دروس مادة الرياضيات للفصل الاول للشعب العلمية سنة ثالثة ثانوي

‫‪.‬‬ ‫‪Z2‬‬ ‫∈‬ ‫ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪\ :‬‬ ‫‪Z-1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬‫ﻨﻌﺘﺒﺭ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ‪f (Z) = Z3 - (2 - 3i) Z2 + 9Z - 18 + 27i :‬‬ ‫‪(1‬ﺍ ﺤﺴﺏ ‪ f (Z) :‬ﺒﺩﻻﻟﺔ ‪. Z‬‬‫‪ (2‬ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ f (Z) = 0‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻨﻬﺎ ﺘﻘﺒل ﺤﻠﻴﻥ ﻤﺘﺭﺍﻓﻘﻴﻥ ‪ Z1‬و ‪. Z1‬‬ ‫‪ (3‬ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪ A‬ﻭ‪ B‬ﻭ‪ C‬ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ‪2 – 3i ; -3i ; 3i :‬‬ ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫= ‪. Z′‬‬ ‫‪ZA - ZB‬‬ ‫‪ -‬ﺍﺤﺴﺏ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪ Z′‬ﺤﻴﺙ ‪:‬‬ ‫‪ZC - ZB‬‬ ‫‪ -‬ﺍﺤﺴﺏ ‪ Z′‬ﻭ ﻋﻤﺩﺓ ‪ . Z′‬ﻤﺎ ﻫﻲ ﻁﺒﻴﻌﺔ ﺍﻟﻤﺜﻠﺙ ‪. ABC‬‬ ‫‪ M (4‬ﻨﻘﻁﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﻻﺤﻘﺘﻬﺎ ‪. Z‬‬‫ﻤﺎ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ‪. AM2 + 2MB2 - 2MC2 = 25 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 25‬‬ ‫‪ (1‬ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z2 - (1 + i) Z - 4i = 0 :‬‬ ‫‪ (2‬ﻨﻌﺘﺒﺭ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬‫‪Z4 - (1 + i) Z3 + (9 - 4i)Z2 - 9 (1 + i) Z - 36 = 0‬‬ ‫‪ -‬ﺒﻴﻥ ﺃﻥ ﻟﻬﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﺘﺨﻴﻠﻴﻴﻥ ﺼﺭﻓﻴﻴﻥ ﻤﺘﻌﺎﻜﺴﻴﻥ ‪ Z3‬و ‪. Z4‬‬ ‫‪ -‬ﺍﺴﺘﻨﺘﺞ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ‪.‬‬‫ﻤﺴﺎﻭﻴﺔ ﺇﻟﻰ‬ ‫‪Z - Z3 - Z4‬‬ ‫‪ -3‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪:‬‬ ‫‪Z - Z1 - Z2‬‬ ‫‪-π‬‬ ‫‪.2‬‬‫ﺤﻴﺙ ‪ Z‬ﻫﻲ ﻻﺤﻘﺔ ‪ M‬ﻭ ‪ Z4 , Z3 , Z2 , Z1‬ﻫﻲ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ ﻓﻲ ﺍﻟﺴﺅﺍل )‪. (2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 26‬‬

‫‪ ABC‬ﻤﺜﻠﺙ‪ .‬ﻨﺭﺴﻡ ﺨﺎﺭﺝ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﺍﻟﻤﺜﻠﺜﺎﺕ ﺍﻟﻤﻘﺎﻴﺴﺔ ﺍﻷﻀﻼﻉ ‪ ABC‬ﻭ ‪ ACE‬ﻭ ‪ BCF‬ﺍﻟﺘﻲ‬ ‫ﻤﺭﺍﻜﺯ ﺜﻘﻠﻬﺎ ‪ K , J , I‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‪.‬‬ ‫ﺒﺭﻫﻥ ﺃﻥ ﻟﻠﻤﺜﻠﺜﺎﺕ ‪ ABC‬ﻭ ‪ FED‬ﻭ ‪ IJK‬ﻨﻔﺱ ﻤﺭﻜﺯ ﺍﻟﺜﻘل‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 27‬‬ ‫ﻋﻴﻥ ﺍﻟﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪. 1 + cosx + i sin x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 28‬‬ ‫‪ ABCD‬ﻤﺭﺒﻊ‪ .‬ﻨﺭﺴﻡ ﺍﻟﻤﺭﺒﻌﻴﻥ ‪ DCEF‬ﻭ ‪ GHBA‬ﻜﻤﺎ ﻴﻅﻬﺭ ﻋﻠﻰ ﺍﻟﺸﻜل ‪:‬‬ ‫‪G AD F‬‬ ‫‪αβ γ‬‬ ‫‪H B CE‬‬ ‫ﺒﺭﻫﻥ ﺃﻥ ‪α + β = γ :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪( )G G . 29‬‬ ‫ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ‪. O ; i , j‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪ C , B , A‬ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ‪ -4 + 2i ; i + 2 , 3 – i :‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫‪ (1‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ N‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺒﺤﻴﺙ ‪Z - 3 + i = 2‬‬‫‪arg‬‬ ‫‪ Z-2-i ‬‬ ‫=‬ ‫‪π‬‬ ‫‪ (2‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪ Z + 4 - 2i ‬‬ ‫‪2‬‬ ‫‪( )JJJG JJJG‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 30‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﺤﻴﺙ ‪ k ∈ Z , AB , AC = α + 2kπ‬ﻭ‬ ‫[‪. α ∈ ]0 ; π‬‬‫ﻨﻨﺸﺊ ﺨﺎﺭﺝ ﻫﺫﺍ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻤﺭﺒﻌﺎﻥ ‪ ABFG‬ﻭ ‪ ACDE‬ﻭ ﻤﺘﻭﺍﺯﻱ ﺍﻷﻀﻼﻉ ‪ AELG‬ﺒﺭﻫﻥ ﺃﻥ‬ ‫)‪ ( AL) ⊥ (BC‬و ‪AL = BC‬‬

‫ﺍﻟﺤـﻠــــــﻭل‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬ ‫‪× (5 × (4‬‬ ‫‪√ (3 √ (2‬‬ ‫‪× (1‬‬ ‫‪× (10 √ (9‬‬ ‫‪√ (8 × (7‬‬ ‫‪× (6‬‬ ‫‪√ (15 × (14 × (13 × (12 × (11‬‬ ‫‪√ (20 √ (19 × (18 × (17 √ (16‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬ ‫‪ (1‬ﻻﺤﻘﺔ ‪ A‬ﻫﻲ ‪ZA = 2 + i :‬‬ ‫ﻻﺤﻘﺔ ‪ B‬ﻫﻲ ‪ZB = -1 + 3i :‬‬ ‫ﻻﺤﻘﺔ ‪ C‬ﻫﻲ ‪ZC = -2 +JJi JG:‬‬‫→‪Z‬‬ ‫=‬ ‫‪ZB‬‬ ‫‪-‬‬ ‫‪ZA‬‬ ‫=‬ ‫‪-3‬‬ ‫‪+‬‬ ‫‪2i‬‬ ‫‪ (2‬ﻻﺤﻘﺔ ﺍﻟﺸﻌﺎﻉ ‪ AB‬ﻫﻲ ‪:‬‬ ‫‪AB‬‬ ‫‪JJJG‬‬ ‫→‪Z‬‬ ‫=‬ ‫‪ZC‬‬ ‫‪-‬‬ ‫‪ZA‬‬ ‫=‬ ‫‪-4‬‬ ‫ﻻﺤﻘﺔ ﺍﻟﺸﻌﺎﻉ ‪ AC‬ﻫﻲ ‪:‬‬ ‫‪AC‬‬ ‫‪JJJG‬‬‫→‪Z‬‬ ‫=‬ ‫‪ZC‬‬ ‫‪-‬‬ ‫‪ZB‬‬ ‫=‬ ‫‪-1‬‬ ‫‪-‬‬ ‫‪2i‬‬ ‫ﻻﺤﻘﺔ ﺍﻟﺸﻌﺎﻉ ‪ BC‬ﻫﻲ ‪:‬‬ ‫‪AB‬‬ ‫‪ (3‬ﺘﻌﻴﻴﻥ ﻻﺤﻘﺔ ‪ D‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ BCD‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ‪.‬‬ ‫ﺘﻜﻭﻥ ‪ ABCD‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬‫‪ZDC = ZC - ZD‬‬ ‫‪, ZJJJG = -3 + 2i‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪Z JJJG‬‬ ‫‪= ZJJJG‬‬ ‫‪AB‬‬ ‫‪AB‬‬ ‫‪DC‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪Z JJJG‬‬ ‫=‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪i‬‬ ‫‪-‬‬ ‫‪ZE‬‬ ‫‪:‬‬ ‫ﺃﻱ‬ ‫‪DC‬‬ ‫‪-2 + i - ZD = -3 + 2i‬‬ ‫ﺇﺫﻥ ‪ZD = 1 - i :‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻻﺤﻘﺔ ‪ I‬ﻤﺭﻜﺯ ‪ I : ABCD‬ﻫﻭ ﻤﻨﺘﺼﻑ ‪[ ]AC‬‬ ‫‪ZI‬‬ ‫=‬ ‫‪ZA‬‬ ‫‪+ ZC‬‬ ‫=‬ ‫‪2+i-2+i‬‬ ‫‪=i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪ZI = i :‬‬

‫‪ -4‬ﺘﻌﻴﻴﻥ ﻻﺤﻘﺔ ‪ E‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ ABCD :‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ ﻴﻜﻭﻥ ‪ DE‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ ﺇﺫﺍ ﻭﻓﻕ ﺇﺫﺍ‬ ‫ﻜﺎﻥ ‪:‬‬‫‪Z JJJG‬‬ ‫=‬ ‫‪ZD‬‬ ‫‪-‬‬ ‫‪ZE‬‬ ‫‪,‬‬ ‫‪Z JJJG‬‬ ‫=‬ ‫‪-3‬‬ ‫‪+‬‬ ‫‪2i‬‬ ‫ﺤﻴﺙ‬ ‫‪Z JJJG‬‬ ‫=‬ ‫‪Z JJJG‬‬ ‫‪ED‬‬ ‫‪AB‬‬ ‫‪AB‬‬ ‫‪ED‬‬ ‫‪Z JJJG‬‬ ‫=‬ ‫‪-‬‬ ‫‪i‬‬ ‫‪-‬‬ ‫‪ZE‬‬ ‫ﺃﻱ‬ ‫‪ED‬‬ ‫ﻭﻋﻠﻴﻪ ‪1 - i - ZE = -3 + 2i :‬‬ ‫ﺇﺫﻥ ‪. ZE = 4 - 3i :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫ﺇﻴﺠﺎﺩ ﺍﻷﻋﺩﺍﺩ ‪ x‬ﻭ‪ y‬ﺒﺤﻴﺙ ‪Z = 7 + 12i :‬‬ ‫‪x + y = 7‬‬ ‫‪‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪.‬‬ ‫‪y‬‬ ‫=‬ ‫‪12‬‬‫ﻭﻤﻨﻪ ‪ x‬ﻭ‪ y‬ﻫﻤﺎ ﺤﻼﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ t1 = 3 , ∆ = 1 , t2 - 7t + 12 = 0 :‬ﻭ‬ ‫‪. t2 = 4‬‬ ‫ﻭﻤﻨﻪ‪ x = 3‬ﻭ ‪ y = 4‬ﺃﻭ ‪ x = 4‬ﻭ ‪y = 3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫ﻁﺒﻴﻌﺔ ﺍﻟﺭﺒﺎﻋﻲ ‪: ABCD‬‬‫ﻟﺩﻴﻨﺎ ‪ ZB = ZA‬ﻭﻤﻨﻪ ‪ A‬ﻭ ‪ B‬ﻤﺘﻨﺎﻅﺭﺘﺎﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪. OA = OB :‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ ZC = - ZA‬ﻓﺈﻥ ‪ A‬ﻭ ‪ C‬ﻤﺘﻨﺎﻅﺭﺘﺎﻥ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﺒﺩﺃ‬ ‫ﺍﻟﻤﻌﻠﻡ ‪ O‬ﻭ ﻟﺩﻴﻨﺎ ‪. OA = OC :‬‬‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ ZD = ZC‬ﻓﺈﻥ ‪ D‬ﻭ‪ C‬ﻤﺘﻨﺎﻅﺭﺘﺎﻥ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪. OC = OD :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ ABCD‬ﻫﻭ ﻤﺴﺘﻁﻴل ﻷﻥ ﻗﻁﺭﺍﻥ ﻤﺘﻘﺎﻴﺴﺎﻥ ﻭ ﻤﺘﻨﺎﺼﻔﺎﻥ ‪.‬‬

‫‪D‬‬ ‫‪β‬‬ ‫‪A‬‬ ‫‪-α‬‬ ‫‪vG‬‬ ‫‪α‬‬ ‫‪O uG‬‬ ‫‪C‬‬ ‫‪B‬‬ ‫‪-β‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫ﺍﻟﻜﺘﺎﺒﺔ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ‪:‬‬‫= ‪Z3‬‬ ‫‪Z1 - Z2‬‬ ‫=‬ ‫‪2 - i + 5 - 3i‬‬ ‫=‬ ‫‪7 - 4i‬‬ ‫×‬ ‫‪3+i‬‬ ‫‪1 + Z1‬‬ ‫‪1+2-i‬‬ ‫‪3-i‬‬ ‫‪3+i‬‬‫‪Z3‬‬ ‫=‬ ‫‪25 - 2i‬‬ ‫=‬ ‫‪5‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪10‬‬ ‫‪2‬‬ ‫‪5‬‬‫‪Z4‬‬ ‫=‬ ‫‪Z12‬‬ ‫=‬ ‫‪(2 - i)2‬‬ ‫=‬ ‫‪4 - 4i - 1‬‬ ‫‪1 - Z2‬‬ ‫‪1 + 5 - 3i‬‬ ‫‪6 - 3i‬‬‫‪Z4‬‬ ‫=‬ ‫‪3 - 4i‬‬ ‫×‬ ‫‪6 + 3i‬‬ ‫=‬ ‫‪30 - 15i‬‬ ‫=‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪6 - 3i‬‬ ‫‪6 + 3i‬‬ ‫‪45‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪i Z1‬‬ ‫)‪i (2 - i‬‬ ‫‪2i + 1‬‬‫‪Z5‬‬ ‫=‬ ‫‪(Z1 + Z2 )2‬‬ ‫=‬ ‫‪(-5 + 3i + 2 - i)2‬‬ ‫=‬ ‫‪(-3 + 2i)2‬‬‫‪Z5‬‬ ‫=‬ ‫‪2i + 1‬‬ ‫×‬ ‫‪5 + 12i‬‬ ‫=‬ ‫‪-19 + 22i‬‬ ‫‪5 - 12i‬‬ ‫‪5 + 12i‬‬ ‫‪169‬‬ ‫‪-19‬‬ ‫‪22‬‬‫‪Z5‬‬ ‫=‬ ‫‪169‬‬ ‫‪+‬‬ ‫‪169‬‬ ‫‪i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫‪ -‬ﺤﺴﺎﺏ ‪ : S‬ﺍﻟﻤﺠﻤﻭﻉ ﻫﻭ ﻤﺠﻤﻭﻉ ‪ n + 1‬ﺤﺩ ﻤﻥ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ‪ 1‬ﻭ ﺃﺴﺎﺴﻬﺎ ‪i‬‬

S1 = 1 × 1 - i2009 = 1 - i . i2008 1-i 1-iS1 = 1 - i . ( ) i 2 1004 = 1 - i ( )-1 1004 = 1-i =1  1-i  1-i 1-i : S2 ‫ ﺤﺴﺎﺏ‬-S2 = 1 - i + i2 - i3 + . . . + (-i)2008S2 = (-i)0 + (-i)1 + (-i)2 + (-i)3 + . . . + (-i)2008 1 ‫ﺤﺩ ﻤﻥ ﺤﺩﻭﺩ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل‬2009 ‫ﻭﻫﻭ ﻤﺠﻤﻭﻉ‬ : ‫ ﺇﺫﻥ‬. –i ‫ﻭ ﺃﺴﺎﺴﻬﺎ‬ S2 = 1 × 1 - (-i)2009 = 1 - (-i) (-i)2008 1 - (-i) 1+i 1+i ( -i )2  1004 1 + i ( )-1 1004 S1 = = 1 +i = 1+i =1 1-i 1+i . 7‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺍﻟﺘﻤﺜﻴل ﺍﻟﻬﻨﺩﺴﻲ‬ ZD = 2 ‫ ؛‬ZD = ZA ZB ‫ ؛‬ZD = ZA′ ZBarg( ZD ) = agr ( ZA ) + arg( ZB ) = π - π = π + 2kπ 3 6 6 arg  1  = -arg( ZA ) + 2kπ : ‫ﻟﺩﻴﻨﺎ‬  ZA   ZE = 1 = 1 = 1 ‫ﻭ‬ arg( ZE ) = - π + 2kπ :‫ﻭ ﻤﻨﻪ‬ ZA ZA 2 3

: ‫ﻭ ﻤﻨﻪ‬ arg  1  = -arg(ZB ) + 2kπ , k ∈ Z :‫ﻟﺩﻴﻨﺎ‬  ZB  ZF = 1 = 1 = 1 =1 ‫ﻭ‬ arg( ZF ) = π + 2kπ ZB ZB 1 6 A C D F EB . 8‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺘﻌﻴﻴﻥ ﺍﻟﻁﻭﻴﻠﺔ ﻭ ﺍﻟﻌﻤﺩﺓ‬• Z1 = (-2)2 + (-2)2 = 8 = 2 2 π cosθ1 = -2 = −2 4 22 2 θ1 =π + + 2kπ :‫ﻭﻤﻨﻪ‬  -2 −2 sin θ1 = 22 = 2 k∈Z ‫ﻭ‬ θ1 = 5π + 2kπ : ‫ﺃﻱ‬ 4( )• Z2 = 3 3 2 + (-3)2 = 36 = 6

-π cosθ2 = 33 = 3 6  = 6 2 k∈Z ; θ2 = + 2kπ : ‫ﻭﻤﻨﻪ‬ sin θ2 -3 1 6 = - 2• Z1 . Z2 = Z1 . Z2 = 2 2 × 6 = 12 2arg ( Z1 . Z2 ) = arg ( Z1 ) + arg( Z2 ) + 2kπ , k ∈ Z = 5π - π + 2kπ = 15π - 2π + 2k 4 6 12 13π = 12 + 2kπ( )• Z12 = Z1 2 = 2 2 2 =8( )arg Z12 = 2 arg( Z1 ) = 2 × 5π + 2kπ 4 5π = 2 + 2kπ , k∈ Z• Z1 = Z1 = 22 = 2 Z2 Z2 6 3arg  Z1  = arg( Z1 ) - arg( Z2 ) + 2kπ , k∈ Z  Z2    = 5π + π + 2kπ , k ∈ Z 4 6 17π = 12 + 2kπ , k∈ Z

• Z42 = Z2 4 = (6)4 = 1296( )arg Z42 = 4arg ( Z2 ) +2kπ , k ∈ Z = 4 -π  + 2kπ , k∈ Z 6  = −2π + 2kπ , k∈ Z 3• 1 = 1 = 1 = 2 Z1 Z1 22 4arg  1  = -arg( Z1 ) + 2kπ , k∈ Z  Z1    = - 5π + 2kπ , k∈Z 4 . 9‫ﺍﻟﺘﻤﺭﻴﻥ‬  1 - 3i 2010  2  ‫(ﺤﺴﺎﺏ‬1 1 - 3i ‫ ﻋﻤﺩﺓ‬θ ‫ﻨﻔﺭﺽ‬ θ= -π + 2kπ ‫ﻭﻤﻨﻪ‬ cosθ = 1 , 1 - 3i = 2 3  2 sin θ = -3 2 1- 3i = 2 cos  -π  + i sin  -π  :‫ﻭﻤﻨﻪ‬  3   3 

1- 3i 2010 cos  -π   -π  2010  2   3   3  = + i sin : ‫ﻭﻋﻠﻴﻪ‬ = cos  -2010π  + i sin  -2010π   3  3  = cos(-670π) + i sin(-670π) =1 ( )1 + i 1962 : ‫ ﺤﺴﺎﺏ‬x( )1 + i 1962 = (1 + i )2  981 = ( 2i)981 = ( 2)981 . ( i )981  ( )= 2981 . i . i980 = 2981 . i . i2 490 = 2981 . i . ( -1)490 = i . ( )2 981  1 + i 1418  1 - i  : ‫ﺤﺴﺎﺏ‬•  1 + i 1418 =  1 + i 2 709 = (1 + i)2   1 - i       1 - i   (1 - i)2  =  2i 709 = ( )-1 709 = -1  -2i  . 10‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺍﻟﺸﻜل ﺍﻷﺴﻲ‬• Z1 = 2 , cosθ1 = 2  2 sinθ1 = -2 2

Z1 = 2 e-i π : ‫ﻭﻋﻠﻴﻪ‬ k∈ Z , θ1 = -π + 2kπ : ‫ﺇﺫﻥ‬ 4 4  cosθ2 = - 1  2• Z2 = 2 ,  sinθ2 -3 = 2 Z2 = 2 e4i π ‫؛‬ θ2 = 4π + 2kπ , k∈ Z : ‫ﻭﻤﻨﻪ‬ 3 3  cosθ 3 = -4 6 = −3  8 2 2• Z3 = 8 2 ,  4 2 1 sinθ12 = 8 2 = 2 Z3 = 8 2 e5i π ‫؛‬ θ3 = 5π + 2kπ , k ∈ Z ‫ﻭﻤﻨﻪ‬ 6 6• Z1 . Z2 = 2 2 ei  -π + 4π  = 2 2 ei1132π  4 3 • Z1 = 2 ei  -π - 4π  = 22 ei  -19π  Z2 2  4 3   12 ( )• Z23 =8 2 3 ei 3× 5π  = 1024 2e 6 ( ) ( ) ( )• Z1Z2Z3 = 2 e = 32 . ei-π+ 4π + 5π  i 23π 2 82  4 3 6  12 . 11‫ﺍﻟﺘﻤﺭﻴﻥ‬

‫‪cosθ + i sinθ = eiθ‬‬ ‫ﻟﺩﻴﻨﺎ ‪cosθ - i sinθ = e-iθ :‬‬ ‫ﻭﺒﺎﻟﺠﻤﻊ ﻨﺠﺩ ‪2cosθ = eiθ + e-iθ :‬‬ ‫ﻭﺒﺎﻟﻁﺭﺡ ﻨﺠﺩ ‪2i sinθ = eiθ - e-iθ :‬‬ ‫‪cosθ‬‬ ‫=‬ ‫‪eiθ‬‬ ‫‪+ e-iθ‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫‪sinθ‬‬ ‫=‬ ‫‪eiθ - e-iθ‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬ ‫‪Z′ = Z +‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪ (1‬ﻓﻲ ﺤﺎﻟﺔ ﺍﻻﻨﺴﺤﺎﺏ ‪: T‬‬ ‫‪2‬‬ ‫‪ (2‬ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﺘﺤﺎﻜﻲ ‪Z′ = kZ : H‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ T(A) = B :‬ﻓﺈﻥ ‪2 + 2i = k (i + 1) :‬‬ ‫ﺇﺫﻥ ‪k = 1 :‬‬ ‫=‪k‬‬ ‫)‪2(i + 1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫=‪k‬‬ ‫‪2 + 2i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪i+1‬‬ ‫‪i+1‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪Z′ = 2Z :‬‬ ‫‪ (3‬ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﺩﻭﺭﺍﻥ ‪Z′ = aZ + b : R‬‬ ‫=‪a‬‬ ‫‪-3‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪a‬‬ ‫=‬ ‫‪cos‬‬ ‫‪5π‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪sin‬‬ ‫‪5π‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪b‬‬ ‫‪b‬‬ ‫‪1-a‬‬ ‫=‬ ‫‪3 +i‬‬ ‫ﻻﺤﻘﺔ ﺍﻟﻤﺭﻜﺯ ﻫﻲ ‪ 1 - a :‬ﻭﻤﻨﻪ ‪:‬‬ ‫ﻭﻤﻨﻪ ‪b = (- 3 + i) a :‬‬ ‫= ‪( )b‬‬ ‫‪- 3‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪- 3 +i‬‬ ‫‪ 2‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪i‬‬ ‫‪‬‬ ‫ﺇﺫﻥ ‪:‬‬‫=‪b‬‬ ‫‪3‬‬ ‫‪-‬‬ ‫‪3‬‬ ‫‪i‬‬ ‫‪-‬‬ ‫‪i‬‬ ‫‪3‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪=1-i‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬

‫‪Z′‬‬ ‫‪- 3‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪‬‬ ‫‪Z+1-i‬‬ ‫‪3‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪ 2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 13‬‬ ‫‪ -1‬ﺘﻌﻴﻴﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ‪:‬‬‫‪ 3 – 4i x‬ﻨﻔﺭﺽ ‪ δ = x + iy‬ﺠﺫﺭ ﺘﺭﺒﻴﻌﻲ ﻟﻠﻌﺩﺩ ‪3 - 4i‬‬‫ﻴﻜﻭﻥ ‪ δ2 = 3 - 4i :‬ﻭﻤﻨﻪ ‪(x + iy)2 = 3 - 4i :‬‬ ‫‪x2 - y2 = 3‬‬ ‫‪‬‬ ‫ﺃﻱ ﺃﻥ ‪:‬‬ ‫‪‬‬ ‫‪2xy‬‬ ‫=‬ ‫‪-4‬‬‫ﻭ ﻟﺩﻴﻨﺎ ﻤﻥ ﺍﻟﻌﻼﻤﺔ ‪δ2 = 3 - 4i : δ2 = 3 - 4i‬‬‫ﺇﺫﻥ ‪ δ 2 = (3)2 + (-4)2‬ﺃﻱ ﺃﻥ ‪x2 + y2 = 5 :‬‬ ‫)‪ x2 - y2 = 3 . . . (1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪xy‬‬ ‫=‬ ‫‪-2‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫)‪(2‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2‬‬ ‫=‬ ‫‪5‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫)‪(3‬‬ ‫‪‬‬ ‫ﺒﺠﻤﻊ )‪ (1‬ﻭ )‪ (3‬ﻨﺠﺩ ‪ 2 x2 = 8 :‬ﻭﻤﻨﻪ ‪x2 = 4‬‬ ‫ﺇﺫﻥ ‪ x = 2 :‬ﺃﻭ ‪ x = -1‬ﻭ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )‪ (2‬ﻨﺠﺩ‪:‬‬ ‫ﻟﻤﺎ ‪ y = -1 : x = 2‬ﻭﻟﻤﺎ ‪y = 1 : x = -2‬‬ ‫ﻭﻤﻨﻪ ‪ δ1 = 2 - i‬ﻭ ‪δ2 = -2 + i‬‬ ‫ﻫﻤﺎ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻟﻠﻌﺩﺩ ‪. 3 – 4i‬‬ ‫• ‪3 + 4i‬‬ ‫ﻟﻴﻜﻥ ‪ δ‬ﺠﺫﺭ ﺘﺭﺒﻴﻌﻲ ﻟﻠﻌﺩﺩ ‪ 3 + 4i‬ﺃﻱ ‪δ2 = 3 + 4i‬‬‫ﻨﻀﻊ ‪ δ = α + iβ‬ﻭﻤﻨﻪ ‪(α + iβ)2 = 3 + 4i‬‬ ‫‪α2 - β2 = 3‬‬ ‫‪2αβ = 4‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬

‫ﻭ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ‪ δ2 = 3 + 4i‬ﻟﺩﻴﻨﺎ ‪δ2 = 3 + 4i :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ δ 2 = (3)2 + (4)2 :‬ﻭﻤﻨﻪ ‪α2 + β2 = 5 :‬‬ ‫)‪α2 - β2 = 3 . . . (1‬‬ ‫)‪α β = 2 . . . (2‬‬ ‫)‪α2 + β2 = 5 . . . (3‬‬ ‫ﺒﺠﻤﻊ )‪ (1‬ﻭ )‪ (3‬ﻨﺠﺩ ‪ 2α2 = 8 :‬ﻭﻤﻨﻪ ‪α2 = 4‬‬ ‫ﺇﺫﻥ ‪ α = 2 :‬ﺃﻭ ‪ α = -2‬ﻭ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ )‪ (2‬ﻨﺠﺩ‪:‬‬ ‫ﻟﻤﺎ ‪ β = 1 : α = 2‬ﻭ ﻟﻤﺎ ‪β = -1 : α = -2‬‬ ‫ﺇﺫﻥ ‪ δ1 = 2 + i :‬ﻭ ‪δ2 = 2 + i‬‬ ‫ﻫﻤﺎ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ ‪3 + 4i‬‬ ‫‪ -2‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z2 - 6Z + 25 = 0 :‬‬‫)‪ ∆′ = (-3)2 - 25(1‬ﻭﻤﻨﻪ ‪ ∆′ = -16‬ﺃﻱ ‪∆′ = 16i2‬‬ ‫ﺇﺫﻥ ﺍﻟﺠﺫﺭﻴﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻫﻤﺎ ‪ 4i :‬ﻭ ‪-4i‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪ Z1 = 3 - 4i :‬ﻭ ‪Z2 = 3 + 4i‬‬ ‫‪ -3‬ﺍﺴﺘﻨﺘﺎﺝ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z4 - 6Z2 + 25 = 0 :‬‬ ‫ﺒﻭﻀﻊ ‪ Z2 = T‬ﻨﺠﺩ ‪T2 - 6T + 25 = 0 :‬‬ ‫ﻭﻟﻬﺎ ﺤﻠﻴﻥ ‪ T1 = 3 - 4i‬ﻭ ‪T2 = 3 - 4i‬‬ ‫ﻭﻤﻨﻪ ‪ Z2 = 3 - 4i :‬ﺃﻭ ‪Z2 = 3 + 4i‬‬ ‫ﺇﺫﻥ ‪ Z = 2 – i :‬ﺃﻭ ‪ Z = -2 + i‬ﺃﻭ ‪Z =-2 - i‬‬ ‫}‪S = {2 - i , -2 + i , 2 + i , -2 - i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬ ‫‪ -1‬ﺘﻤﺜﻴل ﺍﻟﻨﻘﻁ ‪:‬‬

C D B O A .‫ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ‬Z ‫ ﻜﺘﺎﺒﺔ‬-2Z= -1 - i - 2 - i = -3 - 2i × -2 - 3i 4i - 2 - i -2 + 3i -2+3 i 6 + 9i + 4i - 6Z= 4+9 =i arg(Z) = π , Z = 1 ‫ﻭﻤﻨﻪ‬ 2 ( )JJJG JJJG BA ‫ﻭ‬ Z = BC : ‫ﻭﺒﻤﺎ ﺃﻥ‬ arg(Z) = BC , BA ( )JJJG JJJG = π ‫ﻭ‬ BA =1 : ‫ﻓﺈﻥ‬ 2 BC BC , BA ( )JJJG JJJG = π ‫ﻭ‬ BC = BA : ‫ﺇﺫﻥ‬ 2 BC , BA . ‫ ﻭ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ‬B ‫ ﻗﺎﺌﻡ ﻓﻲ‬ABC ‫ﻭﻤﻨﻪ ﺍﻟﻤﺜﻠﺙ‬ . ‫ ﻤﺭﺒﻊ ﺇﺫﺍ ﻭﺍﻓﻕ ﺇﺫﺍ ﻜﺎﻥ‬ABCD ‫ﻲ‬J‫ﻋ‬J‫ﺎ‬JG‫ﻭﻥ ﺍﻟﺭﺒ‬J‫ﻴﻜ‬J-JG3 ZJJJG = ZJJJG : ‫ ﻭﻤﻨﻪ‬BC = AD BC AD ZC - ZB = ZD - ZA : ‫ﻭﻤﻨﻪ‬

ZD = 2i - 3 : ‫ ﺇﺫﻥ‬4i - 2 - i = ZD + 1 + i : ‫ﺇﺫﻥ‬ . 15‫ﺍﻟﺘﻤﺭﻴﻥ‬ Z1 =2. AM ‫ ﻨﺒﻴﻥ ﺃﻥ‬-1 BM 1 - 2Z -2  Z - 1 -2 Z- 1 iZ + 1  2  i 2Z1 = = 1 =  i  Z-i i  Z + Z- 1 2Z1 = 2i Z-i Z- 1 Z- 1 AM 2 2 BMZ1 = 2i . =2. =2. Z-i Z-i : (E) ‫ﺍﺴﺘﻨﺘﺎﺝ‬ AM = BM : ‫ﻭﻤﻨﻪ‬ AM =1 : ‫ ﻤﻌﻨﺎﻩ‬Z1 = 2 BM [ ] [ ]AB ‫ ﻫﻭ ﻤﺤﻭﺭ‬E ‫ﺇﺫﻥ‬. AB ‫ ﻨﻘﻁﺔ ﻤﻥ ﻤﺤﻭﺭ‬M ‫ﻭﻤﻨﻪ‬ Z1 = Z1 : ‫ ﺤﻘﻴﻘﻲ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ‬Z1 ‫ ﻴﻜﻭﻥ‬-2 : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ 1 - 2Z = 1 - 2Z : ‫ﻭﻋﻠﻴﻪ‬ iZ + 1 -iZ + 1(1 - 2Z) (-iZ + 1) = (iZ + 1) (1 - 2Z)-iZ + 1 + 2i ZZ - 2Z = iZ - 2i ZZ + 1 -2 Z-iZ + 2i ZZ - 2Z - iZ + 2i ZZ + 2Z = 0-i (Z + Z) - 2 (Z - Z) + 4i ZZ = 0

‫‪-i (2x) - 2 (2iy) + 4i (x2 + y2 ) = 0‬‬‫‪( )( )2i - x - 2y + 2 x2 + y2 = 0‬‬ ‫ﺇﺫﻥ ‪( )2 x2 + y - x - 2y = 0 :‬‬ ‫‪x2‬‬ ‫‪+ y2‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫‪-y=0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1 2‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪1 2‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪4 ‬‬ ‫‪16‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1 2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪1 2‬‬ ‫=‬ ‫‪5‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫‪4 ‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪16‬‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪ Z ≠ i‬ﻭﻤﻨﻪ )‪(x ; y) ≠ (0 ; 1‬‬‫‪5‬‬ ‫ﻭ ﻨﺼﻑ ﻗﻁﺭﻫﺎ‬ ‫‪W‬‬ ‫‪1‬‬ ‫;‬ ‫‪1‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ F‬ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ‬‫‪4‬‬ ‫‪ 4‬‬ ‫‪2 ‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬ ‫ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪:‬‬ ‫‪ (1‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ w1‬ﺫﻭ ﺍﻟﻼﺤﻘﺔ ‪ 1‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪2‬‬ ‫‪ (2‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ w2‬ﺫﻭ ﺍﻟﻼﺤﻘﺔ ‪ i‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪4‬‬ ‫‪ (3‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ ﻤﺒﺩﺃﻩ ﺍﻟﻨﻘﻁﺔ ‪ I‬ﺫﺍﺕ‬ ‫‪ei‬‬ ‫‪π‬‬ ‫ﻭ ﻻﺤﻘﺔ ﺸﻌﺎﻉ ﺘﻭﺠﻴﻬﻪ‬ ‫ﺍﻟﻼﺤﻘﺔ ‪-1 + i‬‬ ‫‪6‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫)‪iZ + (2i - 1) Z′ = 4 . . . (1‬‬ ‫‪ x‬ﺤل ﺍﻟﺠﻤﻠﺔ‬ ‫)‪-iZ + 3i Z′ =- i . . . (2‬‬ ‫ﺒﺠﻤﻊ )‪ (1‬ﻭ )‪ (2‬ﻨﺠﺩ ‪(5i - 1) Z′ = 4 - i :‬‬

Z′ = (4 - i) (5i + 1) : ‫ﺃﻱ ﺃﻥ‬ Z′ = 4-i ‫ﻭﻤﻨﻪ‬ (5i - 1) (5i + 1) 5i - 1 Z′ = 9 + 19 i : ‫ﻭﻋﻠﻴﻪ‬ 26 26 -iZ + 3i  9 + 19i  = -i : ‫( ﻨﺠﺩ‬2) ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ‬  26  -iZ = -i - 27i - 57 = -53i - 57 : ‫ﻭ ﻤﻨﻪ‬ 26 26 53 57 -53i - 57 iZ= 26 - 26 i :‫ﺃﻱ‬ Z= -26i × i : ‫ﻭ ﻋﻠﻴﻪ‬ S =  59 - 57 i ; 9 + 19 i   : ‫ﺇﺫﻥ‬  26 26 26 26    -2 ×iZ + iZ′ = 3 2iZ - Z′ = 1 + i : ‫ ﺤل ﺍﻟﺠﻤﻠﺔ‬x -2i Z - 2i Z′ = -6 2iZ - Z′ = 1 + i : ‫ﻭ ﻤﻨﻪ‬ -(2i + 1) Z′ = -5 + i : ‫ﺒﺎﻟﺠﻤﻊ ﻨﺠﺩ‬Z′ = -5 + i × -2i + 1 = 10i - 5+ 2 + i : ‫ﺇﺫﻥ‬ -2i - 1 -2i + 1 5 -3 11 -3 11Z′= 5 - 5 i : ‫ﺇﺫﻥ‬ Z′ = 5 + 5 i : ‫ﻭ ﻤﻨﻪ‬iZ + iZ′ = 3 iZ + iZ′ = 3-2Z - iZ′ = i - 1 :‫ﺃﻱ‬ i × 2iZ - Z′ = 1 + i : ‫ﻟﺩﻴﻨﺎ‬ : ‫( ﻭ ﻤﻨﻪ‬i - 2) Z = i + 2 : ‫ﺒﺎﻟﺠﻤﻊ ﻨﺠﺩ‬

Z= i+2 × i+2 = +3 + 4i = - 3 - 4 i i-2 i+2 -5 5 5 S =  - 3 - 4 i ; - 3 - 11 i   5 5 5 5   : ‫ﺇﺫﻥ‬  . 18‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬∆′ = (1 - cosθ)2 - 2 + 2 cosθ∆ = 1 - 2 cosθ + cos2θ - 2 +2 cosθ∆′ = -1 + cos2θ∆′ = -sin2θ = i2 sin2θ -i sinθ ‫ ﺃﻭ‬i sinθ ‫ﻭﻤﻨﻪ ﺠﺫﺭﻱ ∆ ﻫﻤﺎ‬ : ‫ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ‬ Z2 = -1 + cosθ + i sinθ ‫ ﻭ‬Z1 = -1 + cosθ - i sinθ : ‫ﻜﺘﺎﺒﺔ ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ‬Z1 = -1 + cosθ - i sinθZ1 = -1 + 1 - 2sin2 θ -i × 2 sin θ cos θ 2 2 2Z1 = -2 sin2 θ - 2i sin θ . cos θ 2 2 2Z1 = 2 sin θ -sin θ - i cos θ  2 2 2 Z1 = 2 sin θ sin  θ + π  + i cos  θ + π  2  2   2 Z1 = 2 sin θ cos  θ - π  + i sin  -θ - π  2  2   2 2 

sin θ > 0 ‫ ﻭﻤﻨﻪ‬0 ≤ θ ≤ π ‫ ﻓﺈﻥ‬0 ≤ θ ≤ π ‫ﺒﻤﺎ ﺃﻥ‬ 2 2 2 .‫ ﻤﻜﺘﻭﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ‬Z1 : ‫ﻭﻋﻠﻴﻪ‬Z2 = -1 + cosθ + i sinθZ2 = - 1 + 1 - 2sin2 θ + 2i sin θ cos θ 2 2 2Z2 = -2 sin2 θ + 2i θ + 2i sin θ . cos θ 2 2 2 2Z2 = 2 sin θ -sin θ + i cos θ  2 2 2 Z2 = 2 sin θ sin  -θ  + i cos  -θ  2  2  2 Z2 = 2 sin θ cos  π + θ  + i sin  π + θ  2  2 2   2 2  . Z2 ‫ﻭ ﻫﻭ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻠﻌﺩﺩ‬ . 19‫ﺍﻟﺘﻤﺭﻴﻥ‬ Z0 = 8 ‫( ﺘﻘﺒل ﺤل ﺤﻘﻴﻘﻲ‬1) ‫( ﻨﺒﻴﻥ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ‬183 - 12(8)2 + 48(8) - 128 = 0512 - 768 + 384 - 128 = 896 - 896 = 0 .‫ ﺤل ﻟﻬﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ‬8 ‫ﻭﻤﻨﻪ‬ : (1) ‫( ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬2 (Z - 8) (aZ2 + bZ + c) = 0 : ‫( ﺘﻜﺎﻓﺊ‬1) ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ aZ3 + bZ2 + cZ - 8aZ2 - 2bZ - 8c = 0 : ‫ﺃﻱ‬ aZ3 - (-b + 8a) Z2 + (c - 8b) Z - 8c = 0 : ‫ﻭﻋﻠﻴﻪ‬

‫‪a = 1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪a = 1‬‬ ‫ﺇﺫﻥ ‪:‬‬‫‪b = -4‬‬ ‫‪-b + 8a = +12‬‬‫‪c = 16‬‬ ‫‪c - 8b = 48‬‬ ‫‪-8c = -128‬‬‫ﻭﻤﻨﻪ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺘﻜﺎﻓﺊ ‪(Z - 8) (Z2 - 4Z + 16) = 0 :‬‬‫ﻭﻫﺫﺍ ﻴﻜﺎﻓﺊ ‪ Z - 8 = 0 :‬ﺃﻭ ‪Z2 - 4Z + 16 = 0‬‬ ‫ﺇﺫﻥ ‪ Z = 8 :‬ﺃﻭ ‪Z2 - 4Z + 16 = 0‬‬ ‫ﻨﺠﺩ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z2 - 4Z + 16 = 0‬‬‫‪ ∆′ = 4 - 16 = -12‬ﻭﻤﻨﻪ ‪∆′ = 12i2 :‬‬ ‫ﺠﺫﺭﻱ ‪ ∆′‬ﻫﻤﺎ ‪ i 12‬ﺃﻭ ‪-i 12‬‬‫ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪ Z1 = 2 +i 12 :‬ﻭ ‪Z2 = 2 - i 12‬‬‫‪ Z1 = 2 + 2i 3‬ﻭ ‪Z2 = 2 - 2i 3‬‬ ‫‪ -‬ﻜﺘﺎﺒﺔ ‪ Z1‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ‪:‬‬‫‪‬‬ ‫‪cosθ1‬‬ ‫=‬ ‫‪1‬‬‫‪‬‬ ‫‪2‬‬‫‪‬‬ ‫ﻟﺩﻴﻨﺎ ‪ Z1 = 4 :‬ﻭ‬‫= ‪sinθ1‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪θ1‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫‪,‬‬ ‫‪k∈Z‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪3‬‬ ‫‪Z1‬‬ ‫=‬ ‫‪4‬‬ ‫‪cos‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪ -‬ﻜﺘﺎﺒﺔ ‪ Z2‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ‪:‬‬

cosθ2 = 1 2  ‫ ﻭ‬Z2 = 4 : ‫ﻟﺩﻴﻨﺎ‬ sinθ2 - 3 = 2 θ2 = -π + 2kπ , k∈ Z : ‫ﻭﻤﻨﻪ‬ 3 Z2 = 4 cos −π + i sin  −π  3  3  ‫ ﺍﻟﺘﻤﺜﻴل ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ‬- M1 M0 M2 : Z3 ‫ ﺤﺴﺎﺏ ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ‬-4Z3 = Z2 - Z0 = 2-2i 3 - 8 = -6 - 2i 3 Z1 - Z0 2+2i 3 - 8 -6 2i 3

(-6 - 2i 3 ) (-6 - 2 3 )( ) ( )Z3 = -6 + 2i 3 -6 - 2i 3Z3 = 36 + 12i 3 + 12i 3 - 12 36 + 12 Z3 = 24 + 24i 3 = 1 + 1 i 3 : ‫ﻭ ﻤﻨﻪ‬ 48 2 2 Z3 = 1 + 3 =1 : ‫ﺇﺫﻥ‬ 4 4 cosθ = 1 2 arg(Z3 ) = θ :  sin θ = 3 2 θ= π + 2kπ , k ∈ Z : ‫ﻭﻤﻨﻪ‬ 3 : M0 M1 M2 ‫ﻁﺒﻴﻌﺔ ﺍﻟﻤﺜﻠﺙ‬ M0 M2 = M0 M1 : ‫ﻭﻤﻨﻪ‬ Z = M0 M2 =1 M0 M1 JJJJJJG JJJJJJG M0M1 ; M0M2 π ( )argZ1 = = 3 .‫ ﻤﺘﻘﺎﻴﺱ ﺍﻷﻀﻼﻉ‬M0 M1 M2 ‫ﻭﻋﻠﻴﻪ ﺍﻟﻤﺜﻠﺙ‬ . 20‫ﺍﻟﺘﻤﺭﻴﻥ‬ Z = x + iy ‫ ﻨﻔﺭﺽ‬.‫ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ‬Z1 ‫ﻨﻜﺘﺏ‬

Z1 = i (x + iy) + 1 - i = 1 - y + i (x - 1) x + iy - 2 + i x - 2 + i (y + 1)Z1 = 1 - y + i (x - 1) × x - 2 - (y + 1) x - 2 + i (y + 1) x - 2 - i (y + 1)Z1 = (1- y)(x- 2) - i(y+ 1)(1- y)+ i(x- 1)(x- 2) +(x- 1)(y+ 1) (x - 2)2 + (y + 1)2Z1 = x- 2- xy+2y- i(1- y2 )+ i(x2 - 3x+ 2)+ xy+ x- y-1 (x - 2)2 + (y + 1)2Z1 = 2x + y - 3 + i (-1 + y2 + x2 - 3x + 2) (x - 2)2 + (y + 1)2Z1 = 2x + y - 3 +i x2 + y2 - 3x + 1 (x - 2)2 + (y + 1)2 (x - 2)2 + (y + 1)2 : ‫ ﺤﻘﻴﻘﻲ ﺴﺎﻟﺏ‬Z1 ‫ ﺒﺤﻴﺙ ﺘﻜﻭﻥ‬M ‫ ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬-1  x2 + y2 - 3x + 1 = 0  2x + ‫و‬ : ‫ ﺤﻘﻴﻘﻲ ﺴﺎﻟﺏ ﻴﻜﺎﻓﺊ‬Z1 y-3<0  ‫و‬  (x ; y) ≠ (2 ; -1)  x - 3 2 - 9 + y2 + 1 = 0 :‫ﺃﻱ‬ x2 + y2 - 3x + = 0 x  2  4  x - 3 2 + y2 = 5 : ‫ﻭﻤﻨﻪ‬  2  4 (‫)ﺍﻟﺸﻜل‬

‫=‪R‬‬ ‫‪5‬‬ ‫و‬ ‫‪W‬‬ ‫‪3‬‬ ‫;‬ ‫‪0‬‬ ‫‪‬‬ ‫ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ‬ ‫‪2‬‬ ‫‪ 2‬‬ ‫‪‬‬ ‫‪ x‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪2x + y - 3 < 0‬‬ ‫ﻨﺭﺴﻡ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪( )2x + y - 3 = 0 :‬‬ ‫‪x03‬‬ ‫‪y 3 -3‬‬ ‫ﻫل )‪ (0 ; 0‬ﺤل ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪ 2(0) + 0 - 3 < 0 :‬ﺃﻱ ‪-3 < 0 :‬‬ ‫ﻤﺤﻘﻘﺔ ﻭﻤﻨﻪ )‪ (0 ; 0‬ﺤل ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪.‬‬‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ ﻨﺼﻑ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﻔﺘﻭﺡ ﺍﻟﻤﺤﺩﺩ ﺒﺎﻟﻤﺴﺘﻘﻴﻡ ∆ ﻭ ﻴﺸﻤل ‪ . O‬ﻭﻤﻨﻪ) (‬‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ Z1‬ﺤﻘﻴﻘﻲ ﺴﺎﻟﺏ ﻫﻲ ﺍﻟﻘﻭﺱ ﻤﻥ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﺴﺎﺒﻘﺔ ﺍﻟﻭﺍﻗﻊ ﺘﺤﺕ ﺍﻟﻤﺴﺘﻘﻴﻡ‬ ‫)∆( ‪.‬‬ ‫‪:‬‬ ‫‪Z1‬‬ ‫ﻋﻤﺩﺓ ﻟـ‬ ‫‪π‬‬ ‫‪ (2‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺒﺤﻴﺙ ﻴﻜﻭﻥ‬ ‫‪2‬‬ ‫‪2x + y - 3 = 0‬‬ ‫ﻭﻫﺫﺍ ﻴﻜﺎﻓﺊ ‪ Re (Z1 ) = 0 :‬ﻭﻤﻨﻪ ‪:‬‬ ‫)‪(x ; y) ≠ (2 ; -1‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺍﻟﺴﺎﺒﻘﺔ ﺒﺎﺴﺘﺜﻨﺎﺀ) (‬ ‫ﺍﻟﻨﻘﻁﺔ )‪. A (2 ; -1‬‬ ‫‪ -3‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺒﺤﻴﺙ ﺘﻜﻭﻥ ‪. Z1 = 2‬‬‫‪Z1‬‬ ‫=‬ ‫‪iZ + 1 - i‬‬ ‫=‬ ‫)‪1 - y + i (x - 1‬‬ ‫‪Z-2+i‬‬ ‫)‪x - 2 + i (y - 1‬‬‫‪Z1‬‬ ‫=‬ ‫)‪1 - y + i (x - 1‬‬ ‫=‬ ‫‪(1 - y)2 + (x - 1)2‬‬ ‫)‪x - 2 + i (y - 1‬‬ ‫‪( x - 2)2 + (y - 1)2‬‬

‫= ‪(1 - y)2 + (x - 1)2‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪( x - 2)2 + (y - 1)2‬‬ ‫‪(1 −‬‬ ‫‪y)2‬‬ ‫‪+(x‬‬ ‫‪− 1)2‬‬ ‫=‬ ‫‪2‬‬ ‫ﻭ ﺒﺘﺭﺒﻴﻊ ﺍﻟﻁﺭﻓﻲ ﻨﺠﺩ ‪:‬‬ ‫‪(x−‬‬ ‫‪2)2‬‬ ‫‪+(y‬‬ ‫‪− 1)2‬‬ ‫ﺇﺫﻥ ‪(1 - y)2 + (x - 1)2 = 2 (x - 2)2 + (y - 1)2  :‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬‫‪1 - 2y + y2 + x2 - 2x + 1 = 2x2 - 8x + 8 + 2y2 - 4y + 2‬‬‫‪1 - 2y + y2 + x2 - 2x + 1 - 2x2 + 8x - 8 - 2y2 + 4y - 2 = 0‬‬‫‪- x2 - y2 + 6x + 2y - 8 = 0‬‬‫‪(x - 3)2 - 9 + (y - 1)2 - 1 + 8 = 0‬‬‫‪(x - 3)2 + (y - 1)2 = 2‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ )‪ (C‬ﻤﺭﻜﺯﻫﺎ )‪ B(3 ; 1‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪2‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫ﻤﺴﺎﻭﻴﺔ ﺇﻟﻰ‬ ‫‪Z1‬‬ ‫‪ -4‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ﺘﻜﻭﻥ ﻋﻤﺩﺓ‬ ‫‪4‬‬ ‫ﻭ ﻫﺫﺍ ﻴﻜﺎﻓﺊ )‪. IM (Z) Re (Z‬‬ ‫‪2x + y - 3‬‬ ‫=‬ ‫‪x2 + y2 - 3x + 1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪(x - 2)2 + (y + 1)2‬‬ ‫‪(x - 2)2 + (y + 1)2‬‬ ‫‪ x2 + y2 - 3x + 1 = 2x + y - 3‬‬ ‫)‪(x ; y) ≠ (2 ; - 1‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪5 2‬‬ ‫‪-‬‬ ‫‪25‬‬ ‫‪+ y2‬‬ ‫‪=-4‬‬ ‫‪:‬‬ ‫ﻭﻋﻠﻴﻪ‬ ‫‪x2 + y2 -‬‬ ‫‪5x = -4‬‬ ‫‪:‬‬ ‫ﻭﻤﻨﻪ‬‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫≠ )‪(x ; y‬‬ ‫)‪(2 ; - 1‬‬

‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ‬  x - 5 2 + y2 = 9 : ‫ﺃﻱ‬  2  4 . 3 ‫ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ‬ C 5 ; 0  2  2  . 21‫ﺍﻟﺘﻤﺭﻴﻥ‬ : Z ‫ﺘﻌﻴﻴﻥ ﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ‬Z = 1 - cosθ + i sinθZ=1-  1 - 2 sin2 θ + 2i sin θ cos θ  2  2 2Z = +2 sin2 θ + 2i sin θ cos θ 2 2 2Z1 = 2 sin θ sin θ + i cos θ  2 2 2  - π < θ ≤ π : ‫ ﻓﺈﻥ‬-π < θ ≤ π : ‫ﺒﻤﺎ ﺃﻥ‬ 2 2 2 sin θ >0 ‫ﻴﻜﻭﻥ‬ 0 ; π ‫ﻭﻤﻨﻪ ﻓﻲ ﺍﻟﻤﺠﺎل‬ 2 2  Z1 = 2 sin θ cos  π - θ + i sin  π - θ  : ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬ 2  2 2   2 2  arg( Z1 ) = π - θ + 2kπ ; k∈Z‫ﻭ‬ Z1 = 2sin θ : ‫ﺇﺫﻥ‬ 2 2 2 -sin θ >0 : ‫ﺃﻱ‬ sin θ <0 :  -π ; 0 ‫ﻓﻲ ﺍﻟﻤﺠﺎل‬ 2 2  2 Z1 = -sin θ -sin θ - i cos θ  : ‫ﻭﻋﻠﻴﻪ‬ 2 2 2 

Z1 = -sin θ -cos  π - θ + i sin  π - θ  2  2 2   2 2 Z1 = -sin θ cos  π + π - θ + i sin  π + π - θ  2  2 2   2 2 Z1 = -sin θ cos  3π - θ + i sin  3π - θ  2  2 2   2 2  arg( Z1 ) = 3π - θ +2kπ ; k∈Z ‫ﻭ‬ Z1 = - sin θ : ‫ﻭﻤﻨﻪ‬ 2 2 2 . 22‫ﺍﻟﺘﻤﺭﻴﻥ‬‫ ﻭ ﻁﻭﻴﻠﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻭ‬A 3 , A 2 , A1 ‫ ﻫﺫﻩ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ﻭ‬Z3 , Z2 , Z1 ‫ﻨﻔﺭﺽ‬ : ‫ ﻭﻤﻨﻪ‬Z1 . Z2 . Z3 = −27i : ‫ ﻋﻤﺩﺘﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻴﻜﻭﻥ‬θ3 , θ2 , θ1 Z1 . Z2 . Z3 = 27 . A1 . A 2 . A 3 = 27 : ‫ ﺇﺫﻥ‬Z1 . Z2 . Z3 = 27 : ‫ﺃﻱ‬ A 2 = A1 . A3 : ‫ ﻭﻤﻨﻪ‬A 3 ‫ و‬A1 ‫ ﻭﺴﻁ ﻫﻨﺩﺴﻲ ﻟﻠﻌﺩﺩﻴﻥ‬A 2 ‫ﻟﻜﻥ‬ 2 A1 . A3 = 9 : ‫ﻭﻋﻠﻴﻪ‬ A 2 = 3 : ‫ﺃﻱ ﺃﻥ‬ A 3 = 27 : ‫ﻭﻤﻨﻪ‬ 2 A 2 = 1 : ‫ﺇﺫﻥ‬ 9A 2 = 9 : ‫ﻭﻤﻨﻪ‬ A 3 = A1 . 9 : ‫ﻟﻜﻥ‬ 1 1 A 3 = 9 : ‫ ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬A1 = 1 : ‫ﻭﻤﻨﻪ‬ arg ( Z1 . Z2 . Z3 ) = arg( -27i) : ‫ﻭﻜﺫﻟﻙ‬ arg( Z1 ) + arg( Z2 ) + arg( Z3 ) = 3π 2 3π θ3 ‫ و‬θ1 ‫ ﻭﺴﻁ ﺤﺴﺎﺒﻲ ﻟﻠﻌﺩﺩﻴﻥ‬θ2 ‫ﻟﻜﻥ‬ θ1 + θ2 + θ3 = 2 . θ2 = π : ‫ﺇﺫﻥ‬ 3θ2 = 3π :‫ﻭﻤﻨﻪ‬ 2θ2 = θ1 + θ3 : ‫ﺃﻱ ﺃﻥ‬ 2 2

θ1 = θ2 - π = π - π ‫ﺃﻱ‬ θ2 = θ1 + π : ‫ﻟﻜﻥ‬ 3 2 3 3 5π. θ3 = 6 : ‫ﺃﻱ‬ θ3 = θ2 + π : ‫ﻭﻤﻨﻪ‬ θ1 = π : ‫ﺃﻱ ﺃﻥ‬ 3 6Z1 = 3 + 1 i : ‫ﻭﻋﻠﻴﻪ‬ Z1 = cos π + i sin π : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ 2 2 6 6 Z2 = 3i : ‫ﻭﻋﻠﻴﻪ‬ Z2 =3 cos π + i sin π  : ‫ﻭﻤﻨﻪ‬ 2 2 Z3 = -9 3 + 9 i : ‫ﻭﻋﻠﻴﻪ‬ Z3 = 9 cos 5π + i sin 5π  9 2 6 6  . 23‫ﺍﻟﺘﻤﺭﻴﻥ‬ : (C) ‫ ﻨﻘﻁﺔ ﻤﻥ‬B ‫( ﻨﺒﻴﻥ ﺃﻥ‬1 (I AB = ZB - ZA = 1 - ei π -1 = ei π =1 3 3 ( )JJJG JJ.JGB ∈ (C) : ‫ﻭﻤﻨﻪ‬ : AF ; AB ‫( ﺘﻌﻴﻴﻥ ﻗﻴﺴﺎ ﻟﻠﺯﺍﻭﻴﺔ‬2 JJJG JJJG JJJG G JG JJJG( ) ( ) ( )AF ; AB = AF , u + U , AB + 2kπ ; k ∞ Z ( ) ( )JG JJJG JG JJJG = - U ; AF + U , AB + 2kπ ( ) ( )JG JJJG JG JJJG = U ; AB - U ; AF + 2kπ ( ) ( )= arg ZJJJG - arg ZJJJG AB AF( )arg ZJJJG = π : ‫ﻭﻤﻨﻪ‬ Z JJJG = ZB - ZA = ei π : ‫ﻟﻜﻥ‬ AB 3 AB 3

( )arg ZJJJG =0 : ‫ﻭﻤﻨﻪ‬ Z JJJG = ZF - ZA = 2 - 1 = 1 = ei×0 AF AF JJJG JJJG π AF ; AB 3 ( ). = + 2kπ ; k ∈ Z : ‫ﻭﻤﻨﻪ‬ : ‫ ﺘﻌﻴﻴﻥ ﺍﻟﺸﻜل ﺍﻷﺴﻲ‬-3 ZB - ZA : ‫ ﻟﻠﻌﺩﺩ‬xZB - ZA = 1 + ei π - 1 = ei π 3 3ZE - ZA = 1 + ZB2 - 1 = Z 2 =  1 + ei π 2 B  3   =  1 + cos π + i sin π 2 =  1 + 1 +i 6 3 2  3 3   2 2  =  3 +i 3 2 =  3 ei π 2 = 3 ei π  2 2   6  3   JJJG : ‫ﺴﺒﻕ‬J‫ﺎ‬J‫ﻤ‬J‫ﻤ‬G‫ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬E , B , A ‫ ﺍﺴﺘﻨﺘﺎﺝ ﺃﻥ ﺍﻟﻨﻘﻁ‬-4 AE = 3 AB : ‫ ﻭﻤﻨﻪ‬ZE - ZA =JJ3JG ( ZJBJ-JGZA ) . ‫ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬E , B , A ‫ ﻭﻋﻠﻴﻪ ﺍﻟﻨﻘﻁ‬AB // AE : ‫ﺇﺫﻥ‬  Z′ - 1  JJJJG JJJJG  Z - 1  AM , AM′ ( )arg = : ‫( ﻟﺩﻴﻨﺎ‬1 -II

‫‪( )JJJJG JJJJG‬‬ ‫‪Z′ - 1‬‬ ‫ﻭﻤﻨﻪ ﻋﻤﺩﺓ ‪ Z - 1‬ﻫﻲ ﺍﻟﺯﺍﻭﻴﺔ‬ ‫‪AM , AM′‬‬‫‪ (2‬ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁ ‪ A‬ﻭ‪ M‬ﻭ ‪ M′‬ﻋﻠﻰ ﺍﺴﺘ‪G‬ﻘﺎ‪J‬ﻤ‪J‬ﺔ‪J‬ﻭ‪J‬ﺍﺤﺩﺓ ﺇ‪JG‬ﺫﺍ‪JJ‬ﻭﻓ‪J‬ﻘﻁ ﺇﺫﺍ ﻜﺎﻨﺕ ‪( ):‬‬ ‫‪AM , AM′ = kπ , k ∈ Z‬‬‫‪Z′ - 1‬‬ ‫∈‬ ‫\‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪arg‬‬ ‫‪‬‬ ‫‪Z′‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪= kπ‬‬ ‫ﺃﻱ ‪:‬‬‫‪Z-1‬‬ ‫‪‬‬ ‫‪Z‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪Z2‬‬ ‫\∈‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪1 + Z2 - 1‬‬ ‫∈‬ ‫\‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫‪Z-1‬‬ ‫‪Z-1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬ ‫‪ -1‬ﺤﺴﺎﺏ ‪ f Z‬ﺒﺩﻻﻟﺔ ‪( ): Z‬‬ ‫‪f ( Z) = Z2 - (2 + 3i) Z2 + 9Z - 18 - 27i‬‬ ‫‪ -2‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪( ): f Z = 0‬‬ ‫ﻟﺩﻴﻨﺎ ‪ f Z1 = 0 :‬ﻭ ‪( ) ( ). f Z1 = 0‬‬ ‫ﺤﻴﺙ ‪ f Z1 = 0 :‬ﺘﻜﺎﻓﺊ ‪( ) ( )f Z1 = 0 :‬‬‫ﻭﻤﻨﻪ ‪Z13 - (2 + 3i) Z12 + 9Z1 - 18 - 27i = 0 . . . (1) :‬‬ ‫‪ f Z1 = 0‬ﺘﻜﺎﻓﺊ ‪( ):‬‬ ‫)‪Z13 - (2 - 3i) Z12 + 9Z1 - 18 + 27i = 0 . . . (2‬‬ ‫ﺒﻁﺭﺡ )‪ (2‬ﻤﻥ )‪ (1‬ﻨﺠﺩ ‪(-2 - 3i + 2 - 3i) Z12 - 54i = 0 :‬‬ ‫ﻭﻤﻨﻪ ‪ -6iZ12 - 54i = 0 :‬ﺇﺫﻥ ‪ Z12 = -9 :‬ﺃﻱ ‪Z12 = 9i2 :‬‬ ‫ﻭﻤﻨﻪ ‪ Z1 = 3i :‬ﺃﻭ ‪Z1 = -3i‬‬ ‫ﻭﻋﻠﻴﻪ ‪f (Z) = (Z - 3i) (Z + 3i) (aZ + b) :‬‬

‫)‪f (Z) = (Z2 + 9) (aZ + b‬‬‫‪f (Z) = aZ3 + bZ2 + 9aZ + 9b‬‬ ‫‪a = 1‬‬ ‫ﻭﻤﻨﻪ ‪b = -2 + 3i :‬‬ ‫ﻭﻋﻠﻴﻪ ‪f (Z) = (Z - 3i) (Z + 3i) (Z - 2 + 3i) :‬‬‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪ Z - 3i = 0 :‬ﺃﻭ ‪ Z + 3i = 0‬ﺃﻭ ‪Z - 2 + 3i = 0‬‬ ‫ﻭﻤﻨﻪ ‪ Z = 3i :‬ﺃﻭ ‪ Z = -3i‬ﺃﻭ ‪Z = 2 – 3i‬‬ ‫ﺇﺫﻥ ‪S = {3i , -3i , 2 - 3i} :‬‬ ‫= ‪Z′‬‬ ‫‪3i + 3i‬‬ ‫‪= 3i‬‬ ‫‪ (3‬ﺤﺴﺎﺏ ‪: Z′‬‬ ‫‪2 - 3i + 3i‬‬‫= )‪arg(Z′‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫‪,‬‬ ‫‪Z′‬‬ ‫‪=3‬‬ ‫‪; k∈Z‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫ﻁﺒﻴﻌﺔ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻗﺎﺌﻡ ﻓﻲ ‪ B‬ﻭ ﻏﻴﺭ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ‪.‬‬‫‪ (1‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪AM2 + 2MB2 - 2MC2 = 25 :‬‬ ‫ﻨﻔﺭﺽ ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪ M‬ﻫﻲ ‪. Z‬‬ ‫ﻓﻴﻜﻭﻥ ‪:‬‬‫‪Z - 3i 2 + 2 Z + 3i 2 - 2 Z - 2 + 3i 2 = 25‬‬‫‪x2 +(y - 3)2 + 2  x2 +(y+ 3)2  -2 (x- 2)2 +(y+ 3)2  = 25‬‬‫‪x2 +2x2 - 2(x- 2)2 +(y - 3)2 +2(y + 3)2 -2(y + 3)2 = 25‬‬‫‪x2 + 2x2 - 2x2 + 8x - 8 + y2 - 6y + 9 + 2y2 + 12y‬‬‫‪+ 18 - 2y2 - 12y - 18 = 25‬‬ ‫ﻭﻤﻨﻪ ‪x2 + 8x + y2 - 6y = 24 :‬‬‫ﺃﻱ‪( x + 4)2 - 16 + (y - 3)2 - 9 = 24 :‬‬ ‫ﺃﻱ‪(x + 4)2 + (y - 3)2 = 49 :‬‬

‫ﻭﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ )‪ w(-4 ; 3‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪. 7‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 25‬‬ ‫‪ (1‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪Z2 - (1 + i) Z - 4i = 0 :‬‬ ‫)‪ ∆ = (1 + i)2 - 4 (-4i‬ﻭﻤﻨﻪ ‪∆ = 18i‬‬‫ﺇﺫﻥ ‪ ∆ = 9 . 2i :‬ﻭ ﻤﻨﻪ ‪∆ = 9(1 + i)2 = [3(1 + i)]2 :‬‬‫ﻭﻤﻨﻪ ﺠﺫﺭﻱ ∆ ﻫﻤﺎ ‪ - (3 + 3i) , 3 + 3i‬ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪ Z2 , Z1 :‬ﺤﻴﺙ ‪:‬‬‫= ‪Z2‬‬ ‫‪1 + i + 3 + 3i‬‬ ‫ﻭ‬ ‫= ‪Z1‬‬ ‫‪1 + i - 3 - 3i‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ Z1 = -1 - i‬و ‪Z2 = 2 + 2i‬‬ ‫‪ (2‬ﻨﺒﻴﻥ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻘﺒل ﺤﻠﻴﻥ ﺘﺨﻴﻠﻴﻴﻥ ﻤﺘﻌﺎﻜﺴﻴﻥ ‪:‬‬ ‫‪ Z4 , Z3‬ﺃﻱ ‪ Z3 = yi :‬ﻭ ‪Z4 = - yi‬‬ ‫ﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪:‬‬‫‪(Z - yi) (Z + yi) (aZ2 + bZ + c) = 0‬‬‫‪(Z2 + y2 ) (aZ2 + bZ + c) = 0‬‬‫‪aZ4 + bZ3 + cZ2 + ay2 Z2 + by2 Z + cy2 = 0‬‬‫ﺇﺫﻥ ‪aZ4 + bZ3 + (c + ay2 ) Z2 + by2Z + cy2 = 0 :‬‬‫‪a = 1‬‬ ‫‪a = 1‬‬‫‪b = -1 - i‬‬ ‫‪b = -1 - i‬‬‫‪c + y2 = 9 - 4i‬‬ ‫ﻭﻋﻠﻴﻪ ‪ c + ay2 = -9 - n :‬ﺃﻱ ﺃﻥ ‪:‬‬‫‪(-1 - i) y2 = -9 - 9i‬‬ ‫‪ by2 = −9 − 9i‬‬‫ﻭﻋﻠﻴﻪ ‪y2 = 9 :‬‬ ‫‪y2‬‬ ‫=‬ ‫)‪-9 (1 + i‬‬ ‫‪=9‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪- (1 + i‬‬‫ﻭﻤﻨﻪ ‪ y = 3 :‬ﺃﻭ ‪ y = -3‬ﺇﺫﻥ ‪C + 9 = 9 - 4i :‬‬ ‫ﻭﻤﻨﻪ ‪C = - 4i :‬‬

‫ﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪( )(Z - 3i) (Z + 3i) Z2 - (1 + i) Z - 4i = 0 :‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪ Z – 3i = 0 :‬ﺃﻭ ‪ Z + 3i = 0‬ﺃﻭ ‪Z2- (1 + i) Z - 4i = 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ Z = 3i :‬ﺃﻭ ‪ Z = -3i‬ﺃﻭ ‪ Z = -1 - i‬ﺃﻭ ‪) Z = 2 + 2i‬ﻤﻤﺎ ﺴﺒﻕ( ‪ .‬ﺇﺫﻥ ‪:‬‬ ‫}‪. S = {3i ; -3i ; -1 - i ; 2 + 2i‬‬ ‫‪ -3‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪:‬‬‫‪Z - Z3 - Z4‬‬ ‫‪Z - 3i + 3i‬‬ ‫=‬ ‫‪Z‬‬‫‪Z - Z1 - Z2‬‬ ‫‪Z + 1 + i - 2 -2i‬‬ ‫‪Z-1-i‬‬ ‫‪Z‬‬ ‫ﺇﺫﺍ ﻭﺍﻓﻕ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺭﻜﺏ‬ ‫‪-‬‬ ‫‪π‬‬ ‫ﻤﺴﺎﻭﻴﺔ ﺇﻟﻰ‬ ‫‪Z‬‬ ‫ﺘﻜﻭﻥ ﻋﻤﺩﺓ‬‫‪Z-1-i‬‬ ‫‪2‬‬ ‫‪Z-1-i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪Z‬‬ ‫=‬ ‫‪‬‬ ‫‪Z‬‬ ‫‪Z‬‬ ‫‪-‬‬ ‫‪i‬‬ ‫‪‬‬ ‫ﻭ‬ ‫‪Z‬‬ ‫≠‬ ‫‪1-i‬‬ ‫ﺘﺨﻴﻠﻲ ﺼﺭﻑ ﺃﻱ ‪:‬‬ ‫‪Z-1-i‬‬ ‫‪‬‬ ‫‪-1‬‬ ‫‪‬‬ ‫)‪Z(Z - 1 + i) = Z (Z - 1 - i‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪Z‬‬ ‫=‬ ‫‪Z‬‬ ‫‪Z-1-i‬‬ ‫‪Z-1+i‬‬ ‫ﻭﻤﻨﻪ ‪ZZ - Z + iZ = ZZ - Z - iZ :‬‬ ‫ﻭﻤﻨﻪ ‪-Z + Z + iZ + iZ = 0 :‬‬ ‫‪-(Z - Z) + i (Z + Z) = 0‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ -2iy + 2ix = 0 :‬ﺤﻴﺙ ‪Z = x + iy :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ 2i (x - y) = 0 :‬ﺇﺫﻥ ‪. x – y = 0 :‬‬‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻫﻲ ﻨﻘﻁ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪ x – y = 0‬ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﻨﻘﻁﺔ )‪A(1 ; 1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 26‬‬ ‫ﻨﺭﻤﺯ ﺒـ )‪ R(w ; θ‬ﻟﻠﺩﻭﺭﺍﻥ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ w‬ﻭ ﺯﺍﻭﻴﺘﻪ ‪. θ‬‬‫‪D‬‬ ‫‪A‬‬ ‫ﻭ ﺒﺎﻟﺭﻤﺯ ‪ ZM‬ﻟﻼﺤﻘﺔ ‪. M‬‬ ‫ﻟﺩﻴﻨﺎ ‪ E :‬ﺼﻭﺭﺓ ‪C‬‬ ‫‪I‬‬ ‫‪E‬‬ ‫‪R‬‬ ‫‪‬‬ ‫‪A‬‬ ‫;‬ ‫‪π‬‬ ‫‪‬‬ ‫ﺒﺎﻟﺩﻭﺭﺍﻥ ‪3 ‬‬ ‫‪J‬‬ ‫‪BC‬‬

: ‫ﻭﻋﻠﻴﻪ‬ ( )(1) ei π ... ZE - ZA = 3 ZC- ZA A ‫ ﻫﻲ ﺼﻭﺭﺓ‬D ‫ﻭ ﻜﺫﻟﻙ‬ R  B ; π ‫ﺒﺎﻟﺩﻭﺭﺍﻥ‬  3  ( )(2) - ZB = ei π . . . ZD 3 ZA - ZB : ‫ﻭﻋﻠﻴﻪ‬ R  C ; π ‫ ﺒﺎﻟﺩﻭﺭﺍﻥ‬B ‫ ﻫﻲ ﺼﻭﺭﺓ‬F ‫ﻭ ﻟﺩﻴﻨﺎ‬  3  ( )(3) . . . ZF - ZC ei π = 3 ZB - ZC : ‫ﻭﻋﻠﻴﻪ‬ : ‫( ﻁﺭﻓﺎ ﻟﻁﺭﻑ‬3) ‫( ﻭ‬2) ‫( ﻭ‬1) ‫ﺒﺠﻤﻊ‬( )ZEZA+ ZD ZF = ei π- - ZB + - ZC 3 ZC-ZA +ZA -ZB +ZB -ZC ( )ZE + ZD + ZF - ZA + ZB + ZC = 0 : ‫ﻭﻤﻨﻪ‬ (4) . . . ZE + ZD + ZF = ZA + ZB + ZC : ‫ﻭﻤﻨﻪ‬: ‫ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻭ ﻤﻨﻪ‬BCF‫ ﻭ‬ACE ‫ ﻭ‬ABD ‫ ﻤﺭﺍﻜﺯ ﺜﻘل ﻜل ﻤﻥ ﺍﻟﻤﺜﻠﺜﺎﺕ‬K , J , I ‫ﻭﻟﺩﻴﻨﺎ‬ ZI = ZA + ZB + ZD ; ZJ = ZA + ZC + ZE 3 3 : ‫ﻭﻤﻨﻪ‬ ZK = ZB + ZC + ZF 3( )ZI+ ZJ + ZK = 2ZA + 2ZB + 2ZC + ZD + ZE + ZF 3 : ‫ﻭﻋﻠﻴﻪ‬( )ZI 2ZA+ ZJ + ZK = + 2ZB + 2ZC + ZA + ZB + ZC 3

‫=‬ ‫‪3ZA‬‬ ‫‪+ 3ZB‬‬ ‫‪+ 3ZC‬‬ ‫‪3‬‬ ‫‪= ZA + ZB + ZC‬‬‫ﻭﻤﻨﻪ ‪ZI + ZJ + ZK = ZA + ZB + ZC = ZD + ZE + ZF :‬‬‫‪ZI + ZJ + ZK‬‬ ‫=‬ ‫‪ZA + ZB + ZC‬‬ ‫=‬ ‫‪ZD + ZE + ZF‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫ﻭ ﻋﻠﻴﻪ ﺍﻟﻤﺜﻠﺜﺎﺕ ‪ IJK‬ﻭ ‪ ABC‬ﻭ‪ FED‬ﻟﻬﺎ ﻨﻔﺱ ﻤﺭﻜﺯ ﺍﻟﺜﻘل‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 27‬‬ ‫ﺘﻌﻴﻴﻥ ﺍﻟﻁﻭﻴﻠﺔ ﻭ ﻋﻤﺩﺓ ﻟﻠﻌﺩﺩ ‪1 + cosx + i sin x :‬‬ ‫ﻟﺩﻴﻨﺎ ‪( )G G 1 + cosx + i sin x = 1 + eix :‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ﺍﻟﻤﺭﻓﻘﺔ ﺒﺎﻟﻤﻌﻠﻡ ﺍﻟﻤﺘﻌﺎﻤﺩ ﺍﻟﻤﺘﺠﺎﻨﺱ ‪. O ; i , j‬‬ ‫‪H‬‬ ‫‪M‬‬ ‫ﻟﺘﻜﻥ ﺍﻟﻨﻘﻁ ‪M′ , M , I‬‬‫’‪I‬‬ ‫‪I‬‬ ‫ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ‪-1 , eix , 1‬‬ ‫‪O‬‬ ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‪.‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫]‪( )arg eix + 1 =JJ2xJG [2π‬‬ ‫ﺤﻴﺙ ‪eix + 1 = I′M :‬‬ ‫ﻭ ﻟﺘﻜﻥ ‪ H‬ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁﺔ ‪ O‬ﻋﻠﻰ ‪[ ]. I′M‬‬‫ﻭ ﺒﻤﺎ ﺃﻥ ﺍﻟﻤﺜﻠﺙ ‪ OMI′‬ﻤﺘﻘﺎﻴﺱ ﺍﻟﺴﺎﻗﻴﻥ ﻓﺈﻥ ‪ H‬ﻤﻨﺘﺼﻑ ‪ I′M‬ﻭﻤﻨﻪ ‪[ ]:‬‬ ‫‪. I′M = 2I′H‬‬ ‫‪cos‬‬ ‫‪x‬‬ ‫=‬ ‫‪I′H‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ ‪OMI′‬‬ ‫‪2‬‬ ‫‪I′O‬‬ ‫‪x‬‬ ‫‪x‬‬‫‪I′M‬‬ ‫=‬ ‫‪2cos‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪I′O = 1‬‬ ‫ﻷﻥ ‪:‬‬ ‫‪I′H‬‬ ‫=‬ ‫‪cos‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬

( ) [ ]arg eix + 1 = x 2π : ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬ eix +1 = 2 cos x : ‫ﻭﻋﻠﻴﻪ‬ 2 2 . 28‫ﺍﻟﺘﻤﺭﻴﻥ‬ ( ) ( ) ( )JJJG JJJG : ‫ﺠﻬﺔ‬J‫ﻭ‬J‫ﻟﻤ‬JG‫ﺯﻭﺍﻴﺎ ﺍ‬J‫ﺍﻟ‬J‫ﻥ‬G‫ﺎﺱ ﻟﻜل ﻤ‬J‫ﻗﻴ‬J‫ﺃ‬JGγ , βJJJ,Gα ‫ﻨﻌﻠﻡ ﺃﻥ‬ CE ; CF ; BC ; BF ; HB ; HF ( )JJJG JJJG B ; BC ; BA ‫ﻨﺯﻭﺩ ﺍﻟﻤﺴﺘﻭﻯ ﺒﺎﻟﻤﻌﻠﻡ‬ 2 + i , 2 , 1 , 0 , -1 : ‫ﻓﺘﻜﻭﻥ ﻟﺩﻴﻨﺎ‬ .‫ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﻲ ﻫﺫﺍ ﺍﻟﻤﻌﻠﻡ‬F , E , C , B , H ‫ﻟﻭﺍﺤﻕ ﺍﻟﻨﻘﻁ‬ JJJG JJJG  ZF - ZH  HB ; HP ( )arg  ZB - ZH  = α : ‫ﻋﻠﻴﻪ‬ ‫ﻭ‬ =α : ‫ﻟﺩﻴﻨﺎ‬    ZF - ZB  JJJG JJG  ZC - ZB  BC ; BF = α : ‫ﻭﻜﺫﻟﻙ‬ ( )arg   = β : ‫ﻭﻋﻠﻴﻪ‬  ZF - ZC  JJJG JJJG  ZE - ZC  CE ; CF = γ : ‫ﻭﺃﻴﻀﺎ‬ ( )arg  = γ : ‫ﻭﻋﻠﻴﻪ‬ α = arg  2 + i + 1 = arg (3 + i) : ‫ﺇﺫﻥ‬  0 + 1 β = arg  2+i- 0  = arg (2 + i)  1-0 γ = arg  2 +i - 1 = arg (1 + i) = π  2 - 1  4 γ = π : ‫ﻭﻋﻠﻴﻪ‬ 4 : ‫ﻟﻜﻥ‬ α + β = arg(3 + i) + arg(2 + i)

‫)‪= arg [(3 + i) (2 + i)] = arg(6 + 3i + 2i - 1‬‬ ‫])‪= arg(5 + 5i) = arg[5 (1 + i‬‬ ‫‪= arg5 + arg(1 + i) = 0 +‬‬ ‫‪π‬‬ ‫‪4‬‬ ‫ﻭﻤﻨﻪ ‪α + β = γ :‬‬ ‫‪α‬‬ ‫‪+β‬‬ ‫=‬ ‫‪π‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 29‬‬ ‫‪Z-3+i =2‬‬ ‫‪ (1‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ N‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪JJJG‬‬ ‫ﺃﻱ ‪ Z - (3 - i) = 2 :‬ﻭﻤﻨﻪ ‪AN = 2 :‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ N‬ﻫﻲ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ ‪ A‬ﻭ ﻨﺼﻑ ﺍﻟﻘﻁﺭ ‪. 2‬‬‫ﺃﻱ ‪:‬‬ ‫‪arg‬‬ ‫‪‬‬ ‫‪Z-‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪i‬‬ ‫‪‬‬ ‫=‬ ‫‪π‬‬ ‫‪ (2‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ 2‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪‬‬ ‫‪Z+‬‬ ‫‪4‬‬ ‫‪-‬‬ ‫‪2i‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪arg‬‬ ‫‪‬‬ ‫‪Z - (2 + i) ‬‬ ‫=‬ ‫‪π‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪Z‬‬ ‫‪-‬‬ ‫)‪(-4+2i‬‬ ‫‪‬‬ ‫‪( )JJJG JJJG‬‬ ‫=‬ ‫‪π‬‬ ‫ﻭ ﻋﻠﻴﻪ ﻫﺫﺍ ﻴﻜﺎﻓﺊ ‪:‬‬ ‫‪2‬‬ ‫‪MB ; MC‬‬‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﻨﺼﻑ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪ AB‬ﺍﻟﻤﺤﺘﻭﺍﺓ ﻓﻲ ﻨﺼﻑ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﻔﺘﻭﺡ ﺍﻟﺫﻱ] [‬ ‫ﺤﺩﻩ )‪ (AB‬ﻭ ﻴﺸﻤل ﺍﻟﻤﺒﺩﺃ ‪. O‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 30‬‬ ‫ﻨﺒﺭﻫﻥ ﺃﻥ ‪ ( AL) ⊥ ( )BC‬ﻭ ‪AL = BC‬‬ ‫‪π‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﻭﺭﺍﻥ ‪ R‬ﺫﻭ ﺍﻟﻤﺭﻜﺯ ‪ A‬ﻭ ﺍﻟﺯﺍﻭﻴﺔ ‪. 2‬‬ ‫ﻭﻟﺘﻜﻥ ‪ZE , ZG , ZL , ZC , ZB , ZA‬‬ ‫ﻻﺤﻘﺎﺕ ﺍﻟﻨﻘﻁ ‪ F , G , L , C , B , A‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬

( )(1) . . . ZB - ZA = i ZG - ZA : ‫ ﻭﻋﻠﻴﻪ‬R (G) = B : ‫ﻟﺩﻴﻨﺎ‬ R(C) = E : ‫ﻭ ﻟﺩﻴﻨﺎ‬( )(2) . . . ZE - ZJAJJ=G i JZJJCG - ZA : ‫ﻭﻋﻠﻴﻪ‬ AE + AG = AL : ‫ﻭﻟﺩﻴﻨﺎ‬(3) . . . ZE - ZA + ZG - ZA = ZL - ZA : ‫ﻭﻋﻠﻴﻪ‬( )ZG- ZA = i ZB- ZA : ‫ﻭﻤﻨﻪ‬ ZG - ZA = ZB - ZA : (1) ‫ﻤﻥ‬ i‫( ﺒﻘﻴﻤﺘﻴﻬﻤﺎ‬3) ‫ ﻓﻲ‬ZE - ZA ‫ ﻭ‬ZG - ZA ‫ﻨﻌﻭﺽ ﻜل ﻤﻥ‬( ) ( )i ZC - ZA - i ZB - ZA = ZL - ZA : ‫ﻓﻨﺠﺩ‬( ) ( )ZA - ZL = i ZB - ZC : ‫ ﺃﻱ‬ZL - ZA = ZC - ZB : ‫ﺇﺫﻥ‬ZA - ZL = 1 : ‫ﺤﻴﺙ‬ ZA - ZL = i : ‫ﻭﻋﻠﻴﻪ‬ZB - ZC ZB - ZC arg  ZA - ZL  = π ‫ﻭ‬  ZB - ZC  2  ( )JJJG JJJG = π ‫ﻭ‬ LA =1 : ‫ﺇﺫﻥ‬ BC , AL 2 BC( BC) ⊥ ( AL) ‫ ﻭ‬LA = CB : ‫ﻭﻤﻨﻪ‬

LG AEF DBC

‫‪ -7‬ﺍﻟﺘﺸﺎﺒﻬﺎﺕ ﺍﻟﻤﺴﺘﻭﻴﺔ ﺍﻟﻤﺒﺎﺸﺭﺓ‬ ‫ﺍﻟﻜﻔﺎﺀﺓ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‬ ‫‪ -1‬ﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ‪.‬‬ ‫‪ - 2‬ﺍﻟﺘﻌﺒﻴﺭ ﻋﻥ ﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﺒﺎﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ‪.‬‬ ‫‪ -3‬ﺘﺭﻜﻴﺏ ﺘﺸﺎﺒﻬﻴﻥ ﻤﺒﺎﺸﺭ ‪.‬‬ ‫‪ -4‬ﺘﻌﻴﻴﻥ ﺍﻟﺘﺤﻠﻴل ﺍﻟﻘﺎﻨﻭﻨﻲ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﺒﻭﺍﺴﻁﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‪.‬‬‫‪ -5‬ﺘﻭﻅﻴﻑ ﺍﻟﺘﺤﻠﻴل ﺍﻟﻘﺎﻨﻭﻨﻲ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﺒﻭﺍﺴﻁﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ‪.‬‬ ‫‪ -6‬ﺘﻭﻅﻴﻑ ﺨﻭﺍﺹ ﺍﻟﺘﺸﺎﺒﻬﺎﺕ ﺍﻟﻤﺒﺎﺸﺭﺓ ﻟﺤل ﻤﺴﺎﺌل ﻫﻨﺩﺴﻴﺔ‪.‬‬ ‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫ﺃﻨﺸﻁﺔ‬‫‪ -2‬ﺤﺎﻻﺕ ﺨﺎﺼﺔ‬ ‫‪ -1‬ﺘﻌﺭﻴﻑ‬ ‫‪ -3‬ﺘﻌﺭﻴﻑ ‪2‬‬‫‪ -4‬ﺍﻟﺸﻜل ﺍﻟﻤﻤﻴﺯ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺭﻜﺏ‬‫‪ -5‬ﺍﻟﺒﺤﺙ ﻋﻥ ﺍﻟﻨﻘﻁ ﺍﻟﺼﺎﻤﺩﺓ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ‬‫‪ -7‬ﻤﺭﻜﺏ ﺘﺤﺎﻜﻲ ﻭﺩﻭﺭﺍﻥ‬ ‫‪ -6‬ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺼﺭ ﻟﺘﺸﺎﺒﻪ‬‫‪ -9‬ﻤﺭﻜﺏ ﺘﺸﺎﺒﻬﻴﻥ ﻤﺒﺎﺸﺭﻴﻥ‬ ‫‪ -8‬ﻨﻅﺭﻴﺔ ﻭ ﺘﻌﺭﻴﻑ‬ ‫‪ -10‬ﺍﻟﺨﺎﺼﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ‬ ‫‪ -11‬ﺼﻭﺭ ﺒﻌﺽ ﺍﻷﺸﻜﺎل ﺍﻟﻬﻨﺩﺴﻴﺔ‬‫ﺍﻟﺤـﻠــــــﻭل‬ ‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬

‫ﺃﻨﺸﻁﺔ‬‫‪( )G G‬‬ ‫ﺍﻟﻨﺸﺎﻁ ‪:‬‬‫ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﻭﺠﻪ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ﻤﺒﺎﺸﺭ ‪O; i , j‬‬ ‫‪ ω‬ﻨﻘﻁﺔ ﻻﺤﻘﺘﻬﺎ ‪Z0 = 1 + i‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫‪ ω‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫ﺍﻟﺩﻭﺭﺍﻥ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ‬ ‫‪R‬‬ ‫‪2‬‬ ‫‪ H‬ﺍﻟﺘﺤﺎﻜﻲ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ ω‬ﻭﻨﺴﺒﺘﻪ ‪. 2‬‬ ‫‪ -1‬ﺃﻜﺘﺏ ﻋﺒﺎﺭﺘﻲ ‪ R‬ﻭ ‪ H‬ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ ‪.‬‬ ‫‪ -2‬ﻋﻴﻥ ‪ HOR‬ﻭ ‪. ROH‬‬‫‪ -3‬ﻟﺘﻜﻥ ‪ A‬ﻭ ‪ B‬ﻨﻘﻁﺘﺎﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻻﺤﻘﺘﻴﻬﻤﺎ ‪ 3i ،1 − i‬ﻋﻠﻰ‬ ‫ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬‫ﻋﻴﻥ ﻻﺤﻘﺘﻲ ﺍﻟﻨﻘﻁﺘﻴﻥ ‪ A′‬ﻭ ‪ B′‬ﺼﻭﺭﺘﻲ ‪ A‬ﻭ ‪ B‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‬ ‫ﺒﺎﻟﺘﺤﻭﻴل ‪. HOR‬‬ ‫‪ -4‬ﻗﺎﺭﻥ ﺒﻴﻥ ‪ AB‬ﻭ ‪. A′B′‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪ -1‬ﺘﻌﻴﻴﻥ ﻋﺒﺎﺭﺓ ‪: R‬‬‫ﻟﺘﻜﻥ ‪ M‬ﻨﻘﻁﺔ ﻜﻴﻔﻴﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻻﺤﻘﺘﻬﺎ ‪ Z‬ﻭﻟﺘﻜﻥ ‪ M ′‬ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z ′‬ﻭﺼﻭﺭﺓ‬ ‫‪ M‬ﺒﻭﺍﺴﻁﺔ ‪. R‬‬ ‫ﻟﺩﻴﻨﺎ ‪ Z′ = aZ + b‬ﺤﻴﺙ ‪ a‬ﻭ ‪ b‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻤﻊ ‪a = 1‬‬‫ﻭﻋﻠﻴﻪ ‪Z′ = iZ + b :‬‬ ‫‪a =i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫= )‪arg(a‬‬ ‫‪π‬‬ ‫[‬ ‫]‪2π‬‬ ‫ﻭ‬ ‫‪2‬‬‫ﻭﻟﺩﻴﻨﺎ ‪ R(ω) = ω :‬ﻭﻤﻨﻪ ‪Z0 = iZ0 + b :‬‬‫ﺇﺫﻥ ‪ b = (1 − i)Z0 :‬ﺃﻱ ‪b = (1 − i)(1 + i) = 2 :‬‬‫ﺇﺫﻥ ‪ Z ′ = iZ + 2 :‬ﻫﻲ ﻋﺒﺎﺭﺓ ﺍﻟﺩﻭﺭﺍﻥ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺭﻜﺏ ‪.‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻋﺒﺎﺭﺓ ‪: H‬‬

‫ﻟﺘﻜﻥ ‪ M‬ﻨﻘﻁﺔ ﻜﻴﻔﻴﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻻﺤﻘﺘﻬﺎ ‪ Z‬ﻭﺼﻭﺭﺘﻬﺎ ﺒﻭﺍﺴﻁﺔ ‪ H‬ﺍﻟﻨﻘﻁﺔ ‪ M ′‬ﺫﺍﺕ‬ ‫ﺍﻟﻼﺤﻘﺔ ‪. Z ′‬‬ ‫ﻟﺩﻴﻨﺎ ‪ Z′ = aZ + b‬ﺤﻴﺙ ‪ a = 2‬ﻭ ‪ b‬ﻋﺩﺩ ﻤﺭﻜﺏ ﺃﻱ ‪Z′ = 2Z + b‬‬ ‫ﻟﻜﻥ ‪ H (ω) = ω‬ﻭﻤﻨﻪ ‪ Z0 = 2Z0 + b :‬ﻭﻋﻠﻴﻪ ‪b = − Z0 :‬‬‫ﺃﻱ ﺃﻥ ‪ b = − 1 − i‬ﺇﺫﻥ ‪ Z′ = 2Z − 1 − i‬ﻭﻫﻲ ﻋﺒﺎﺭﺓ ﺍﻟﺘﺤﺎﻜﻲ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ‬ ‫ﺍﻟﻤﺭﻜﺏ ‪.‬‬ ‫‪ -2‬ﺘﻌﻴﻴﻥ ‪: HOR‬‬ ‫ﻟﺘﻜﻥ )‪M ( Z ) R→ M1 ( Z1 ) H→ M′( Z′‬‬ ‫‪ Z′ = 2Z1 − 1 − i‬؛ ‪Z1 = iZ + 2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ Z′ = 2( iZ + 2) − 1 − i‬ﺇﺫﻥ ‪Z′ = 2iZ + 4 − 1 − i‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ Z′ = 2iZ + 3 − i‬ﻭﻫﻲ ﻋﺒﺎﺭﺓ ‪ HOR‬ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺭﻜﺏ ‪.‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ‪: ROH‬‬ ‫)‪M ( Z ) H→ M1 ( Z1 ) R→ M′( Z′‬‬ ‫‪ Z′ = iZ1 + 2‬؛ ‪Z1 = 2Z − 1 − i‬‬ ‫ﻭﻋﻠﻴﻪ ‪ Z′ = i ( 2Z − 1 − i ) + 2 :‬ﺇﺫﻥ ‪Z′ = 2iZ − i + 1 + 2 :‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ Z′ = 2iZ + 3 − i‬ﻨﻼﺤﻅ ﺃﻥ ‪. HOR = ROH‬‬ ‫‪ -3‬ﺘﻌﻴﻴﻥ ﻻﺤﻘﺔ ‪ A′‬ﻭﻟﺘﻜﻥ ‪: Z1‬‬ ‫ﻟﺩﻴﻨﺎ ‪ ( HOR)( A) = A' :‬ﻭﻤﻨﻪ ‪Z1 = 2i (1 − i ) + 3 − i :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ Z1 = 2i + 2 + 3 − i :‬ﺇﺫﻥ ‪Z1 = 5 + i :‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻻﺤﻘﺔ ‪ B′‬ﻭﻟﺘﻜﻥ ‪: Z2‬‬ ‫ﻟﺩﻴﻨﺎ ‪ ( HOR)( B) = B′ :‬ﻭﻤﻨﻪ ‪Z2 = 2i ( 3i ) + 3 − i :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ Z2 = − 6 + 3 − i :‬ﺇﺫﻥ ‪Z2 = − 3 − i :‬‬ ‫‪ -4‬ﺍﻟﻤﻘﺎﺭﻨﺔ ﺒﻴﻥ ‪ AB‬ﻭ ‪: A′B′‬‬ ‫‪AB = ZB − ZA = 3i − 1 + i = −1 + 4i = 17‬‬

‫‪A′B′ = ZB′ − ZA′ = Z2 − Z1 = −3 − i − 5 − i = −8 − 2i‬‬ ‫‪= 64 + 4 = 68‬‬ ‫‪= 22.17‬‬ ‫‪= 2 17‬‬ ‫ﻭﻋﻠﻴﻪ ‪. A′B′ = 2.AB‬‬ ‫‪ -1‬ﺘﻌﺭﻴﻑ ‪:‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻭﺠﻪ ﻤﺭﻓﻕ ﺒﻭﺤﺩﺓ ﺃﻗﻴﺎﺱ ‪.‬‬‫ﻨﻘﻭل ﻋﻥ ﺘﺤﻭﻴل ﻨﻘﻁﻲ ‪ S‬ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺃﻨﻪ ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻯ ﻤﺒﺎﺸﺭ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﺘﺤﻘﻕ ﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﻤﻬﻤﺎ ﺘﻜﻥ ﺍﻟﻨﻘﻁ ‪ N ، M ، B ، A‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺤﻴﺙ ‪ A≠ B‬ﻭ ‪M ≠ N‬‬ ‫ﻭﺍﻟﺘﻲ ﺼﻭﺭﻫﺎ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪ N′ ، M ′ ، B′ ، A′‬ﺒﺎﻟﺘﺤﻭﻴل ‪ S‬ﻓﺈﻥ ‪:‬‬‫‪‬‬ ‫‪A′ ≠ B′ ; M′ ≠ N ′‬‬‫‪‬‬ ‫‪AB‬‬ ‫)‪=JJMJAJG′′BN′′JJJ.J.JG. (1‬‬ ‫‪JJJJGMN‬‬ ‫= ‪MN‬‬ ‫]‪A′B′ ; M′N ′ [2π‬‬‫‪( ) ( )‬‬‫‪JJJG‬‬‫;‬ ‫)‪. . . (2‬‬ ‫‪AB‬‬‫‪‬‬‫‪‬‬‫‪‬‬‫‪‬‬ ‫ﻤﻥ )‪ (1‬ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ ﻴﺤﺎﻓﻅ ﻋﻠﻰ ﻨﺴﺏ ﺍﻟﻤﺴﺎﻓﺎﺕ ‪.‬‬ ‫ﻤﻥ )‪ (2‬ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ ﻴﺤﺎﻓﻅ ﻋﻠﻰ ﺍﻟﺯﻭﺍﻴﺎ ﺍﻟﻤﻭﺠﻬﺔ ‪.‬‬

: ‫ ﺤﺎﻻﺕ ﺨﺎﺼﺔ‬-2 : ‫( ﺘﻜﺎﻓﺌﺎﻥ‬2) ‫( ﻭ‬1) ‫ﺍﻟﻤﺴﺎﻭﺍﺘﻴﻥ‬ M′N ′ J=JJAJGA′BBJ′ JJJJG JJJJGMN( ) ( ) JJJG AB ; A′B′ = MN ; M′N ′ [2π]( )JJJJG JJJJJG = 0[2πJ]J‫ﻭ‬JGMM'NN ' = 1 ‫ﺃ( ﺇﺫﺍ ﻜﺎﻥ‬ MN ; M′N ′ . AA′ ‫ ﺍﻨﺴﺤﺎﺏ ﺸﻌﺎﻋﻪ‬S ‫ﻓﺈﻥ‬ ( )JJJJG JJJJJG MN ; M ′N ′ [ ]= 0 π ‫ﺏ( ﺇﺫﺍ ﻜﺎﻥ‬ . ‫ ﺘﺤﺎﻜﻲ‬S ‫ﻓﺈﻥ‬( )JJJJG JJJJJG ≠ 0[π] ‫ﻭ‬ M′N ′ =1 ‫ﺝ( ﺇﺫﺍ ﻜﺎﻥ‬ MN ; M′N ′ MN . ‫ ﺩﻭﺭﺍﻥ‬S ‫ﻓﺈﻥ‬

‫‪ -3‬ﺘﻌﺭﻴﻑ ‪: 2‬‬ ‫ﻟﻴﻜﻥ ‪ S‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻱ ﻤﺒﺎﺸﺭ ‪.‬‬‫‪ N ، M‬ﻨﻘﻁﺘﺎﻥ ﻤﺘﻤﺎﻴﺯﺘﺎﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺤﻴﺙ ‪ N ′ ، M ′‬ﺼﻭﺭﺘﻴﻬﻤﺎ ﺒﺎﻟﺘﺸﺎﺒﻪ ‪. S‬‬‫ﻫﻭ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﻭﻴﺴﻤﻰ ﻨﺴﺒﺔ ﺍﻟﺘﺸﺎﺒﻪ ‪. S‬‬ ‫‪M′N′‬‬ ‫‪ -‬ﺍﻟﻨﺴﺒﺔ‬ ‫‪JJJJG JJJJJG‬‬ ‫‪MN‬‬‫‪ -‬ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ‪ MN ; M ′N ′‬ﻫﻲ ﺯﺍﻭﻴﺔ ﺜﺎﺒﺘﺔ ﻭ ﺘﺴﻤﻰ ﺯﺍﻭﻴﺔ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ ‪( )S‬‬ ‫ﻭﻨﻁﺎﺒﻘﻬﺎ ﻜﻤﺎ ﺠﺭﺕ ﺍﻟﻌﺎﺩﺓ ﻤﻊ ﺃﺤﺩ ﺃﻗﻴﺎﺴﻬﺎ ‪. θ‬‬‫‪ -4‬ﺍﻟﺸﻜل ﺍﻟﻤﻤﻴﺯ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺭﻜﺏ ‪:‬‬ ‫‪ S‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻱ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬‫ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ﻤﺒﺎﺸﺭ ﻟﺘﻜﻥ ‪ O′‬ﺼﻭﺭﺓ ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ‪ O‬ﺒﺎﻟﺘﺸﺎﺒﻪ‬ ‫‪ S‬ﺤﻴﺙ ‪ b‬ﻻﺤﻘﺔ ﺍﻟﻨﻘﻁﺔ ‪. O′‬‬ ‫ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺤﻴﺙ ‪. M ≠ O‬‬‫ﻓﺈﻥ ﺍﻟﻨﻘﻁﺔ ‪ M ′‬ﺼﻭﺭﺓ ‪ M‬ﺒﺎﻟﺘﺸﺎﺒﻪ ‪JGS‬ﺘﻌ‪J‬ﺭ‪JJ‬ﻑ‪J‬ﻜﻤﺎ ﻴﻠ‪JG‬ﻲ‪( ) [ ]JJ:J‬‬‫)‪ OM ; O′M′ = θ 2π . . . (1‬‬‫‪‬‬‫)‪ O′M′ = k.OM . . . (2‬‬ ‫ﻓﺈﺫﺍ ﻜﺎﻨﺕ ‪ Z ′‬ﻻﺤﻘﺔ ‪ M ′‬ﻓﺈﻥ ‪:‬‬ ‫‪arg‬‬ ‫‪‬‬ ‫‪Z′ − b‬‬ ‫‪‬‬ ‫=‬ ‫]‪θ[2π‬‬ ‫ﻤﻥ )‪: (1‬‬ ‫‪‬‬ ‫‪Z −0‬‬ ‫‪‬‬ ‫‪Z′ − b = k Z‬‬ ‫ﻤﻥ )‪: (2‬‬‫]‪arg ( Z′ − b) − arg ( Z ) = θ[2π‬‬‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬‫‪‬‬ ‫‪Z′ − b = k Z‬‬

‫]‪arg ( Z′ − b) = arg ( Z ) + θ[2π‬‬‫‪‬‬ ‫ﻭﺘﻜﺎﻓﺊ ‪:‬‬‫‪‬‬ ‫‪Z′ − b = k Z‬‬‫ﻭﺘﻜﺎﻓﺊ ‪ Z′ − b = a.z :‬ﺤﻴﺙ ‪ a = k‬ﻭ ]‪arg (a ) = θ[2π‬‬‫ﺃﻱ ﺃﻥ ‪ a = keiθ :‬ﻭﻋﻠﻴﻪ ‪Z′ − b = k.eiθ Z :‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪Z′ = keiθ Z + b :‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ ‪ S‬ﻴﺭﻓﻕ ﺒﻜل ﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪. Z‬‬‫ﺍﻟﻨﻘﻁﺔ ‪ M ′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z′‬ﺒﺤﻴﺙ ‪Z′ = k.eiθ .Z + b :‬‬ ‫‪ -5‬ﺍﻟﺒﺤﺙ ﻋﻥ ﺍﻟﻨﻘﻁ ﺍﻟﺼﺎﻤﺩﺓ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ‪:‬‬ ‫‪ S‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻱ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬ ‫ﻟﺘﻜﻥ ‪ ω‬ﻨﻘﻁﺔ ﺼﺎﻤﺩﺓ ﺒﺎﻟﺘﺤﻭﻴل ‪ S‬ﺃﻱ ‪( )S ω = ω‬‬ ‫ﻓﺈﺫﺍ ﻜﺎﻨﺕ ‪ Z0‬ﻻﺤﻘﺔ ‪ ω‬ﻓﺈﻥ ‪Z0 = keiθ .Z0 + b‬‬ ‫ﻭﻤﻨﻪ )‪( )1 − keiθ Z0 = b . . . (1‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ k = 1‬ﻭ ‪: θ = 0‬‬‫ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺤﻠﻭل ﻭﻋﻠﻴﻪ ﻻ ﺘﻭﺠﺩ ﻨﻘﻁ ﺼﺎﻤﺩﺓ ﻷﻥ ﺍﻟﺘﺤﻭﻴل ‪S‬‬ ‫ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﻨﺴﺤﺎﺏ ‪.‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ k ≠ 1‬ﻭ ‪: θ ≠ 0‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪b‬‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪1 − keiθ‬‬ ‫ﻭﻫﻭ ﺤل ﻭﺤﻴﺩ ﻭﻋﻠﻴﻪ ‪ S‬ﻴﻘﺒل ﻨﻘﻁﺔ ﺼﺎﻤﺩﺓ ﻭﺤﻴﺩﺓ ‪.‬‬ ‫ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺘﺩﻋﻰ ﻤﺭﻜﺯ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺒﺎﺸﺭ ‪. S‬‬

‫‪ -6‬ﺍﻟﺸﻜل ﺍﻟﻤﺨﺘﺼﺭ ﻟﺘﺸﺎﺒﻪ ‪:‬‬ ‫ﻟﻴﻜﻥ ‪ S‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻱ ﻤﺒﺎﺸﺭ ﻤﺭﻜﺯﻩ ‪ ω‬ﻨﺴﺒﺘﻪ ‪ k‬ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬ ‫ﻟﺘﻜﻥ ‪ Z0‬ﻻﺤﻘﺔ ﺍﻟﻤﺭﻜﺯ ‪.‬‬ ‫ﻟﻜﻥ ﺍﻟﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺼﻭﺭﺘﻬﺎ ﺒﻭﺍﺴﻁﺔ ‪ S‬ﻫﻲ ﺍﻟﻨﻘﻁﺔ ‪M ′‬‬ ‫ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z′‬ﺤﻴﺙ ‪ Z′ = keiθ Z + b‬ﻟﻜﻥ ‪Z0 = keiθ Z0 + b :‬‬ ‫ﻭﺒﺎﻟﻁﺭﺡ ﻨﺠﺩ ‪( )Z′ − Z0 = k.eiθ Z − Z0 :‬‬‫ﻭﻫﻲ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﻤﺨﺘﺼﺭﺓ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ‪ S‬ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z0‬ﻭﻨﺴﺒﺘﻪ ‪k‬‬ ‫ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬ ‫‪ -7‬ﻤﺭﻜﺏ ﺘﺤﺎﻜﻲ ﻭﺩﻭﺭﺍﻥ ‪:‬‬ ‫‪ H‬ﺘﺤﺎﻜﻲ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪. Z0‬‬ ‫‪ R‬ﺩﻭﺭﺍﻥ ﻤﺭﻜﺯﻩ ‪ ω‬ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬ ‫ﻨﻔﺭﺽ ‪( )( ) ( )M Z H(ω;k)→ M1 Z1 R(ω;θ)→ M ′ Z′ :‬‬ ‫‪Z1 − Z0 = k ( Z − Z0 ) ، ( )Z′ − Z0 = eiθ Z1 − Z0‬‬ ‫ﻭﻋﻠﻴﻪ ‪( )Z′ − Z0 = keiθ Z − Z0 :‬‬ ‫ﺇﺫﻥ ‪R(ω,θ)OH (ω,k ) = S (ω,k,θ) :‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺭﻜﺏ ﺘﺤﺎﻜﻲ ﻭ ﺩﻭﺭﺍﻥ ﻫﻭ ﺘﺸﺎﺒﻪ ‪.‬‬ ‫ﻭﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ ‪H (ω, k )OR(ω,θ) = S (ω, k,θ) :‬‬

‫‪ -8‬ﻨﻅﺭﻴﺔ ﻭ ﺘﻌﺭﻴﻑ ‪:‬‬‫ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺒﺎﺸﺭ ‪ S‬ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ ω‬ﻭﻨﺴﺒﺘﻪ ‪ k‬ﻭﺯﺍﻭﻴﺘﻪ ‪ θ‬ﻫﻭ ﺍﻟﺘﺤﻭﻴل ﺍﻟﻨﻘﻁﻲ ﺍﻟﺫﻱ‬ ‫ﻴﺭﻓﻕ ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺒﻨ‪G‬ﻔ‪J‬ﺴ‪J‬ﻬ‪J‬ﺎ‪J‬ﻭ ﻴﺭﻓ‪JG‬ﻕ‪J‬ﺒ‪J‬ﻜل‪ J‬ﻨﻘﻁﺔ ‪ M‬ﺘﺨﺘﻠﻑ ﻋﻥ ‪ ω‬ﺍﻟﻨﻘﻁﺔ ‪ M ′‬ﺤﻴﺙ ‪:‬‬ ‫‪( ) [ ] OM ; OM′ = θ 2π‬‬ ‫‪‬‬ ‫‪ OM′ = k.OM‬‬ ‫ﻓﺈﺫﺍ ﻜﺎﻨﺕ ‪ Z0‬ﻭ ‪ Z‬ﻭ ‪ Z ′‬ﻟﻭﺍﺤﻕ ﺍﻟﻨﻘﻁ ‪ M ′ ، M ، ω‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻓﺈﻥ ‪:‬‬ ‫) ‪Z′ − Z0 = keiθ ( Z − Z0‬‬ ‫‪ -9‬ﻤﺭﻜﺏ ﺘﺸﺎﺒﻬﻴﻥ ﻤﺒﺎﺸﺭﻴﻥ ‪:‬‬ ‫ﻟﻴﻜﻥ ﺍﻟﺘﺸﺎﺒﻬﻴﻥ ﺍﻟﻤﺒﺎﺸﺭﻴﻥ ‪( ) ( )S2 ω2 , k2 ,θ2 ، S1 ω1, k1, θ1‬‬ ‫ﺤﻴﺙ ‪ Z0′ ، Z0‬ﻻﺤﻘﺘﻲ ‪ ω2 ، ω1‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫ﻨﻔﺭﺽ ‪M ( Z ) S1→ M1 ( Z1 ) S2→ M ′( Z′) :‬‬ ‫‪( ) ( )Z1 − Z0 = k1eiθ1 Z − Z0 ، Z′ − Z0′ = k2eiθ2 Z1 − Z0‬‬ ‫ﺇﺫﻥ ‪( )Z1 = k1eiθ1 Z − Z0 + Z0 :‬‬ ‫‪( )Z′ = k2eiθ2 Z1 − Z0′ + Z0′‬‬ ‫ﻭﻤﻨﻪ ‪( )Z′ = k2eiθ2 k1eiθ1 Z − Z0 + Z0 − Z0′  + Z0′ :‬‬ ‫‪( ) ( )Z′ = k1k2ei(θ1+θ2) Z − Z0 + k2eiθ2 Z0 − Z0′ + Z0′‬‬ ‫ﻭﻫﻭ ﻤﻥ ﺍﻟﺸﻜل ‪Z ′ = k1k2ei(θ1+θ2 ) Z + b :‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ k1 .k2 = 1‬ﻭ ‪ θ1 + θ2 = 0‬ﻓﺈﻥ ‪ S2oS1‬ﻫﻭ ﺍﻨﺴﺤﺎﺏ ‪.‬‬

‫* ﺇﺫﺍ ﻜﺎﻥ ‪ k.k2 = 1‬ﻭ ‪ θ1 + θ2 ≠ 0‬ﻓﺈﻥ ‪ S2oS1‬ﻫﻭ ﺩﻭﺭﺍﻥ ‪.‬‬‫* ﺇﺫﺍ ﻜﺎﻥ ‪ k1 .k2 ≠ 1‬ﻭ ‪ θ1 + θ2 = 0‬ﻓﺈﻥ ‪ S2oS1‬ﻫﻭ ﺘﺤﺎﻜﻲ ‪.‬‬‫‪ -10‬ﺍﻟﺨﺎﺼﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻟﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ‪:‬‬ ‫ﻤﺒﺭﻫﻨﺔ ‪:‬‬ ‫‪ k‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ‪ θ ،‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪.‬‬‫ﻴﻜﻭﻥ ﺍﻟﺘﺤﻭﻴل ﺍﻟﻨﻘﻁﻲ ‪ S‬ﺘﺸﺎﺒﻬﺎ ﻤﺒﺎﺸﺭﺍ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﺯﺍﻭﻴﺘﻪ ‪ θ‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻤﻥ ﺃﺠل ﻜل‬‫ﻨﻘﻁﺘﻴﻥ ‪ A‬ﻭ ‪ B‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺼﻭﺭﺘﻴﻬﻤﺎ ‪ A′‬ﻭ ‪ B′‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﺒﻭﺍﺴﻁﺔ ‪ S‬ﻓﺈﻥ ‪:‬‬ ‫‪A′B′ = k.AB‬‬‫‪JJJG JJJJG‬‬‫]‪AB ; A′B′ = θ[2π‬‬‫‪( )‬‬‫‪‬‬‫‪‬‬ ‫ﺤﺎﻻﺕ ﺨﺎﺼﺔ ‪:‬‬‫* ﺇﺫﺍ ﻜﺎﻥ ‪ : k = 1‬ﺍﻟﺘﺸﺎﺒﻪ ﻫﻭ ﺇﺯﺍﺤﺔ ﺃﻱ ﺇﻤﺎ ﺩﻭﺭﺍﻥ ﺃﻭ ﺍﻨﺴﺤﺎﺏ ﻭﻫﻭ‬ ‫ﺘﻘﺎﻴﺱ ‪.‬‬‫* ﺇﺫﺍ ﻜﺎﻥ ‪ : θ = 0‬ﺍﻟﺘﺸﺎﺒﻪ ﻫﻭ ﺘﺤﺎﻜﻲ ‪.‬‬ ‫‪ -11‬ﺼﻭﺭ ﺒﻌﺽ ﺍﻷﺸﻜﺎل ﺍﻟﻬﻨﺩﺴﻴﺔ ‪:‬‬ ‫‪ S‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻱ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﻤﺭﻜﺯﻩ ‪ ω‬ﻭﺯﺍﻭﻴﺘﻪ ‪. θ‬‬‫* ﺼﻭﺭﺓ ﻗﻁﻌﺔ ﻤﺴﺘﻘﻴﻤﺔ ‪ AB‬ﺒﻭﺍﺴﻁﺔ ‪ S‬ﻫﻲ ﺍﻟﻘﻁﻌﺔ ﺍﻟﻤﺴﺘﻘﻴﻤﺔ ‪ A′B′‬ﺤﻴﺙ ‪[ ] [ ]:‬‬ ‫)‪ A′ = S ( A‬ﻭ )‪ B′ = S ( B‬ﻭ ‪A′B′ = k.AB‬‬ ‫* ﺼﻭﺭﺓ ﻤﺴﺘﻘﻴﻡ ∆ ﻫﻭ ﻤﺴﺘﻘﻴﻡ ‪ ∆′‬ﺒﻭﺍﺴﻁﺔ ‪ S‬ﺤﻴﺙ ‪[ ]( ) ( ) ( )∆;∆′ = θ π‬‬ ‫* ﺼﻭﺭﺓ ﺯﺍﻭﻴﺔ ﻤﻭﺠﻬﺔ ﻫﻲ ﺯﺍﻭﻴﺔ ﻤﻭﺠﻬﺔ ‪.‬‬ ‫* ﺼﻭﺭﺓ ﺩﺍﺌﺭﺓ )‪ C ( F , R‬ﻫﻲ ﺩﺍﺌﺭﺓ )‪C′( F ′, R′‬‬