻴﺙ ‪. k 2 = Z‬‬ ‫ﻨﻔﺭﺽ ‪ Z = x + iy‬ﻭ ‪k = α + iβ‬‬ ‫‪(α + iβ)2 = x + iy‬‬ ‫ﻟﺩﻴﻨﺎ ‪ k 2 = Z :‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪k‬‬ ‫‪2‬‬ ‫=‬ ‫‪Z‬‬ ‫‪α2α2 β-‬‬ ‫= ‪β2‬‬ ‫‪x‬‬ ‫‪...‬‬ ‫)‪(1‬‬ ‫‪=y.‬‬ ‫‪.‬‬ ‫)‪. (2‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪β2‬‬ ‫=‬ ‫)‪x2 + y2 . . . (3‬‬ ‫‪α‬‬ ‫ﻨﺠﻤﻊ )‪ (1‬ﻭ )‪ (3‬ﻨﺠﺩ ‪2α2 + β2 = x+ x2 + y2 :‬‬ ‫= ‪ α2‬ﻟﻜﻥ ‪x2 + y2 > -x‬‬ ‫‪x2 + y2 + x‬‬ ‫ﻭﻋﻠﻴﻪ ‪2 :‬‬‫‪α=-‬‬ ‫‪x2 + y2 + x‬‬ ‫= ‪ α‬ﺃﻭ‬ ‫‪x2 + y2 + x‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪2‬‬‫= ‪α2‬‬ ‫‪x2 + y2 + x‬‬ ‫ﻭ‬ ‫= ‪α1‬‬ ‫‪x2 + y2 + x‬‬ ‫‪2‬‬ ‫ﻨﻔﺭﺽ ‪2 :‬‬ ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )‪: (2‬‬ ‫‪β2‬‬ ‫=‬ ‫‪y‬‬ ‫‪: α = α2‬‬ ‫‪ ،‬ﻟﻤﺎ‬ ‫= ‪β1‬‬ ‫‪y‬‬ ‫ﻟﻤﺎ ‪: α = α1‬‬ ‫‪2α2‬‬ ‫‪2α1‬‬ ‫ﺤﻴﺙ ‪p2 = −β1 ، α2 = -α1 :‬‬

‫* ﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ﻤﻥ ﺍﻟﺸﻜل ‪:‬‬ ‫)‪aZ2 + bZ + C = 0 . . . (1‬‬ ‫ﺤﻴﺙ ‪ b , c , a‬ﺃﻋﺩﺍﺩ ﻤﺭﻜﺒﺔ ﻭ ‪a ≠ 0‬‬ ‫‪ x‬ﻨﻜﺘﺏ ‪ aZ2 + bC + C‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ ﻓﻨﺠﺩ ‪:‬‬ ‫‪aZ2‬‬ ‫‪+ bZ + c = a‬‬ ‫‪ x‬‬ ‫‪+‬‬ ‫‪b 2‬‬ ‫‪-‬‬ ‫‪∆‬‬ ‫‪2a ‬‬ ‫‪‬‬ ‫‪4a‬‬ ‫‪2‬‬ ‫‪‬‬ ‫ﻤﻊ ‪ ∆ = b2 - 4ac‬ﻜﻤﺎ ﺴﺒﻕ ‪.‬‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ∆ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ‪ :‬ﻟﻠﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺤﻠﻴﻥ‬‫‪Z2‬‬ ‫=‬ ‫‪-b +‬‬ ‫∆‬ ‫ﻭ‬ ‫‪Z1‬‬ ‫=‬ ‫‪-b -‬‬ ‫∆‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ∆ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﺴﺎﻟﺏ ‪ :‬ﻟﻠﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺤﻠﻴﻥ‬‫‪Z2‬‬ ‫=‬ ‫‪-b + i‬‬ ‫∆‪−‬‬ ‫ﻭ‬ ‫‪Z1‬‬ ‫=‬ ‫‪-b - i‬‬ ‫∆‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ∆ ﻋﺩﺩ ﻤﺭﻜﺏ ﻏﻴﺭ ﺤﻘﻴﻘﻲ ‪:‬‬ ‫ﻨﺒﺤﺙ ﻋﻥ ﺠﺫﺭﻴﻪ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻭ ﻟﻴﻜﻥ ‪ k‬ﺃﺤﺩﻫﻤﺎ ﻭﻤﻨﻪ )‪ (1‬ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪a‬‬ ‫‪ x +‬‬ ‫‪b 2‬‬ ‫‪-‬‬ ‫‪k2 ‬‬ ‫‪=0‬‬ ‫ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬ ‫‪2a ‬‬ ‫‪‬‬ ‫‪4a2‬‬ ‫‪‬‬‫‪Z2‬‬ ‫=‬ ‫‪-b + k‬‬ ‫‪,‬‬ ‫‪Z1‬‬ ‫=‬ ‫‪-b - k‬‬ ‫‪ Z1‬ﻭ ‪ Z2‬ﺤﻴﺙ ‪:‬‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z2 - (3 - 2i) Z + 5 - i = 0 :‬‬ ‫‪ ∆ = b2 - 4a C‬ﻭﻤﻨﻪ ‪∆ = (3 - 2i)2 - 4 (5 - i) :‬‬‫ﺇﺫﻥ ‪ ∆ = 9 - 12i - 4 - 20 + 4i :‬ﻭﻤﻨﻪ ‪∆ = -15 - 8i :‬‬

‫ﻨﺒﺤﺙ ﻋﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻟﻠﻌﺩﺩ ∆ ‪.‬‬ ‫ﻟﻴﻜﻥ ‪ k‬ﺠﺫﺭ ﺘﺭﺒﻴﻌﻲ ﻟﻠﻌﺩﺩ ∆ ‪.‬‬ ‫∆ = ‪k2‬‬ ‫ﻨﻔﺭﺽ ‪k = x + iy :‬‬ ‫‪‬‬ ‫ﻓﻴﻜﻭﻥ ‪:‬‬ ‫‪‬‬ ‫‪k‬‬ ‫‪2‬‬ ‫∆=‬ ‫‪( x + iy)2 = -15 - 8i‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2‬‬ ‫=‬ ‫‪(-15)2 + (-8)2‬‬ ‫)‪ x - y2 = -15 . . . (1‬‬ ‫)‪2xy = -8 . . . (2‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2‬‬ ‫=‬ ‫‪17‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫)‪(3‬‬ ‫‪‬‬ ‫ﺒﺠﻤﻊ )‪ (1‬ﻭ )‪ (2‬ﻨﺠﺩ ‪ 2 x2 = 2 :‬ﻭﻤﻨﻪ ‪x2 = 1 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ x = 1 :‬ﺃﻭ ‪. x = -1‬‬ ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ)‪ (2‬ﻨﺠﺩ ‪ :‬ﻟﻤﺎ ‪y = -4 : x = 1‬‬ ‫ﻟﻤﺎ ‪. y = 4 : x = 1‬‬ ‫ﺇﺫﻥ ‪ k = 1 - 4i‬ﺃﻭ ‪k = -1 + 4i‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻫﻤﺎ ‪:‬‬‫‪Z2‬‬ ‫=‬ ‫‪3 - 2i + 1 - 4i‬‬ ‫‪= 2 - 3i‬‬ ‫‪,‬‬ ‫‪Z1‬‬ ‫=‬ ‫‪3 - 2i - 1 + 4i‬‬ ‫‪=1+i‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ -9‬ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﻭ ﺍﻟﺘﺤﻭﻴﻼﺕ ﺍﻟﻨﻘﻁﻴﺔ ‪:‬‬ ‫* ﻤﺒﺭﻫﻨﺔ ‪:‬‬‫ﻟﻴﻜﻥ ‪ f‬ﺍﻟﺘﺤﻭﻴل ﺍﻟﻨﻘﻁﻲ ﺍﻟﺫﻱ ﻴﺭﻓﻕ ﺒﻜل ﻨﻘﻁﺔ ‪ M‬ﻻﺤﻘﺘﻬﺎ ‪ Z‬ﺍﻟﻨﻘﻁﺔ ‪ M′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪. Z′‬‬ ‫‪β‬‬ ‫∈‬ ‫‪ ZvG′‬ﺫﻭﺤﻴﺍﻟﺙﻼ ‪:‬ﺤﻘﺔ^‪β‬‬ ‫‪=Z‬‬ ‫‪+β‬‬ ‫‪:‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ‬ ‫‪x‬‬ ‫‪f‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪.‬‬ ‫ﺸﻌﺎﻋﻪ‬ ‫ﺍﻨﺴﺤﺎﺏ‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ‪ Z′ - Z0 = k (Z - Z0 ) :‬ﺤﻴﺙ ^ ∈ ‪ Z0‬ﻭ‬ ‫*\ ∈ ‪ k‬ﻓﺈﻥ ‪ f‬ﺘﺤﺎﻜﻲ ﻨﺴﺒﺘﻪ ‪ k‬ﻭ ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ M0‬ﺫﺍﺕ‬

‫ﺍﻟﻼﺤﻘﺔ ‪. Z0‬‬‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ‪ Z′ - Z0 = eiθ (Z - Z0 ) :‬ﺤﻴﺙ ^ ∈ ‪ Z0‬ﻭ \ ∈ ‪θ‬‬ ‫ﻓﺈﻥ ‪ f‬ﺩﻭﺭﺍﻥ ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ M0‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z0‬ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬ ‫ﺍﻟﺒﺭﻫﺎﻥ ‪:‬‬‫= ‪Z′ - Z‬‬ ‫‪Z JJJJJG‬‬ ‫=‬ ‫‪β‬‬ ‫ﻓﺈﻥ‬ ‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ‪Z′ = Z + β‬‬ ‫‪JJJJJG‬‬ ‫‪MM′‬‬‫ﻭﻤﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻓﺈﻥ ﺍﻟﺸﻌﺎﻉ ‪ MM′‬ﺜﺎﺒﺕ‪.‬‬ ‫ﻭﻋﻠﻴﻪ ﻓﻬﻭ ﻴﻌﺭﻑ ﺍﻨﺴﺤﺎﺏ ‪.‬‬‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ) ‪ Z′ - Z0 = k (Z - Z0‬ﻓﺈﻥ ﺼﻭﺭﺓ ﺍﻟﻨﻘﻁﺔ ‪ M0‬ﺫﺍﺕ‬‫ﺍﻟﻼﺤﻘﺔ ‪ Z0‬ﺒﻭﺍﺴﻁﺔ ‪ f‬ﻫﻲ ﻨﻔﺴﻬﺎ‪ .‬ﻭﻤﻥ ﺃﺠل ‪Z ≠ Z0‬‬ ‫‪.‬‬ ‫‪Z′ - Z0‬‬ ‫‪=k‬‬ ‫‪,‬‬ ‫*\ ∈ ‪k‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪Z - Z0‬‬ ‫‪M0M′‬‬ ‫=‬ ‫‪k‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪M0M‬‬‫‪( )JJJJJG JJJJJJG‬‬ ‫ﻓﺈﺫﺍ ﻜﺎﻥ ‪ k > 0‬ﻓﺈﻥ ‪[ ]= 0 2π :‬‬ ‫‪M0M ; M0M′‬‬‫‪( )JJJJJG JJJJJJG‬‬‫ﻭ ﺇﺫﺍ ﻜﺎﻥ ‪ k < 0‬ﻓﺈﻥ ‪[ ]M0M ; M0M′ = π 2π :‬‬‫ﻭﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻥ ﺍﻟﻨﻘﻁ ‪ M′ , M , M0‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‪.‬‬‫ﺘﻤﻴﺯ ﺘﺤﺎﻜﻲ ﻤﺭﻜﺯﻩ ‪ M0‬ﻭ ﻨﺴﺒﺘﻪ ‪. k‬‬ ‫‪M0M′‬‬ ‫=‬ ‫‪k‬‬ ‫ﻭ ﺒﺎﻟﻨﺴﺒﺔ‬ ‫‪M0M‬‬‫‪ x‬ﺇﺫﺍ ﻜﺎﻥ ) ‪ Z′ - Z0 = eiθ (Z - Z0‬ﻓﺈﻥ ﺼﻭﺭﺓ ﺍﻟﻨﻘﻁﺔ ‪ M0‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪Z0‬‬ ‫ﺒﻭﺍﺴﻁﺔ ‪ f‬ﻫﻲ ﻨﻔﺴﻬﺎ‪ .‬ﻭﻤﻥ ﺃﺠل ‪Z ≠ Z0 :‬‬ ‫‪Z′ - Z0‬‬ ‫‪= eiθ‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪Z - Z0‬‬

‫‪( )JJJJJG JJJJJJG‬‬ ‫ﻭ ]‪= θ [2π‬‬ ‫‪M0M′‬‬ ‫‪=1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪M0M‬‬ ‫‪M0M ; M0M′‬‬ ‫ﻫﺎﺘﻴﻥ ﺍﻟﻌﻼﻗﺘﻴﻥ ﺘﻤﻴﺯﺍﻥ ﺩﻭﺭﺍﻥ ﻤﺭﻜﺯﻩ ‪ M0‬ﻭ ﺯﺍﻭﻴﺘﻪ ‪. θ‬‬ ‫ﺃﻤﺜﻠﺔ ‪:‬‬‫ﺍﺩﺭﺱ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺤﻭﻴل ‪ f‬ﺍﻟﺫﻱ ﻴﺭﻓﻕ ﺒﺎﻟﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺍﻟﻨﻘﻁﺔ ‪ M′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z′‬ﻓﻲ‬ ‫ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬‫‪1) Z′ = Z - 1 2) Z′ = Z + i + 1 3) Z′ = 3Z‬‬‫‪4) Z′ = -2Z + i + 2‬‬ ‫‪5) Z′ = iZ‬‬ ‫= ‪6) Z′‬‬ ‫‪2‬‬ ‫‪(i‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫‪Z‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪2‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫=‪ ′wG‬ﺫ‪Z‬ﻭ‬ ‫ﻟﺩﻴﻨﺎ ‪Z - 1 :‬‬ ‫‪(1‬‬ ‫‪.‬‬ ‫‪-1‬‬ ‫ﺍﻟﻼﺤﻘﺔ‬ ‫‪ f‬ﺍﻨﺴﺤﺎﺏ ﺸﻌﺎﻋﻪ‬ ‫‪ZwG+‬ﺫﻭ ﺍ=ﻟ ‪′‬ﻼﺤﻘ‪Z‬ﺔ‬ ‫ﻟﺩﻴﻨﺎ ‪i + 1 :‬‬ ‫‪(2‬‬ ‫‪.‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪ f‬ﺍﻨﺴﺤﺎﺏ ﺸﻌﺎﻋﻪ‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪Z′ = 3Z :‬‬ ‫‪ f‬ﺘﺤﺎﻙ ﻨﺴﺒﺘﻪ ‪ 3‬ﻭ ﻤﺭﻜﺯﻩ ‪. O‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪Z′ = -2Z + i + 2 :‬‬‫‪.‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪i‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪ f‬ﺘﺤﺎﻙ ﻨﺴﺒﺘﻪ ‪ -2‬ﻭ ﻤﺭﻜﺯﻩ ‪ I‬ﺤﻴﺙ ﻻﺤﻘﺘﻪ‬ ‫‪1-i‬‬ ‫ﺇﺫﻥ ‪. Z0 = -1‬‬ ‫‪Z′ = iZ‬‬ ‫‪,‬‬ ‫‪i‬‬ ‫=‬ ‫‪ei‬‬ ‫‪π‬‬ ‫‪ (5‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪π‬‬ ‫‪ f‬ﺩﻭﺭﺍﻥ ﻤﺭﻜﺯﻩ ‪ O‬ﻭﺯﺍﻭﻴﺘﻪ ﻋﻤﺩﺓ ‪ i‬ﺃﻱ ‪. 2‬‬ ‫= ‪Z′‬‬ ‫‪2‬‬ ‫‪(i + 1) Z + i‬‬ ‫‪ (6‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬

‫‪2‬‬ ‫‪(i‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫‪ei‬‬ ‫‪π‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪Z0‬‬ ‫ﻭ ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪π‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﺩﻭﺭﺍﻥ ﺯﺍﻭﻴﺘﻪ‬ ‫‪4‬‬ ‫‪2i‬‬ ‫‪i‬‬‫‪ Z0 = 2 -‬ﺃﻱ ‪:‬‬ ‫‪2-‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫= ‪Z0‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪2i‬‬ ‫‪1-‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫)‪(1 + i‬‬ ‫‪( )2i 2 - 2 + 2i‬‬ ‫‪Z0 = 2 - 2 - 2i 2 - 2 + 2i‬‬ ‫= ‪. Z0‬‬ ‫‪( )-2 2 + 2 2 - 2 i‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪( )2‬‬ ‫‪2- 2 +2‬‬

‫ﺘﻜﻨﻭﻟﻭﺠﻴﺎ ﺍﻹﻋﻼﻡ ﻭ ﺍﻻﺘﺼﺎل‬ ‫ﺍﻟﺘﻁﺒﻴﻕ ‪:‬‬‫‪ . z‬‬ ‫‪1  3i 2‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪5  i‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ‪:‬‬ ‫‪ (1‬ﻋﻴﻥ ﻤﺭﺍﻓﻕ ﺍﻟﻌﺩﺩ ‪ (2 . z‬ﻋﻴﻥ ﺍﻟﺠﺯﺀ ﺍﻟﺤﻘﻴﻘﻲ ﻟﻠﻌﺩﺩ ‪. z‬‬ ‫‪ (3‬ﻋﻴﻥ ﺍﻟﺠﺯﺀ ﺍﻟﺘﺨﻴﻠﻲ ﻟﻠﻌﺩﺩ ‪ (4 . z‬ﻋﻴﻥ ﻁﻭﻴﻠﺔ ﺍﻟﻌﺩﺩ ‪. z‬‬‫‪ (5‬ﻋﻴﻥ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ‪ (6 . z‬ﺃﻜﺘﺏ ﺍﻟﻌﺩﺩ ‪ z‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ‪.‬‬ ‫‪ (7‬ﺃﻜﺘﺏ ﺍﻟﻌﺩﺩ ‪ z‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻷﺴﻲ‪.‬‬ ‫ﺍﻟﺤل‬ ‫‪ (1‬ﺘﻌﻴﻴﻥ ﺍﻟﻤﺭﺍﻓﻕ ‪:‬‬‫ﻭﻨﺤﺭﻙ ﺍﻟﺯﺍﻟﻘﺔ‬ ‫ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬‫ﺍﻟﻰ ‪ CPX‬ﻜﻤﺎﻴﻅﻬﺭ ﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﻓﺘﻅﻬﺭ‬ ‫ﻗﺎﺌﻤﺔ ﻜﻤﺎ ﻓﻲ ﺍﻟﺸﺎﺸﺔ ﺍﻟﻤﻭﺍﻟﻴﺔ‪.‬‬ ‫ﻨﺨﺘﺎﺭ ﺍﻟﺭﻗﻡ ‪ 1‬ﻟﻨﺤﺼل ﻋﻠﻰ ‪:‬‬ ‫(‪1:Conj‬‬ ‫ﻨﻜﺘﺏ ﻋﺒﺎﺭﺓ ‪ z‬ﻜﻤﺎﻴﻠﻲ ‪:‬‬ ‫))‪Conj((1-3i)^2)/(5+i‬‬‫ﻭﻨﻨﻘﺭ ﻋﻠﻰ ‪ Enter‬ﻓﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻨﺘﻴﺠﺔ‬ ‫‪-1,77+0,85i‬‬ ‫ﻭﺍﻟﺘﻲ ﺘﻤﺜل ﻤﺭﺍﻓﻕ ‪z‬‬

‫‪ ( 2‬ﺘﻌﻴﻴﻥ ﺍﻟﺠﺯﺀ ﺍﻟﺤﻘﻴﻘﻲ ‪:‬‬‫ﻭﻨﺤﺭﻙ ﺍﻟﺯﺍﻟﻘﺔ‬ ‫ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬ ‫ﺍﻟﻰ ‪ CPX‬ﺜﻡ ﻋﻠﻰ ﺍﻟﺭﻗﻡ ‪2‬‬‫ﻓﺘﻅﻬﺭﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪real( :‬‬‫ﻨﺩﺨل ﻋﺒﺎﺭﺓ ‪real((1-3i)^2)/(5+i)) : :‬‬ ‫ﻨﻨﻘﺭ ﻋﻠﻰ ‪ Enter‬ﻓﻨﺠﺩ ‪−1,77 :‬‬ ‫‪ ( 3‬ﺘﻌﻴﻴﻥ ﺍﻟﺠﺯﺀ ﺍﻟﺘﺨﻴﻠﻲ ‪:‬‬‫ﻭﻨﺤﺭﻙ‬ ‫ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬‫ﺍﻟﺯﺍﻟﻘﺔ ﺍﻟﻰ ‪ CPX‬ﺜﻡ ﻋﻠﻰ ﺍﻟﺭﻗﻡ ‪3‬‬‫ﻓﺘﻅﻬﺭﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪Imag( :‬‬‫ﻨﺩﺨل ﻋﺒﺎﺭﺓ ‪Imag((1-3i)^2)/(5+i)) : :‬‬‫ﻨﻨﻘﺭ ﻋﻠﻰ ‪ Enter‬ﻓﻨﺠﺩ‪- 0,8 5 :‬‬‫ﻭﻨﺤﺭﻙ ﺍﻟﺯﺍﻟﻘﺔ‬ ‫‪(4‬ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬‫ﺇﻟﻰ ‪ CPX‬ﺜﻡ ﻋﻠﻰ ﺍﻟﺭﻗﻡ ‪ 5‬ﻓﺘﻅﻬﺭﻋﻠﻰ‬ ‫ﺍﻟﺸﺎﺸﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪abs ( :‬‬‫ﻨﺩﺨل ﻋﺒﺎﺭﺓ ‪abs((1-3i)^2)/(5+i)) :‬‬‫ﻨﻨﻘﺭ ﻋﻠﻰ ‪ Enter‬ﻓﻨﺠﺩ‪1,96 :‬‬‫ﻭﻨﺤﺭﻙ ﺍﻟﺯﺍﻟﻘﺔ‬ ‫‪(5‬ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬‫ﺇﻟﻰ ‪ CPX‬ﺜﻡ ﻋﻠﻰ ﺍﻟﺭﻗﻡ ‪ 4‬ﻓﺘﻅﻬﺭﻋﻠﻰ‬‫ﺍﻟﺸﺎﺸﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪angle ( :‬‬‫ﻨﺩﺨل ﻋﺒﺎﺭﺓ ‪angle((1-3i)^2)/(5+i)) :‬‬‫ﻨﻨﻘﺭ ﻋﻠﻰ ‪ Enter‬ﻓﻨﺠﺩ‪– 2,70 :‬‬

‫‪(6‬ﻜﺘﺎﺒﺔ ‪ Z‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ‪:‬‬ ‫ﻨﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﻋﺒﺎﺭﺓ ‪ Z‬ﻜﻤﺎﻴﻠﻲ ‪:‬‬ ‫)‪(1-3i)^2/(5+i‬‬ ‫ﻭﻨﺤﺭﻙ ﺍﻟﺯﺍﻟﻘﺔ‬ ‫ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬ ‫ﺇﻟﻰ ‪ CPX‬ﺜﻡ ﻨﻨﻘﺭﻋﻠﻰ ﺍﻟﺭﻗﻡ ‪6‬‬ ‫ﻭ ﻨﻨﻘﺭ ﻋﻠﻰ ‪Enter‬‬ ‫ﻓﺘﻅﻬﺭﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪:‬‬ ‫‪-1,77- 0,85 i‬‬ ‫‪ (7‬ﻜﺘﺎﺒﺔ ‪ Z‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ‪:‬‬ ‫ﻨﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﻋﺒﺎﺭﺓ ‪ Z‬ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫)‪(1-3i)^2/(5+i‬‬ ‫ﻭﻨﺤﺭﻙ ﺍﻟﺯﺍﻟﻘﺔ‬ ‫ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬ ‫ﺇﻟﻰ ‪ CPX‬ﺜﻡ ﻨﻨﻘﺭﻋﻠﻰ ﺍﻟﺭﻗﻡ ‪ 7‬ﻭ ﻨﻨﻘﺭ‬‫ﻋﻠﻰ ‪ Enter‬ﻓﺘﻅﻬﺭﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪1,96e2,70i :‬‬ ‫ﻤﻼﺤﻅﺔ ‪: 1‬‬ ‫ﻟﻜﺘﺎﺒﺔ ﺍﻟﺤﺭﻑ ‪ i‬ﻨﻘﻭﻡ ﺒﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﺜﻡ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬ ‫ﻨﻨﻘﺭﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬ ‫ﻤﻼﺤﻅﺔ ‪: 2‬‬‫ﻭﻨﺤﺩﺩ ﻋﺩﺩ‬ ‫ﻟﺘﻐﻴﻴﺭ ﻋﺩﺩ ﺍﻷﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ ﻨﻨﻘﺭ ﻋﻠﻰ ﺍﻟﻠﻤﺴﺔ‬‫ﺍﻷﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺯﺍﻟﻘﺔ ﻭﻫﺫﺍ ﻓﻲ ﺍﻟﺴﻁﺭ ﺍﻟﺜﺎﻨﻲ ﺜﻡ ﻨﻨﻘﺭ‬‫ﻋﻠﻰ‪ Enter‬ﻭﻗﺩ ﺍﺨﺘﺭﻨﺎ ﻓﻲ ﺍﻟﺸﻜل ﺜﻼﺜﺔ ﺃﺭﻗﺎﻡ ﺒﻌﺩ ﺍﻟﻔﺎﺼﻠﺔ ‪.‬‬

‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬‫ﻀﻊ ﺍﻟﻌﻼﻤﺔ √ ﺃﻤﺎﻡ ﻜل ﺠﻤﻠﺔ ﺼﺤﻴﺤﺔ ﻭ ﺍﻟﻌﻼﻤﺔ×ﺃﻤﺎﻡ ﻜل ﺠﻤﻠﺔ ﺨﺎﻁﺌﺔ‪:‬‬ ‫‪ (1‬ﺍﻟﻌﺩﺩ ‪ 0‬ﻟﻴﺱ ﻋﺩﺩ ﻤﺭﻜﺏ ‪.‬‬ ‫ﻤﺭﻜﺏ‪.‬‬ ‫ﻋﺩﺩ‬ ‫ﻫﻭ‬ ‫‪cos‬‬ ‫‪π‬‬ ‫‪(2‬‬ ‫‪6‬‬ ‫‪ (3‬ﻜل ﻋﺩﺩ ﺤﻘﻴﻕ ﻫﻭ ﻋﺩﺩ ﻤﺭﻜﺏ‪.‬‬ ‫‪ (4‬ﻜل ﻋﺩﺩ ﻤﺭﻜﺏ ﻫﻭ ﻋﺩﺩ ﺤﻘﻴﻘﻲ‪.‬‬‫‪ (5‬ﺇﺫﺍ ﻜﺎﻥ ‪ x + 2i = -3 + yi :‬ﺤﻴﺙ ‪ x‬ﻭ‪ y‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻏﻴﺭ ﺤﻘﻴﻘﻴﻴﻥ ﻓﺈﻥ ‪y = 2 :‬‬ ‫ﻭ ‪x = -3‬‬ ‫‪Z = - Z (6‬‬ ‫‪‬‬ ‫‪i+Z‬‬ ‫=‬ ‫‪i‬‬ ‫‪-‬‬ ‫‪Z‬‬ ‫‪(7‬‬ ‫‪‬‬ ‫‪i - Z ‬‬ ‫‪i‬‬ ‫‪-‬‬ ‫‪Z‬‬ ‫=‪Z‬‬ ‫‪1‬‬ ‫ﺇﺫﺍ ﻜﺎﻨﺕ ‪ Z = 1‬ﻓﺈﻥ ‪:‬‬ ‫‪(8‬‬ ‫‪Z‬‬‫‪ (9‬ﺇﺫﺍ ﻜﺎﻨﺕ ‪ Z = 4‬ﺤﻴﺙ ‪ Z‬ﻋﺩﺩ ﻤﺭﻜﺏ ﻭﻫﻭ ﻻﺤﻘﺔ )‪ M(x ; y‬ﻓﺈﻨﻪ ﻋﻨﺩﻤﺎ ﻴﺘﻐﻴﺭ ﻜل ﻤﻥ ‪x‬‬ ‫ﻭ‪ y‬ﻓﻲ \ ﻴﺸﻜﻼﻥ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ O‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪. 4‬‬‫ﻫﻲ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ﻁﻭﻴﻠﺘﻪ ‪ 4‬ﻭ ﻋﻤﺩﺘﻪ‬ ‫‪4‬‬ ‫‪‬‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪cos‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪ (10‬ﺍﻟﻌﺒﺎﺭﺓ‬ ‫‪‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪.2‬‬ ‫‪ (11‬ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪Z‬‬ ‫‪π‬‬ ‫‪Z=2‬‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪+i‬‬ ‫‪cos‬‬ ‫‪π‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪:‬‬‫ﻫﻤﺎ‪ 2‬ﻭ ‪ 4‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪‬‬

‫ﻫﻤﺎ ‪ 3‬ﻭ‬ ‫‪Z=3‬‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪cos‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪ (12‬ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z‬ﺤﻴﺙ ‪:‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪ 6‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫ﻫﻲ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ﻁﻭﻴﻠﺘﻪ ‪ 4‬ﻭ ﻋﻤﺩﺘﻪ‬ ‫‪Z‬‬ ‫=‬ ‫‪-4‬‬ ‫‪ei‬‬ ‫‪π‬‬ ‫‪ (13‬ﺍﻟﻌﺒﺎﺭﺓ ‪:‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪π+‬‬ ‫‪π‬‬ ‫ﻫﻲ ﻟﻌﺩﺩ ﻤﺭﻜﺏ ﻁﻭﻴﻠﺘﻪ ‪ 4‬ﻭ ﻋﻤﺩﺘﻪ‬ ‫‪Z‬‬ ‫=‬ ‫‪-4‬‬ ‫‪ei‬‬ ‫‪π‬‬ ‫‪ (14‬ﺍﻟﻌﺒﺎﺭﺓ ‪:‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪Z = -2 +‬‬ ‫‪i‬‬ ‫‪+‬‬ ‫‪k‬‬ ‫‪ei‬‬ ‫‪π‬‬ ‫‪ (15‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺤﻴﺙ‬ ‫‪4‬‬ ‫ﻫﻲ ﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ ﻤﺒﺩﺃﻩ ﺍﻟﻨﻘﻁﺔ ‪ I‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ -2 + i‬ﻭ ‪ k‬ﻤﺘﻐﻴﺭ ﻓﻲ \‬‫‪ (16‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺤﻴﺙ ‪ Z = i + 3 eiθ‬ﺤﻴﺙ ‪ θ‬ﻤﺘﻐﻴﺭ ﻓﻲ \ ﻫﻲ‬ ‫ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ i‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ i‬ﻭ ﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪3‬‬ ‫‪ (17‬ﺘﻭﺠﺩ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻓﻲ ^ ﻻ ﺘﻘﺒل ﺤﻠﻭل‬ ‫‪ (18‬ﺇﺫﺍ ﻜﺎﻥ ﻤﻤﻴﺯ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﺴﺎﻟﺒﺎ ﻓﺈﻨﻬﺎ ﺘﻘﺒل ﺤﻠﻴﻥ ﻤﺘﺭﺍﻓﻘﻴﻥ‪.‬‬ ‫‪ (19‬ﺇﺫﺍ ﻜﺎﻥ‪ ABC‬ﻤﺜﻠﺙ ﺤﻴﺙ ﻟﻭﺍﺤﻕ ‪ A‬ﻭ ‪ B‬ﻭ‪ C‬ﻫﻲ ‪ ZA‬ﻭ ‪ ZB‬ﻭ ‪ ZC‬ﺒﺤﻴﺙ‪:‬‬ ‫‪ZC - ZA‬‬ ‫‪=1‬‬ ‫‪ZB - ZA‬‬ ‫‪ (20‬ﻓﺈﻥ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ ‪.‬‬ ‫‪ (21‬ﺇﺫﺍ ﻜﺎﻥ ‪ ABC‬ﻤﺜﻠﺙ ﺤﻴﺙ ‪ ZC , ZB , ZA‬ﻫﻲ ﻟﻭﺍﺤﻕ ‪C, B ,‬‬ ‫ﻓﺈﻥ ﺍﻟﻤﺜﻠﺙ‬ ‫‪π‬‬ ‫ﻫﻲ‬ ‫‪ZC‬‬ ‫‪- ZA‬‬ ‫ﻭ ﻜﺎﻨﺕ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‬ ‫‪A‬‬ ‫‪2‬‬ ‫‪ZB‬‬ ‫‪- ZA‬‬ ‫‪ ABC‬ﻗﺎﺌﻡ ﻓﻲ ‪. A‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬ ‫ﺇﻟﻴﻙ ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ ‪:‬‬‫‪B‬‬‫‪C‬‬ ‫‪A‬‬ ‫‪vG‬‬ ‫‪O uG‬‬ ‫‪ (1‬ﻋﻴﻥ ﻟﻭﺍﺤﻕ ﺍﻟﻨﻘﻁ ‪GA‬ﻭ‪JBJJ‬ﻭ ‪JJJG JJ.JGC‬‬ ‫‪ (2‬ﻋﻴﻥ ﻟﻭﺍﺤﻕ ﺍﻷﺸﻌﺔ ‪(3 . BC , AC , AB‬‬ ‫ﻋﻴﻥ ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪ D‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﺭﺒﺎﻋﻲ‬ ‫‪ ABCD‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ‪ .‬ﺜﻡ ﻋﻴﻥ ﻻﺤﻘﺔ ﻤﺭﻜﺯﻩ ‪. I‬‬ ‫‪ (4‬ﻋﻴﻥ ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪ E‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ ABDE‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬‫ﻫل ﺘﻭﺠﺩ ﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ‪ x‬ﻭ‪ y‬ﻴﻜﻭﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z = (x + y) + xyi‬ﻤﺴﺎﻭﻴﺎ ﺇﻟﻰ ﺍﻟﻌﺩﺩ‬ ‫ﺍﻟﻤﺭﻜﺏ ‪7 + 12i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪ D , C , B , A‬ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ‪ ZD , ZC , ZB , ZA‬ﺤﻴﺙ ‪:‬‬ ‫‪ZD = ZC ; ZC = -ZA , ZB = ZA ; ZA = α + iβ‬‬ ‫ﺤﻴﺙ ‪ α‬و ‪ β‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪.‬‬

‫ﻤﺎ ﻫﻲ ﻁﺒﻴﻌﺔ ﺍﻟﺭﺒﺎﻋﻲ ‪. ABCD‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﻤﺭﻜﺒﺎﻥ ‪ Z1 = 2 - i :‬ﻭ ‪Z2 = -5 + 3i‬‬ ‫ﺍﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﺍﻷﻋﺩﺍﺩ ﺍﻵﺘﻴﺔ‪.‬‬‫‪Z3‬‬ ‫=‬ ‫‪Z1 - Z2‬‬ ‫‪Z4‬‬ ‫=‬ ‫‪Z12‬‬ ‫‪iZ1‬‬ ‫‪1 + Z1‬‬ ‫‪1 - Z2‬‬ ‫‪Z1 + Z2 2‬‬ ‫‪( ),‬‬ ‫= ‪, Z5‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫ﺍﺤﺴﺏ ﺍﻟﻤﺠﻤﻭﻉ ‪S1 = 1 + i + i2 + . . . + i2008 :‬‬ ‫ﺜﻡ ﺍﻟﻤﺠﻤﻭﻉ ‪S2 = 1 - i + i2 - i3 + . . . + (-i)2008 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬‫ﻤﺜل ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺭﻜﺏ ﺍﻟﻨﻘﻁ ‪ F , E , D , C , B , A‬ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪:‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪ZA‬‬ ‫‪.ZB‬‬ ‫‪,‬‬ ‫‪ZA‬‬ ‫‪+ ZB ,‬‬ ‫‪ZB‬‬ ‫‪,‬‬ ‫‪ZA‬‬ ‫‪ZB‬‬ ‫‪ZA‬‬ ‫‪ZA‬‬ ‫= ) ‪ arg ( ZA‬ﻭ ‪= 2‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫‪,‬‬ ‫]∈‪k‬‬ ‫‪3‬‬ ‫‪ZB‬‬ ‫= ) ‪ arg ( ZB‬ﻭ ‪= 1‬‬ ‫‪−π‬‬ ‫‪+ 2kπ‬‬ ‫‪,‬‬ ‫∈‪k‬‬ ‫]‬ ‫‪6‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﻤﺭﻜﺒﺎﻥ ‪ Z2 = 3 3 - 3i ; Z1 = -2 - 2i :‬ﻋﻥ ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ‬ ‫ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻵﺘﻴﺔ ‪:‬‬ ‫ﻭ ‪ Z12‬ﻭ ‪Z1 . Z2‬‬ ‫‪Z1‬‬ ‫ﻭ‬ ‫‪Z42‬‬ ‫ﻭ‬ ‫‪1‬‬ ‫‪, Z2 , Z1‬‬ ‫‪Z2‬‬ ‫‪Z1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬

‫‪( ) 1 -‬‬‫‪3i 2010‬‬ ‫;‬ ‫‪1+i‬‬ ‫‪1962‬‬ ‫;‬ ‫‪ 1 + i 1418‬‬ ‫ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪:‬‬ ‫‪‬‬ ‫‪ 1 - i ‬‬‫‪ 2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‬‫‪Z3 = -4 6 + 4 2i ; Z2 = -1 - 3i ; Z1 = 1 - i‬‬ ‫ﺍﻜﺘﺏ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻜل ﻤﻥ ‪:‬‬ ‫‪Z1‬‬ ‫‪, Z2‬‬ ‫‪, Z3‬‬ ‫‪, Z1‬‬ ‫‪, Z2‬‬ ‫‪,‬‬ ‫‪Z1‬‬ ‫‪, Z32‬‬ ‫‪, Z1‬‬ ‫‪Z2‬‬ ‫‪Z3‬‬ ‫‪Z2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻌﻼﻗﺘﻴﻥ ‪:‬‬‫‪ cosθ + i sinθ = eiθ‬ﻭ ‪cosθ - i sinθ = e-iθ‬‬ ‫ﺒﻴﻥ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ θ‬ﻓﺈﻥ ‪:‬‬‫‪cosθ‬‬ ‫=‬ ‫‪eiθ‬‬ ‫‪+ e-iθ‬‬ ‫ﻭ‬ ‫‪sin θ‬‬ ‫=‬ ‫‪eiθ‬‬ ‫‪- e-iθ‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬‫ﺍﻜﺘﺏ ﺍﻟﻼﺤﻘﺔ ‪ Z′‬ﻟﻠﻨﻘﻁﺔ ‪ M′‬ﺒﺩﻻﻟﺔ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﻟﻠﻨﻘﻁﺔ ‪ M‬ﺇﺫﺍ ﻜﺎﻨﺕ ‪ M ′‬ﻫﻲ ﺼﻭﺭﺓ ‪M‬‬ ‫ﺒﻭﺍﺴﻁﺔ ‪:‬‬ ‫‪.‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪ wG‬ﻫﻲ‬ ‫ﺍﻟﺫﻱ ﻻﺤﻘﺔ ﺸﻌﺎﻋﻪ‬ ‫‪t‬‬ ‫‪ (1‬ﺍﻻﻨﺴﺤﺎﺏ‬ ‫‪2‬‬‫‪ (2‬ﺍﻟﺘﺤﺎﻜﻲ ‪ H‬ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ O‬ﻭﻴﺤﻭل ﺍﻟﻨﻘﻁﺔ ‪ A‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ 1 + i‬ﺇﻟﻰ ﺍﻟﻨﻘﻁﺔ ‪ B‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪2‬‬ ‫‪. + 2i‬‬‫‪.-‬‬ ‫‪3‬‬ ‫‪+i‬‬ ‫ﻭ ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ C‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪5π‬‬ ‫‪ (3‬ﺍﻟﺩﻭﺭﺍﻥ ‪ R‬ﺍﻟﺫﻱ ﺯﺍﻭﻴﺘﻪ‬ ‫‪6‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 13‬‬ ‫‪ -1‬ﻋﻴﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻟﻜل ﻤﻥ ‪3 + 4i ; 3 – 4i‬‬ ‫‪ -2‬ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z2 - 6Z + 25 = 0 :‬‬

‫‪ -3‬ﺍﺴﺘﻨﺘﺞ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z4 - 6Z2 + 25 = 0 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬‫‪ A,B,C‬ﺜﻼﺙ ﻨﻘﻁ ﻟﻭﺍﺤﻘﻬﺎ ‪a = -1 - i , b = 2 + i , c = 4i :‬‬ ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫‪ -1‬ﻤﺜل ﺍﻟﻨﻘﻁ ‪. A,B,C‬‬ ‫ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ‪.‬‬ ‫=‪Z‬‬ ‫‪a-b‬‬ ‫‪ -2‬ﺍﻜﺘﺏ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪:‬‬ ‫‪c-b‬‬ ‫ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺔ ﺍﻟﻤﺜﻠﺙ ‪. ABC‬‬ ‫‪ -3‬ﻋﻴﻥ ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪ D‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﺭﺒﺎﻋﻲ ‪ ABCD‬ﻤﺭﺒﻊ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 15‬‬‫‪Z‬‬ ‫≠‬ ‫‪i‬‬ ‫‪ Z , i ,‬ﺤﻴﺙ‪:‬‬ ‫‪1‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪ M , B , A‬ﺫﺍﺕ ﺍﻟﻠﻭﺍﺤﻕ‬ ‫‪2‬‬ ‫‪1 - 2Z‬‬ ‫‪.‬‬ ‫‪Z1‬‬ ‫=‬ ‫‪iZ + 1‬‬ ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪ .‬ﻨﻀﻊ‪:‬‬ ‫‪.‬‬ ‫‪Z1‬‬ ‫‪=2.‬‬ ‫‪AM‬‬ ‫‪ -1‬ﺒﻴﻥ ﺃﻥ ‪:‬‬ ‫‪BM‬‬‫ﺍﺴﺘﻨﺘﺞ ﺍﻟﻤﺠﻤﻭﻋﺔ )‪ (E‬ﻟﻠﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪. Z1 = 2‬‬ ‫‪ -2‬ﻋﻴﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ )‪ (F‬ﻟﻠﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ Z1‬ﺤﻘﻴﻘﻲ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬ ‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪ Z = 1 + 2eiθ (1‬ﺤﻴﺙ ‪ θ‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻴﻤﺴﺢ ﻜل \ ‪.‬‬ ‫‪ Z = i + 4eiθ (2‬ﺤﻴﺙ ‪ θ‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻴﻤﺴﺢ ﻜل \ ‪.‬‬‫\‪.‬‬ ‫ﺤﻴﺙ ‪ k‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻴﻤﺴﺢ ﻜل‬ ‫‪Z‬‬ ‫=‬ ‫‪-1‬‬ ‫‪+i‬‬ ‫‪+‬‬ ‫‪kei‬‬ ‫‪π‬‬ ‫‪(3‬‬ ‫‪6‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫ﺤل ﻓﻲ ^ ﺍﻟﺠﻤﻠﺘﻴﻥ ‪:‬‬

‫‪iZ + iZ′ = 3‬‬ ‫‪iZ + (2i - 1) Z′ = 4‬‬‫‪ -iZ + 3iZ′ = -i‬؛ ‪2iZ - Z′ = 1 + i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬ ‫ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪Z2 + 2 (1 - cosθ) Z + 2 - 2 cosθ = 0‬‬ ‫ﺤﻴﺙ ‪ . 0 < θ ≤ π‬ﻨﻜﺘﺏ ﻜل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 19‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻵﺘﻴﺔ ﻓﻲ ^ ‪:‬‬‫)‪Z3 - 12Z2 + 48Z - 128 = 0 . . . (1‬‬ ‫‪ -1‬ﺒﻴﻥ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺘﻘﺒل ﺤﻼ ﺤﻘﻴﻘﻴﺎ ‪ Z0‬ﺤﻴﺙ ‪. Z0 = 8‬‬ ‫‪ -2‬ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪. (1‬‬‫‪ -3‬ﻟﻴﻜﻥ ‪ Z2 , Z1 , Z0‬ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺤﻴﺙ ‪ :‬ﺍﻟﻘﺴﻡ ﺍﻟﺘﺨﻴﻠﻲ ﻟﻠﻌﺩﺩ ‪ Z2‬ﺴﺎﻟﺏ‪.‬‬ ‫‪ -‬ﺍﻜﺘﺏ ﻜل ﻤﻨﻬﺎ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ‪.‬‬‫‪ -‬ﻤﺜل ﺍﻟﻨﻘﻁ ‪ M2 , M1 , M0‬ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ‪ Z2 , Z1 , Z0‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺭﻜﺏ ‪.‬‬ ‫‪Z3‬‬ ‫=‬ ‫‪Z2 - Z0‬‬ ‫‪ -4‬ﻨﻔﺭﺽ ‪:‬‬ ‫‪Z1 - Z0‬‬‫‪ -‬ﺍﺤﺴﺏ ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ‪ . Z3‬ﻤﺎ ﻫﻲ ﻁﺒﻴﻌﺔ ﺍﻟﻤﺜﻠﺙ ‪. M0M1M2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 20‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z‬ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪. M‬‬ ‫‪Z1‬‬ ‫=‬ ‫‪iZ+1-i‬‬ ‫ﻭ ﻨﻀﻊ ‪:‬‬ ‫‪Z-2+i‬‬ ‫‪ -1‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﺘﻜﻭﻥ ‪ Z1‬ﺤﻘﻴﻘﻴﺎ ﻤﻭﺠﺒﺎ‪.‬‬‫‪.‬‬ ‫‪π‬‬ ‫ﺘﺴﺎﻭﻱ‬ ‫‪Z1‬‬ ‫‪ -2‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﺘﻜﻭﻥ ‪ :‬ﻋﻤﺩﺓ‬ ‫‪2‬‬

‫‪ -3‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﺘﻜﻭﻥ ‪. Z1 = 2‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫ﻤﺴﺎﻭﻴﺔ ﺇﻟﻰ‬ ‫‪Z1‬‬ ‫‪ -4‬ﻋﻴﻥ ﻤﺠﻤﻭﻉ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﻋﻤﺩﺓ‬ ‫‪4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 21‬‬ ‫ﻋﻴﻥ ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪:‬‬ ‫‪ Z = 1 - cosθ + i sinθ‬ﻤﻊ ‪. -π < θ ≤ π‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 22‬‬‫ﻋﻴﻥ ﺜﻼﺙ ﺃﻋﺩﺍﺩ ﻤﺭﻜﺒﺔ ﺠﺩﺍﺅﻫﺎ ‪ -27i‬ﻭ ﻁﻭﻴﻼﺘﻬﺎ ﺘﺸﻜل ﺤﺩﻭﺩ ﻤﺘﺘﺎﺒﻌﺔ ﻤﻥ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﺎﻫﺎ ‪3‬‬ ‫‪π‬‬ ‫‪ .‬ﻭ ﻋﻤﺩﺘﻬﺎ ﺘﺸﻜل ﺤﺩﻭﺩ ﻤﺘﺘﺎﺒﻌﺔ ﻤﻥ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﺎﻫﺎ ‪. 3‬‬‫‪.(2 cm‬‬ ‫)ﺍﻟﻭﺤﺩﺓ‬ ‫؛‬ ‫‪(O‬‬ ‫;‬ ‫‪uG‬‬ ‫‪,‬‬ ‫) ‪vG‬‬ ‫ﻤﺘﺠﺎﻨﺱ‬ ‫ﻭ‬ ‫ﻤﺘﻌﺎﻤﺩ‬ ‫ﻤﻌﻠﻡ‬ ‫ﺇﻟﻰ‬ ‫ﺍﻟﻤﻨﺴﻭﺏ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 23‬‬ ‫ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ‬‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁﺔ ‪ A‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ ZA = 1‬ﻭ ﺍﻟﺩﺍﺌﺭﺓ )‪ (C‬ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ ‪ A‬ﻭ ﻨﺼﻑ ﺍﻟﻘﻁﺭ ‪. R = 1‬‬‫ﻭ ﺍﻟﻨﻘﻁﺔ ُ‪ E‬ﺫﺍﺕ‬ ‫‪ZB‬‬ ‫=‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪ei‬‬ ‫‪π‬‬ ‫‪ (I‬ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁﺔ ‪ F‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ 2‬ﻭ ﺍﻟﻨﻘﻁﺔ ‪ B‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪3‬‬ ‫ﺍﻟﻼﺤﻘﺔ ‪. ZE = 1 + ZB2‬‬ ‫‪ -1‬ﺒﻴﻥ ﺃﻥ ‪ B‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﺩ‪G‬ﺍ‪J‬ﺌ‪J‬ﺭﺓ‪( )JJ.JG(C) J‬‬ ‫‪ -2‬ﻋﻴﻥ ﻗﻴﺱ ﻟﻠﺯﺍﻭﻴﺔ ‪. AF , AB‬‬ ‫‪ -3‬ﻋﻴﻥ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻠﻌﺩﺩ ‪ ZB - ZA‬ﻭ ‪. ZE - ZA‬‬ ‫‪ -4‬ﺍﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻨﻘﻁ ‪ A‬ﻭ ‪ B‬ﻭ‪ E‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ‪.‬‬ ‫‪ (II‬ﻟﺘﻜﻥ ﺍﻟﻨﻘﻁﺘﺎﻥ ‪ M‬ﻭ ‪ M′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺘﻴﻥ ‪ Z‬ﻭ ‪ Z′‬ﺤﻴﺙ ‪:‬‬ ‫‪ Z′ = 1 + Z2‬ﻭ ‪Z ≠ 0 , Z ≠ 1‬‬ ‫≠‪. Z‬‬ ‫‪Z‬و‪1‬‬ ‫≠‬ ‫‪0‬‬ ‫ﺤﻴﺙ‬ ‫‪Z′ - 1‬‬ ‫‪ -1‬ﻤﺎ ﻫﻭ ﺍﻟﺘﻔﺴﻴﺭ ﺍﻟﻬﻨﺩﺴﻲ ﻟـ ‪Z - 1‬‬ ‫‪ -2‬ﺍﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻨﻘﻁ ‪ A‬ﻭ ‪ M‬ﻭ ‪ M′‬ﺘﻜﻭﻥ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬