دروس مادة الرياضيات للفصل الاول للشعب العلمية سنة ثالثة ثانوي

‫ﺤﻴﺙ )‪R′ = k.k ، F = S ( R‬‬ ‫‪ -12‬ﺩﺭﺍﺴﺔ ﺍﻟﺘﺤﻭﻴﻼﺕ ﺍﻟﻨﻘﻁﻴﺔ ‪: f‬‬ ‫‪ M → M ′‬ﺤﻴﺙ ‪Z′ = aZ + b‬‬ ‫‪ Z‬ﻭ ‪ Z ′‬ﻫﻤﺎ ﻻﺤﻘﺘﻲ ‪ M‬ﻭ ‪ M ′‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫‪ a‬ﻭ ‪ b‬ﻋﺩﺩﺍﻥ ﻤﺭﻜﺒﺎﻥ ﻤﻊ ‪. a ≠ 0‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪Z′ = Z G+ b : a = 1‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﺍﻨﺴﺤﺎﺏ ﺸﻌﺎﻋﻪ ‪ v‬ﺫﻭ ﺍﻟﻼﺤﻘﺔ ‪b‬‬‫* ﺇﺫﺍ ﻜﺎﻥ ‪ f : a ∈ R* − 1‬ﺘﺤﺎﻙ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ} {‬ ‫‪b‬‬ ‫‪. 1−a‬‬‫* ﺇﺫﺍ ﻜﺎﻥ ‪ a = 1‬ﻭ ‪ a‬ﻏﻴﺭ ﺤﻘﻴﻘﻲ ‪ f :‬ﺩﻭﺭﺍﻥ ﺯﺍﻭﻴﺘﻪ ﻋﻤﺩﺓ ‪ a‬ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺫﺍﺕ‬ ‫‪b‬‬ ‫ﺍﻟﻼﺤﻘﺔ ‪. 1 − a‬‬‫* ﺇﺫﺍ ﻜﺎﻥ ‪ a = k‬ﻭ ‪ a‬ﻏﻴﺭ ﺤﻘﻴﻘﻲ ‪ f :‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻱ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪k‬‬‫‪.‬‬ ‫‪b‬‬ ‫ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪ω‬‬ ‫ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬ ‫‪a‬‬ ‫ﻭﺯﺍﻭﻴﺘﻪ ﻋﻤﺩﺓ‬ ‫‪1−a‬‬

‫ﺘﻤـﺎﺭﻴﻥ ﻭ ﻤﺸﻜﻼﺕ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬‫ﻀﻊ ﺍﻟﻌﻼﻤﺔ√ﺃﻤﺎﻡ ﻜل ﺠﻤﻠﺔ ﺼﺤﻴﺤﺔ ﻭ ﺍﻟﻌﻼﻤﺔ × ﺃﻤﺎﻡ ﻜل ﺠﻤﻠﺔ ﺨﺎﻁﺌﺔ ‪:‬‬ ‫‪ (1‬ﺇﺫﺍ ﻜﺎﻨ‪JG‬ﺕ‪ MJ′JJ‬ﺼﻭ‪JG‬ﺭﺓ‪ MJJJ‬ﺒﺘﺸﺎﺒﻪ ﻨﺴﺒﺘﻪ ‪ k‬ﻭﻤﺭﻜﺯﻩ ‪ω‬‬ ‫ﻓﺈﻥ ‪. ωM ′ = kωM‬‬ ‫‪ (2‬ﺼﻭﺭﺓ ﺯﺍﻭﻴﺔ ﻤﻭﺠﻬﺔ ﺒﺘﺸﺎﺒﻪ ﻫﻲ ﺯﺍﻭﻴﺔ ﻤﻭﺠﻬﺔ ‪. .‬‬ ‫‪ (3‬ﺼﻭﺭﺓ ﺯﺍﻭﻴﺔ ﺒﺘﺸﺎﺒﻪ ﻫﻲ ﺯﺍﻭﻴﺔ ﺘﻘﺎﻴﺴﻬﺎ ‪.‬‬‫‪ (4‬ﺼﻭﺭﺓ ﻗﻁﻌﺔ ﻤﺴﺘﻘﻴﻤﺔ ﺒﺘﺸﺎﺒﻪ ﻫﻲ ﻗﻁﻌﺔ ﻤﺴﺘﻘﻴﻤﺔ ﺘﻘﺎﻴﺴﻬﺎ ‪.‬‬‫‪( )JJJG JJJG‬‬ ‫‪AB , AC‬‬ ‫=‬ ‫‪π‬‬ ‫ﻤﺜﻠﺙ ﺒﺤﻴﺙ‬ ‫‪ABC‬‬ ‫‪ (5‬ﺇﺫﺍ ﻜﺎﻥ‬ ‫‪2‬‬ ‫ﻓﺈﻨﻪ ﻴﻭﺠﺩ ﺘﺸﺎﺒﻪ ﻴﺤﻭل ‪ B‬ﺇﻟﻰ ‪. C‬‬‫‪ (6‬ﺍﻟﻌﺒﺎﺭﺓ ‪ Z′ = aZ + b‬ﻫﻲ ﻟﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻯ ﻤﺒﺎﺸﺭ ‪.‬‬‫‪ (7‬ﻜل ﺘﺸﺎﺒﻪ ﻤﺭﻜﺯﻩ ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ‪ O‬ﻟﻪ ﻋﺒﺎﺭﺓ ﻤﻥ ﺍﻟﺸﻜل ‪:‬‬ ‫‪ Z′ = aZ‬ﺤﻴﺙ ‪. a ≠ 1 :‬‬‫‪ (8‬ﺘﻭﺠﺩ ﻨﻘﻁﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﻟﻴﺱ ﻟﻬﺎ ﺼﻭﺭﺓ ﺒﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻯ ﻤﺒﺎﺸﺭ ‪.‬‬‫‪ (9‬ﺼﻭﺭﺓ ﺃﻴﺔ ﻨﻘﻁﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﻫﻲ ﻨﻘﻁﺔ ﺘﺨﺘﻠﻑ ﻋﻨﻬﺎ ‪.‬‬‫ﻭﻨﺴﺒﺘﻪ ‪2‬‬ ‫‪O‬‬ ‫‪Z′‬‬ ‫=‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪Z‬‬ ‫‪ (10‬ﺍﻟﻌﺒﺎﺭﺓ‬ ‫‪2‬‬ ‫ﻫﻲ ﻟﺘﺸﺎﺒﻪ ﻤﺭﻜﺯﻩ‬ ‫‪π‬‬ ‫ﻭ ﺯﺍﻭﻴﺘﻪ ‪. 2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬‫ﺍﺩﺭﺱ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺤﻭﻴل ‪ f‬ﺍﻟﺫﻱ ﻴﺭﻓﻕ ﺒﺎﻟﻨﻘﻁﺔ ‪ M‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ Z‬ﺍﻟﻨﻘﻁﺔ ‪ M ′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪ Z ′‬ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬

‫‪1) Z′ = 2iZ‬‬ ‫‪2) Z′ = (−i + 1) Z + 2‬‬ ‫‪( )3) Z′ = −2 3 − 2 Z‬‬ ‫‪( )4) Z′ = 1 − 3i Z − 3 − i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺘﺤﻭﻴل ﺍﻟﻨﻘﻁﻲ ‪ S‬ﺍﻟﺫﻱ ﻴﺭﻓﻕ ﺒﺎﻟﻨﻘﻁﺔ ‪ M x ; y‬ﺍﻟﻨﻘﻁﺔ ‪( ) ( )M′ x′; y′‬‬ ‫‪ x′ = x + y + 1‬‬ ‫‪‬‬ ‫ﺒﺤﻴﺙ ‪:‬‬ ‫‪‬‬ ‫‪y′‬‬ ‫=‬ ‫‪−‬‬ ‫‪x+‬‬ ‫‪y+1‬‬ ‫‪ -‬ﺍﺩﺭﺱ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺤﻭﻴل ‪. S‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺒﺎﺸﺭ ‪ S‬ﺍﻟﺫﻱ ﻴﺭﻓﻕ ﺒﺎﻟﻨﻘﻁﺔ ‪ A‬ﺫﺍﺕ‬ ‫ﺍﻟﻼﺤﻘﺔ ‪. 1 + 2i‬‬‫ﺍﻟﻨﻘﻁﺔ ‪ A′‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ −4 + 2i‬ﻭﺒﺎﻟﻨﻘﻁﺔ ‪ B‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪ 3 + i‬ﺍﻟﻨﻘﻁﺔ ‪ B′‬ﺫﺍﺕ‬ ‫ﺍﻟﻼﺤﻘﺔ ‪. −2 + 2i 3‬‬ ‫‪ -1‬ﻋﺭﻑ ﻋﺒﺎﺭﺓ ﺍﻟﺘﺸﺎﺒﻪ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ ﻤﻌﻴﻨﺎ ﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫‪ -2‬ﻋﻴﻥ ﻋﺒﺎﺭﺓ ﺍﻟﺘﺸﺎﺒﻪ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺸﻜل ﺍﻷﺴﻲ ‪.‬‬ ‫‪ -3‬ﺍﻜﺘﺏ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﺘﺤﻠﻴﻠﻴﺔ ﻟﻬﺫﺍ ﺍﻟﺘﺸﺎﺒﻪ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺘﺸﺎﺒﻬﻴﻥ ‪ S1‬ﻭ ‪ S2‬ﺤﻴﺙ ‪:‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫ﻭﻨﺴﺒﺘﻪ ‪ 2‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫‪O‬‬ ‫ﻫﻭ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ‬ ‫‪S1‬‬ ‫‪3‬‬ ‫‪2π‬‬ ‫‪1‬‬‫‪.‬‬ ‫‪3‬‬ ‫ﻭﺯﺍﻭﻴﺘﻪ‬ ‫‪2‬‬ ‫ﻭﻨﺴﺒﺘﻪ‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪2i‬‬ ‫ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪A‬‬ ‫ﻫﻭ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬ ‫‪S2‬‬ ‫‪ -1‬ﻋﻴﻥ ﺍﻟﺸﻜل ﺍﻟﻤﺭﻜﺏ ﻟﻜل ﻤﻥ ﺍﻟﺘﺸﺎﺒﻬﻴﻥ ‪.‬‬ ‫‪ -2‬ﻋﻴﻥ ﺍﻟﺘﺤﻭﻴل ﺍﻟﻤﺭﻜﺏ ‪ S2oS1‬ﻤﻊ ﺇﻋﻁﺎﺀ ﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫‪ -1‬ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪2Z 2 − (1 + 5i ) Z + 2( i − 1) = 0‬‬

‫‪،‬‬ ‫‪Z2‬‬ ‫=‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫‪،‬‬ ‫‪Z1‬‬ ‫=‬ ‫ﺫﺍﺕ ﺍﻟﻠﻭﺍﺤﻕ ‪2i‬‬ ‫‪ -2‬ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪C ، B ، A‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫= ‪ Z3‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫‪2‬‬ ‫‪i‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2Z2‬‬ ‫‪ 2008‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪2007‬‬ ‫‪‬‬ ‫‪2Z3‬‬ ‫‪1961‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪.‬‬ ‫‪−‬‬ ‫‪Z1‬‬ ‫‪+‬‬ ‫=‬ ‫‪1‬‬ ‫ﺃﻥ‬ ‫ﺒﻴﻥ‬ ‫‪-‬‬ ‫‪ -‬ﻋﻴﻥ ﺍﻟﺘﺸﺎﺒﻪ ‪ S‬ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ O‬ﻭﻴﺤﻭل ‪ B‬ﺇﻟﻰ ‪. A‬‬ ‫‪ -‬ﻋﻴﻥ ﺍﻟﺩﻭﺭﺍﻥ ‪ r‬ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ O‬ﻭﻴﺤﻭل ‪ B‬ﺇﻟﻰ ‪. C‬‬ ‫* ﻋﻴﻥ ‪ Sor‬ﻤﻌﻴﻨﺎ ﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫‪ -‬ﻋﻴﻥ ‪ SoS‬ﻤﻌﻴﻨﺎ ﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬ ‫ﻨﻌﺘﺒﺭ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ‪ P Z‬ﺤﻴﺙ ‪( ):‬‬ ‫‪P ( Z ) = Z 3 + (3 − 2i ) Z 2 + (1 − 5i) Z − 2i − 2‬‬ ‫‪ (1‬ﺍﺤﺴﺏ ‪ . P −1 + i‬ﻤﺎﺫﺍ ﺘﺴﺘﻨﺘﺞ ؟) (‬ ‫‪ (2‬ﺤل ﻓﻲ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪( ). P Z = 0‬‬ ‫‪ (3‬ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪ C ، B ، A‬ﺍﻟﺘﻲ ﻟﻭﺍﺤﻘﻬﺎ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪Z3 ، Z2 ، Z1‬‬ ‫ﺤﻴﺙ ‪. Z3 = i ، Z2 = −2 ، Z1 = − 1 + i‬‬‫‪ -‬ﻟﻴﻜﻥ ﺍﻟﺘﺸﺎﺒﻪ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺒﺎﺸﺭ ‪ S‬ﺍﻟﺫﻱ ﻴﺭﻓﻕ ﺒﺎﻟﻨﻘﻁﺔ ‪ A‬ﺍﻟﻨﻘﻁﺔ ‪ B‬ﻭﺒﺎﻟﻨﻘﻁﺔ ‪ C‬ﺍﻟﻨﻘﻁﺔ ‪، A‬‬ ‫ﺍﻜﺘﺏ ﻋﺒﺎﺭﺘﻪ ﺍﻟﻤﺭﻜﺒﺔ ﺜﻡ ﻋﻴﻥ ﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫‪ -‬ﺍﻜﺘﺏ ﻋﺒﺎﺭﺓ ﺍﻟﺘﺸﺎﺒﻪ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ‪.‬‬ ‫‪ -‬ﻋﻴﻥ ‪ SoS‬ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺘﻪ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫ﻨﻌﺘﺒﺭ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻜﺒﺔ ^ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬ ‫‪Z 3 − (4 − 3i ) Z 2 + (9 − 2i ) Z − 2 + 11i = 0‬‬ ‫‪ -1‬ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻘﺒل ﺤﻼ ﺘﺨﻴﻠﻴﺎ ﺼﺭﻓﺎ ‪. Z0‬‬

‫‪ -‬ﺍﺴﺘﻨﺘﺞ ﺍﻟﺤﻠﻴﻥ ﺍﻵﺨﺭﻴﻥ ‪ Z2 ، Z1‬ﺤﻴﺙ ‪. Z1 < Z2‬‬‫‪ -2‬ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪ D ، C ، B ، A‬ﺼﻭﺭ ﺍﻷﻋﺩﺍﺩ ‪Z − 4i ، Z2 ، Z1 ، Z0‬‬ ‫ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‪ - .‬ﺃﺜﺒﺕ ﺃﻨﻪ ﻴﻭﺠﺩ ﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ‪ S‬ﻟﻠﻤﺴﺘﻭﻯ ﻓﻲ ﻨﻔﺴﻪ‬‫ﺒﺤﻴﺙ ‪ S C = D ، S A = B :‬ﻤﻌﻴﻨﺎ ﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪( ) ( ).‬‬ ‫‪ -3‬ﻋﻴﻥ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻬﺫﺍ ﺍﻟﺘﺸﺎﺒﻪ ‪.‬‬‫‪ -4‬ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﺍﺤﺴﺏ ‪ . SoS‬ﻤﺎ ﻫﻲ ﻁﺒﻴﻌﺘﻪ ؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁﺘﺎﻥ ‪ B ، A‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺘﻴﻥ ‪ 3i ،1 − i‬ﻭﻟﻴﻜﻥ ‪ H‬ﺍﻟﺘﺤﺎﻜﻲ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪A‬‬‫‪−π‬‬ ‫ﺍﻟﺩﻭﺭﺍﻥ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ B‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫ﻭ‪r‬‬ ‫‪3‬‬ ‫ﻭﻨﺴﺒﺘﻪ‬ ‫‪3‬‬ ‫‪2‬‬‫‪ (1‬ﺍﻜﺘﺏ ﻋﺒﺎﺭﺓ ﻜل ﻤﻥ ‪ H‬ﻭ ‪ r‬ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ ‪.‬‬ ‫‪ (2‬ﺍﺤﺴﺏ ‪ roH‬ﻭﺍﺩﺭﺱ ﻁﺒﻴﻌﺘﻪ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬‫‪ f‬ﺘﺤﻭﻴل ﻨﻘﻁﻲ ﻴﺭﻓﻕ ﺒﺎﻟﻨﻘﻁﺔ ‪ M x; y‬ﺍﻟﻨﻘﻁﺔ ‪( ) ( )M ′ x′; y′‬‬ ‫‪ x′ = 3x − y‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪‬‬ ‫‪y′‬‬ ‫=‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪3y‬‬ ‫ﻨﻔﺭﺽ ‪ Z′ ، Z‬ﻻﺤﻘﺘﻲ ‪ M‬ﻭ ‪. M ′‬‬‫‪ (1‬ﺍﻜﺘﺏ ‪ Z ′‬ﺒﺩﻻﻟﺔ ‪ . Z‬ﻤﺎ ﻫﻲ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺤﻭﻴل ‪. f‬‬ ‫‪ -‬ﺍﻜﺘﺏ ﻋﺒﺎﺭﺘﻪ ﺍﻷﺴﻴﺔ ‪.‬‬ ‫‪ (2‬ﻨﻔﺭﺽ ﺍﻟﻨﻘﻁﺔ ‪ M0 1 ; 0‬ﻭﻨﻀﻊ ‪( ) ( )Mi+1 = f Mi‬‬ ‫ﻤﻥ ﺃﺠل ‪. i ∈{0 ; 1 ; ... ;n} :‬‬ ‫‪ -‬ﻋﻴﻥ ﻟﻭﺍﺤﻕ ﻜل ﻤﻥ ‪ Z2 ، Z1 ، Z0‬ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﻋﺒﺎﺭﺓ ‪. Zn‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬‫‪ ABC‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪ A‬ﺤﻴﺙ ‪ AB = 2 :‬ﻭ ‪. BC = 4‬‬ ‫ﻟﺘﻜﻥ ‪ H‬ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁﺔ ‪ A‬ﻋﻠﻰ ‪( ). BC‬‬

‫ﺒﻴﻥ ﺃﻥ ﺍﻟﻤﺜﻠﺙ ‪ HBA‬ﻫﻭ ﺼﻭﺭﺓ ﺍﻟﻤﺜﻠﺙ ‪ HAC‬ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻯ ﻤﺒﺎﺸﺭ ﻴﻁﻠﺏ ﺇﻋﻁﺎﺀ ﻋﻨﺎﺼﺭﻩ‬ ‫ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬‫∆ ﻤﺴﺘﻘﻴﻡ ﻤﻌﻁﻰ ‪ A .‬ﻨﻘﻁﺔ ﺜﺎﺒﺘﺔ ﻻ ﺘﻨﺘﻤﻲ ﺇﻟﻰ ∆ ‪( ) ( ).‬‬ ‫‪ M‬ﻨﻘﻁﺔ ﻜﻴﻔﻴﺔ ﻤﻥ ∆ ‪( ).‬‬‫‪( )‬‬ ‫‪MA = MB‬‬‫‪‬‬‫‪‬‬‫‪JJJG JJJG‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫;‬ ‫‪k∈Z‬‬ ‫‪ B‬ﻨﻘﻁﺔ ﺤﻴﺙ ‪:‬‬‫‪MA ; MB‬‬ ‫‪2‬‬ ‫‪ -‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪. B‬‬

‫ﺍﻟﺤـﻠــــــﻭل‬ ‫‪. √ (3‬‬ ‫‪. √ (2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬ ‫‪× (1‬‬ ‫× ‪(6 . √ (5‬‬ ‫‪(4‬‬ ‫×‪.‬‬‫×‪.‬‬ ‫× ‪(9‬‬ ‫√ ‪(8 .‬‬ ‫‪(7‬‬ ‫‪(12 × (11 × (10‬‬ ‫√‪..‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬ ‫ﺩﺭﺍﺴﺔ ﻁﺒﻴﻌﺔ ﺍﻟﺘﺤﻭﻴل ‪: f‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪Z′ = 2iZ :‬‬ ‫‪π‬‬ ‫ﻭﻋﻤﺩﺓ ‪ a‬ﻫﻲ‬ ‫‪a‬‬ ‫‪ Z′ = aZ‬ﺤﻴﺙ ‪= 2‬‬ ‫ﻭﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪:‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫‪ O‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻯ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪ 2‬ﻭﻤﺭﻜﺯﻩ‬ ‫‪f‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪2‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪Z′ = ( −i + 1) Z + 2 :‬‬ ‫ﻭﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪ Z′ = aZ + b‬ﺤﻴﺙ ‪ a = 2‬ﻭﻋﻤﺩﺓ ‪ a‬ﻫﻲ ‪θ‬‬

‫‪−π‬‬ ‫‪‬‬ ‫‪cos‬‬ ‫‪θ‬‬ ‫=‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪θ‬‬ ‫=‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫‪, k ∈ Z/‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪sin‬‬ ‫‪−2‬‬ ‫‪θ‬‬ ‫=‬ ‫‪2‬‬‫ﺤﻴﺙ ‪:‬‬ ‫‪Z0‬‬ ‫ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪M0‬‬ ‫ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬ ‫‪−π‬‬ ‫ﻭﺯﺍﻭﻴﺘﻪ‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﺘﺸﺎﺒﻪ ﻨﺴﺒﺘﻪ ‪2‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪b‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪1+ i −1‬‬ ‫=‬ ‫‪i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪1−a‬‬ ‫‪.‬‬ ‫‪Z0‬‬ ‫=‬ ‫)‪2(−i‬‬ ‫=‬ ‫‪− 2i‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪i(−i‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪( )Z′ = −2 3 − 2 Z :‬‬ ‫ﻭﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪ Z′ = aZ‬ﺤﻴﺙ ‪ a = 4‬ﻭﻋﻤﺩﺓ ‪ a‬ﻫﻲ ‪θ‬‬ ‫]∈‪k‬‬ ‫ﻤﻊ‬ ‫‪θ‬‬ ‫=‬ ‫‪7π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫= ‪cosθ‬‬ ‫‪−3‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪6‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪−1‬‬ ‫‪‬‬ ‫= ‪sin θ‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪O‬‬ ‫ﻭﻤﺭﻜﺯﻩ‬ ‫‪7π‬‬ ‫ﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪ 4‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫‪f‬‬ ‫ﺇﺫﻥ‬ ‫‪6‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪( )Z′ = 1 − 3i Z − 3 − i :‬‬ ‫ﻭﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪ Z′ = aZ + b‬ﺤﻴﺙ ‪ a = 2‬ﻭﻋﻤﺩﺓ ‪ a‬ﻫﻲ ‪θ‬‬

‫‪−π‬‬ ‫‪‬‬ ‫= ‪cos θ‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪2‬‬ ‫]∈‪k‬‬ ‫ﻤﻊ‬ ‫=‪θ‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪‬‬ ‫ﺤﻴﺙ‬ ‫‪sin θ‬‬ ‫‪−3‬‬ ‫=‬ ‫‪2‬‬‫‪Z0‬‬ ‫‪ M0‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬ ‫‪−π‬‬ ‫ﺘﺸﺎﺒﻪ ﻤﺴﺘﻭﻯ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪ 2‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫‪f‬‬ ‫ﻭﻋﻠﻴﻪ‬ ‫‪3‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪− 3−i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪b‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪1+ i −1‬‬ ‫‪1−a‬‬ ‫‪Z0 = − 1 + i‬‬ ‫‪3‬‬ ‫‪ Z0 = −‬ﻭﻋﻠﻴﻪ‬ ‫‪3‬‬ ‫‪−‬‬ ‫‪i‬‬ ‫×‬ ‫‪−i‬‬ ‫=‬ ‫‪−‬‬ ‫‪1+‬‬ ‫‪i‬‬ ‫ﺃﻱ ‪3 :‬‬ ‫‪i‬‬ ‫‪−i‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫‪ -‬ﻨﻔﺭﺽ ‪ Z‬ﻻﺤﻘﺔ ‪ M‬ﻭ ‪ Z′‬ﻻﺤﻘﺔ ‪. M ′‬‬ ‫ﺇﺫﻥ ‪Z′ = x′ + iy′ = ( x + y + 1) + i ( − x + y + 1) :‬‬ ‫‪= x + y + 1 − ix + iy + i‬‬ ‫‪= x + iy + y − ix + i‬‬ ‫‪= Z + y − ix + i‬‬ ‫‪= Z − i2 y − ix + i‬‬ ‫‪= Z − i ( x + iy) + i‬‬ ‫‪Z′ = Z − iZ + i = (1 − i ) Z + i‬‬ ‫‪Z′ = (1− i) Z + i‬‬ ‫ﻭﻫﻭ ﻤﻥ ﺍﻟﺸﻜل ‪ Z′ = aZ + b‬ﺤﻴﺙ ‪ a = 2‬ﻭﻋﻤﺩﺓ ‪ a‬ﻫﻲ ‪θ‬‬

‫‪−π‬‬ ‫‪‬‬ ‫‪cos‬‬ ‫‪θ‬‬ ‫=‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪2‬‬ ‫]∈‪k‬‬ ‫ﻤﻊ‬ ‫‪θ‬‬ ‫=‬ ‫‪+ 2kπ‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪sin θ‬‬ ‫‪−2‬‬ ‫=‬ ‫‪2‬‬ ‫ﺫﺍﺕ‬ ‫‪M0‬‬ ‫ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬ ‫‪−π‬‬ ‫ﻭﺯﺍﻭﻴﺘﻪ‬ ‫ﺇﺫﻥ ‪ S‬ﺘﺸﺎﺒﻪ ﻤﺒﺎﺸﺭ ﻨﺴﺒﺘﻪ ‪2‬‬ ‫‪4‬‬ ‫‪i‬‬ ‫‪b‬‬‫‪.‬‬ ‫‪Z0‬‬ ‫‪=1‬‬ ‫ﺇﺫﻥ‬ ‫‪Z0‬‬ ‫=‬ ‫‪1−1+ i‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪Z0‬‬ ‫=‬ ‫‪1−a‬‬ ‫ﺤﻴﺙ‬ ‫‪Z0‬‬ ‫ﺍﻟﻼﺤﻘﺔ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫‪ -1‬ﻋﺒﺎﺭﺓ ﺍﻟﺘﺸﺎﺒﻪ ‪ S‬ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻤﺭﻜﺏ ‪Z′ = aZ + b :‬‬‫ﻭﺒﻤﺎ ﺃﻥ ‪ S = ( A) = A′‬ﻓﺈﻥ ‪−4 + 2i = a (1 + 2i ) + b . . . (1) :‬‬‫ﻭﺒﻤﺎ ﺃﻥ ‪ S = ( B) = B′‬ﻓﺈﻥ ‪( )−2 + 2i 3 = a 3 + i + b . . . (2) :‬‬ ‫ﻤﻥ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﺍﻟﺴﺎﺒﻘﺘﻴﻥ ‪:‬‬‫‪( ) ( )(−4 + 2i ) − −2 + 2i 3 = a(1 + 2i ) − a 3 + i‬‬ ‫‪( ) ( )−2 + 2i 1 − 3 = a 1 − 3 + i‬‬ ‫‪( )−2 + 2i 1 −‬‬ ‫‪3 ×1−‬‬ ‫‪3−i‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫=‪a‬‬ ‫)‪3‬‬ ‫‪1− 3 +i 1− 3 −i‬‬ ‫‪( ) (2‬‬ ‫‪−2 + 2 3 + 2i + 2i 1 − 3 + 2 1 −‬‬‫‪( )a = 1 − 3 2 + 1‬‬‫‪a = −2 + 2 3 + 2i + 2i − 4i 3 + 6i + 2 − 2 3‬‬ ‫‪1−2 3 +3+1‬‬

‫‪( )a = 10i − 4i 3 = 2i 5 − 2 3 = 2i‬‬ ‫‪5−2 3‬‬ ‫‪5−2 3‬‬ ‫ﻭﻤﻨﻪ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (1‬ﻓﺈﻥ ‪:‬‬ ‫‪−4 + 2i = 2i (1 + 2i ) + b‬‬ ‫‪−4 + 2i = 2i − 4‬‬ ‫ﻭﻤﻨﻪ ‪ b = 0 :‬ﺇﺫﻥ ‪Z′ = 2i Z :‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫‪,‬‬ ‫‪k‬‬ ‫∈‬ ‫‪Z‬‬ ‫ﻭﻋﻤﺩﺓ ‪ 2i‬ﻫﻲ ‪:‬‬ ‫‪2i‬‬ ‫‪=2‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪O‬‬ ‫ﻭﻤﺭﻜﺯﻩ‬ ‫‪π‬‬ ‫ﺇﺫﻥ ﻨﺴﺒﺔ ﺍﻟﺘﺸﺎﺒﻪ ﻫﻲ ‪ 2‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫‪2‬‬ ‫‪ -2‬ﺍﻟﺘﻌﺒﻴﺭ ﻋﻥ ﺍﻟﺘﺸﺎﺒﻪ ﺒﺎﻟﺸﻜل ﺍﻷﺴﻲ ‪:‬‬ ‫‪Z′‬‬ ‫=‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪Z‬‬ ‫‪Z′ = 2i Z‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪ -3‬ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﺘﺤﻠﻴﻠﻴﺔ ﻟﻠﺘﺸﺎﺒﻪ ‪:‬‬ ‫ﺒﻭﻀﻊ ‪ Z = x + iy‬ﻭ ‪Z′ = x′ + iy′‬‬ ‫ﻨﺠﺩ )‪x′ + iy′ = 2i ( x + iy‬‬ ‫‪ x′ = − 2 y‬‬ ‫ﻭﻋﻠﻴﻪ ‪x′ + iy′ = 2i x − 2 y :‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪y′ = 2x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫‪ -1‬ﺘﻌﻴﻴﻥ ﺍﻟﺸﻜل ﺍﻟﻤﺭﻜﺏ ﻟﻜل ﻤﻥ ﺍﻟﺘﺸﺎﺒﻬﻴﻥ ‪:‬‬‫‪a‬‬ ‫=‬ ‫‪2‬‬ ‫‪cos‬‬ ‫‪π‬‬ ‫‪+i‬‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪S1 : Z′ = aZ :‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪ a = 1 + i 3 :‬ﺇﺫﻥ ‪( )Z′ = 1 + i 3 Z :‬‬

a= 1  cos 2π + i sin 2π  : ‫ﺤﻴﺙ‬ S2 : Z′ = aZ + b : ‫ﻟﺩﻴﻨﺎ‬ 2  3 3  a= −1 + i 3 a = 1  −1 + i 3 4 4 : ‫ﻭﻤﻨﻪ‬ 2  2  : ‫ﻭﻋﻠﻴﻪ‬  2  ( )Z0 b Z0 1 = 1−a : ‫ﺤﻴﺙ‬ ‫ﻭﻻﺤﻘﺘﻪ‬ a = 4 −1 + i 3 ‫ﺃﻱ ﺃﻥ‬ b = (1 − a)(1 − 2i) : ‫ﺃﻱ‬ b = 1 − 2i : ‫ﺇﺫﻥ‬ 1−a b =  1+ 1 − i 3  (1 − 2i ) : ‫ﺇﺫﻥ‬  4 4    ( )b  5 3 (1 − 2i ) 1 (1 − 2i) =  4 − i 4 = 4 5−i 3   ( )b 1 = 4 5 − 10i − i 3−2 3 ( )b = 1 5  4 − 2 3−i 10 + 3 ( ) ( )Z′ =1 1 5  4 −1 + i 3 Z + 4 − 2 3−i 10 + 3  : ‫ﻭﻤﻨﻪ‬ : S2oS1 ‫ ﺘﻌﻴﻴﻥ ﺍﻟﺘﺤﻭﻴل ﺍﻟﻤﺭﻜﺏ‬-2 M ( Z ) S1→ M1 ( Z1 ) S2→ M′( Z′) ( )Z1 = 1 + i 3( ) ( )Z′ =1 3 Z1 1 5 2 3  4 −1 + i + 4 − 3−i 10 + 

‫=‬‫‪( )( ) ( )Z′‬‬‫‪1‬‬ ‫‪1‬‬ ‫‪5 −‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪−1+ i‬‬ ‫‪3‬‬ ‫‪1+ i‬‬ ‫‪3‬‬ ‫‪Z‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪3−i‬‬ ‫‪10 +‬‬ ‫‪3‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫‪( )Z′‬‬ ‫=‬ ‫‪1‬‬ ‫(‬ ‫)‪−4‬‬ ‫‪1‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪Z‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪3−i‬‬ ‫‪10 +‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪( )Z′‬‬ ‫=‬ ‫‪Z‬‬ ‫‪1‬‬ ‫‪5‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪3−i‬‬ ‫‪10 +‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪ S2oS1‬ﺘﺤﺎﻙ ﻨﺴﺒﺘﻪ ‪ -1‬ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ ‪ ω‬ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ ‪Z0‬‬ ‫‪.‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪b‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪1−a‬‬ ‫‪145−2‬‬‫‪( ) ( )Z0 = 185−2‬‬ ‫‪‬‬ ‫‪3−i‬‬ ‫‪10+‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪3 −i 10+‬‬ ‫‪3‬‬ ‫ﺃﻱ ‪:‬‬ ‫= ‪Z0‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫‪ -1‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪2Z 2 − (1 + 5i ) Z + 2( i − 1) = 0 :‬‬ ‫)‪∆ = (1 + 5i)2 − 4(2i − 2)(2‬‬ ‫‪∆ = 1 + 10i − 25 − 16i + 16‬‬ ‫‪∆ = − 8 − 6i‬‬ ‫ﻨﺤﺴﺏ ﺠﺫﺭﻴﻪ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ‪:‬‬ ‫ﻨﻔﺭﺽ ‪ δ = α + iβ‬ﺠﺫﺭﺍ ﺘﺭﺒﻴﻌﻴﺎ ﻟـ ∆ ﺃﻱ ‪δ2 = ∆ :‬‬ ‫‪α2 − β2 = − 8‬‬ ‫ﻭﻤﻨﻪ ‪ ( α + iβ)2 = − 8 − 6i :‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2αβ = − 6‬‬ ‫ﻭﻜﺫﻟﻙ ﻟﺩﻴﻨﺎ ‪ δ2 = ∆ :‬ﻫﻲ ‪α2 + β2 = 10 :‬‬ ‫)‪α2 − β2 = − 8 . . . (1‬‬ ‫‪‬‬ ‫‪2α2‬‬ ‫=‬ ‫‪2‬‬ ‫‪:‬‬ ‫ﺒﺠﻤﻊ )‪ (1‬ﻭ )‪ (3‬ﻨﺠﺩ‬ ‫‪‬‬ ‫)‪αβ = − 3 . . . (2‬‬ ‫ﺇﺫﻥ‬ ‫‪‬‬ ‫‪α2‬‬ ‫‪+ β2‬‬ ‫=‬ ‫‪10‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫)‪(3‬‬ ‫‪‬‬

α = − 1 ‫ ﺃﻭ‬α = 1 : ‫ ﻭﻋﻠﻴﻪ‬α2 = 1 : ‫ﺇﺫﻥ‬ : (2) ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ‬β = 3 : ‫ ﻓﺈﻥ‬α = − 1 ‫ ﻭﻟﻤﺎ‬β = − 3 :‫ ﻓﺈﻥ‬α = 1 ‫ﻟﻤﺎ‬ : ‫ﺇﺫﻥ ∆ ﻟﻪ ﺠﺫﺭﻴﻥ ﺘﺭﺒﻴﻌﻴﻴﻥ ﻫﻤﺎ‬ δ2 = − 1 + 3i ‫ ﻭ‬δ1 = 1 − 3i : ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ‬Z ′′ = 1+ 5i + 1− 3i ‫ﻭ‬ Z′ = 1+ 5i −1+ 3i 4 4 Z ′′ = 1 + 1 i ‫ ﻭ‬Z′ = 2i 2 2 : ‫ ﻨﺒﻴﻥ ﺃﻥ‬-  2Z2 2008 − 1  Z1 2007 +  2Z3 1961 = 1  2  2  2   2  : ‫ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ‬Z3 ‫ ﻭ‬Z2 ‫ ﻭ‬Z1 ‫ﻨﻜﺘﺏ‬ Z1 = 2 ، arg ( Z1 ) = π 2 Z1 = 2 cos π + i sin π  2 2   1 Z1 2007 =  cos π + i sin π  2007 : ‫ﻭﻤﻨﻪ‬  2   2 2  = cos 2007π + i sin 2007π 2 2 = cos  2008π − π  + i sin  2008π − π  2 2   2 2 

= cos  1004π − π  + i sin  1004π − π  2   2  = cos  −π  + i sin  −π   2   2  = −i Z2 = 2 ، arg( Z2 ) = π 2 4 Z2 = 2 cos π + i sin π 2 4 4   2Z2 2008 =  cos π + i sin π 2008  2   4 4  2Z2  2008 = cos 2008π + i sin 2008π 2  4 4 = cos 502π + i sin 502π =1 Z3 = 2 ، arg( Z3 ) = π 2 2 Z3 = 2 cos π + i sin π  2 2 2  2Z3 1961 =  cos π +i sin π 1961 2   2 2  = cos 1961π + i sin 1961π 2 2

= cos  980π + π  + i sin  980π + π   2   2  = cos π + i sin π = i 2 2 2Z2 2008 − 1 Z1 2007 +  2 Z3 1961 = − i + 1+ i = 1 : ‫ﻭﻤﻨﻪ‬ 2  2  2   2  : S ‫ ﺘﻌﻴﻴﻥ ﺍﻟﺘﺸﺎﺒﻪ‬- Z′ = aZ : ‫ ﻫﻲ ﻤﻥ ﺍﻟﺸﻜل‬S ‫ﻋﺒﺎﺭﺓ‬ 2i = a  1 + 1 i  : ‫ﻓﺈﻥ‬ S ( B) = A : ‫ﻭﺒﻤﺎ ﺃﻥ‬  2 2  a = 4(1− i) : ‫ﺃﻱ‬ a = 2i : ‫ﺇﺫﻥ‬ (1+ i)(1− i) 1 + 1 i 2 2 Z′ = 2(1 + i ) Z : ‫ ﻭﺒﺎﻟﺘﺎﻟﻲ‬a = 2 + 2i : ‫ﺇﺫﻥ‬ : r ‫ ﺘﻌﻴﻴﻥ ﺍﻟﺩﻭﺭﺍﻥ‬- Z ′ = a Z : ‫ﻋﺒﺎﺭﺓ ﺍﻟﺩﻭﺭﺍﻥ ﻤﻥ ﺍﻟﺸﻜل‬ 2 i = a  1 + 1 i  : ‫ﻓﺈﻥ‬ r(B) = C : ‫ﻭﺒﻤﺎ ﺃﻥ‬ 2  2 2  2 2 1+ i a = : ‫ﺃﻱ‬ a = 1 2 : ‫ﻭﻤﻨﻪ‬ 2 (1 + i ) a= 2 (1 − i ) : ‫ﺃﻱ‬ a = 2(1− i) : ‫ﻭﻤﻨﻪ‬ 2 (1+ i)(1− i)

Z′ = 2 (1 − i ) Z : ‫ﻭﻤﻨﻪ‬ 2 : Sor ‫ ﺘﻌﻴﻴﻥ‬- M ( Z ) r→ M1 ( Z1 ) → M′( Z′) Z1 = 2 (1 − i ) Z ، Z′ = (2 + 2i) Z1 2Z′ = 2 2Z : ‫ﺇﺫﻥ‬ Z′ = 2(1+ i)(1− i) 2 Z : ‫ﻭﻤﻨﻪ‬ 2 . O ‫ ﻭﻤﺭﻜﺯﻩ‬2 2 ‫ ﻫﻭ ﺘﺤﺎﻙ ﻨﺴﺒﺘﻪ‬Sor ‫ﺇﺫﻥ‬ : SoS ‫ ﺘﻌﻴﻴﻥ‬- M ( Z ) S→ M1 ( Z1 ) S→ M′( Z′) Z1′ = ( 2 + 2i ) Z ، Z′ = ( 2 + 2i ) Z1Z′ = 4i Z : ‫ ﺇﺫﻥ‬Z′ = ( 2 + 2i )( 2 + 2i ) Z : ‫ﻭﻤﻨﻪ‬ . π ‫ ﻭﺯﺍﻭﻴﺘﻪ‬4 ‫ ﻭﻨﺴﺒﺘﻪ‬O ‫ﻫﻭ ﺘﺸﺎﺒﻪ ﻤﺭﻜﺯﻩ‬ SoS ‫ﻭﻋﻠﻴﻪ‬ 2 . 7‫ﺍﻟﺘﻤﺭﻴﻥ‬ : p( −1 + i ) ‫( ﺤﺴﺎﺏ‬1p(−1+ i) = (−1+ i)3 +(3−2i)(−1+ i)2 +(1−5i)(−1+ i) − 2i − 2= (2i + 2) + (−6i − 4) + 4 + 6i − 2i − 2 = 0 ( ). p Z ‫ ﺠﺫﺭ ﻟـ‬−1 + i : ‫ﺇﺫﻥ‬ ( ): p Z = 0 ‫( ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬2p( Z ) = ( Z + 1 − i )(aZ 2 + bZ + c)= aZ 3 + bZ2 + cZ + aZ2 + bZ + c − iaZ2 − ibZ − ic

‫‪p( Z ) = aZ 3 + (b + a − ia) Z 2 + (c + b − ib) Z − ic + c‬‬‫‪ a=1‬‬ ‫‪ a=1‬‬‫‪b = 2 − i‬‬ ‫‪b + a − ia = 3 − 2i‬‬‫‪c = − 2i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪c‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪−‬‬ ‫‪ib‬‬ ‫=‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪5i‬‬ ‫‪−ic + c = − 2i − 2‬‬ ‫ﺇﺫﻥ ‪( )p( Z ) = ( Z + 1 − i ) Z 2 + ( 2 − i ) Z − 2i :‬‬‫‪ p( Z ) = 0‬ﺘﻜﺎﻓﺊ ‪ Z + 1 − i = 0‬ﺃﻱ ‪Z 2 + ( 2 − i ) Z − 2i = 0‬‬ ‫‪ Z + 1 − i = 0‬ﻤﻌﻨﺎﻩ ‪Z = − 1 + i‬‬ ‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪( )Z 2 + 2 − i Z − 2i = 0 :‬‬ ‫) ‪ ∆ = ( 2 − i )2 − 4( −2i‬ﻭﻤﻨﻪ ‪∆ = 3 + 4i :‬‬ ‫ﻟﻨﺤﺴﺏ ﺠﺫﺭﻱ ∆ ‪ :‬ﻨﻔﺭﺽ ‪ δ = x + iy‬ﺠﺫﺭ ﺘﺭﺒﻴﻌﻲ ﻟﻠﻌﺩﺩ ‪δ‬‬ ‫)‪ x2 − y2 = 3 . . . (1‬‬ ‫‪‬‬‫‪2x2‬‬ ‫=‬ ‫‪8‬‬ ‫ﻨﺠﺩ‬ ‫)‪(2‬‬ ‫ﻭ‬ ‫)‪(1‬‬ ‫‪ ‬ﺒﺠﻤﻊ‬ ‫)‪xy = 2 . . . (2‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2‬‬ ‫=‬ ‫‪5‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫‪.‬‬ ‫)‪(3‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪ x2 = 4 :‬ﻭﻋﻠﻴﻪ ‪ x = 2 :‬ﺃﻭ ‪x = − 2‬‬ ‫ﻟﻤﺎ ‪ y = 1 : x = 2‬ﻭﻟﻤﺎ ‪y = − 1 : x = − 2‬‬ ‫ﻴﻭﺠﺩ ﺠﺫﺭﺍﻥ ﺘﺭﺒﻴﻌﻴﺎﻥ ﻫﻤﺎ ‪δ2 = − 2 − i ، δ1 = 2 + i :‬‬‫= ‪Z′′‬‬ ‫‪−2 + i − 2 − i‬‬ ‫‪،‬‬ ‫‪Z′‬‬ ‫=‬ ‫‪−2 + i + 2 + i‬‬ ‫ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪Z′′ = − 2 ، Z′ = i‬‬ ‫}‪S = {−2 , i , − 1 + i‬‬ ‫‪ (3‬ﻜﺘﺎﺒﺔ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﻤﺭﻜﺒﺔ ﻟﻠﺘﺸﺎﺒﻪ ‪Z′ = aZ + b : S‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪ S ( A) = B :‬ﻭﻤﻨﻪ ‪−2 = a ( −1 + i ) + b :‬‬

‫ﻭﻟﺩﻴﻨﺎ ‪ S (C ) = A :‬ﻭﻤﻨﻪ ‪−1 + i = a ( i ) + b :‬‬ ‫ﺇﺫﻥ ‪−2 − ( −1 + i ) = a ( −1 + i ) + b − ai − b :‬‬ ‫‪a‬‬ ‫=‬ ‫‪−1 − i‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ‪−1 − i = a ( −1) :‬‬ ‫‪−1‬‬ ‫ﺃﻱ ﺃﻥ ‪ a = 1 + i :‬ﻭﻤﻨﻪ ‪−1 + i = (1 + i ) i + b :‬‬ ‫ﺇﺫﻥ ‪ b = − 1 + i − i + 1 :‬ﻭﻋﻠﻴﻪ ‪b = 0 :‬‬ ‫ﺇﺫﻥ ‪Z′ = (1 + i ) Z :‬‬ ‫ﻋﺒﺎﺭﺓ ﺍﻟﺘﺸﺎﺒﻪ ﻤﻥ ﺍﻟﺸﻜل ‪ Z′ = aZ‬ﺤﻴﺙ ‪a = 2 :‬‬ ‫=‪θ‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫= ‪cosθ‬‬ ‫‪2‬‬ ‫ﻭﻋﻤﺩﺓ ‪ θ‬ﺤﻴﺙ ‪:‬‬ ‫‪4‬‬ ‫‪ ‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫= ‪sin θ‬‬ ‫‪2‬‬ ‫‪π‬‬ ‫ﺇﺫﻥ ﻤﺭﻜﺯ ﺍﻟﺘﺸﺎﺒﻪ ﻫﻭ ‪ O‬ﻭﻨﺴﺒﺘﻪ ‪2‬‬ ‫ﻭﺯﺍﻭﻴﺘﻪ ‪. 4‬‬ ‫‪ -‬ﻋﺒﺎﺭﺓ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ‪:‬‬ ‫= ‪Z′‬‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪Z‬‬ ‫‪4‬‬ ‫‪ -‬ﺤﺴﺎﺏ ‪: SoS‬‬ ‫)‪M ( Z ) S→ M1 ( Z1 ) S→ M′( Z′‬‬ ‫= ‪Z1‬‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪Z‬‬ ‫= ‪Z′‬‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪Z1‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪،‬‬‫‪O‬‬ ‫‪Z‬‬ ‫=‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫= ‪Z′‬‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫×‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫ﻫﻭ ﺘﺸﺎﺒﻪ ﻤﺭﻜﺯﻩ‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪π‬‬ ‫ﻭﻨﺴﺒﺘﻪ ‪ 2‬ﻭﺯﺍﻭﻴﺘﻪ ﻫﻲ ‪. 2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬

: Z0 ‫ ﺇﺜﺒﺎﺕ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻘﺒل ﺤل ﺘﺨﻴﻠﻲ ﺼﺭﻑ‬-1 : ‫ ﻓﻨﺠﺩ‬Z0 = iy ‫ﻨﻔﺭﺽ‬( i y)3 − (4 − 3i )( iy)2 + ( y − 2i ) iy − 2 + 11i = 0 −iy3 + 4 y2 − 3iy2 + 9iy + 2 y − 2 + 11i = 0 ( )4 y2 + 2 y − 2 + i − y3 − 3 y2 + 9 y + 11 = 0  4 y2 + 2 y − 2 = 0 . . . (1) − y3 − 3 y2 + 9 y + 11 = 0 . . . (2) : ‫ﻭﻤﻨﻪ‬y2 = − 1 : ‫ﺃﻱ‬ y1 = 1 : ‫∆ ﻭﻤﻨﻪ‬′ = 1 + 8 = y : 1 ‫ﻨﺤل‬ 2 . (2) ‫ ﻴﺤﻘﻕ‬y = − 1 : ‫( ﻨﺠﺩ‬2) ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ‬ Z0 = − i : ‫ﻭﻤﻨﻪ‬( )( )Z + i aZ 2 + bZ + c = 0 : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﺼﺒﺢ‬aZ 3 + bZ 2 + cZ + i aZ 2 + i bZ + ic = 0aZ 3 + (b + ia) Z 2 + (c + ib) Z + ic = 0 a=1  a=1b = − 4 + 2i b + ia = − 4 + 3i c = 11 + 2i : ‫ﻭﻋﻠﻴﻪ‬  : ‫ﺇﺫﻥ‬  c + ib = 9 − 2i  ic = − 2 + 11i( Z + i )  Z 2 + ( −4 + 2i ) Z + 11 + 2i  : ‫ﻭﻤﻨﻪ‬Z 2 + ( −4 + 2i ) Z + 11 + 2i = 0 ‫ ﺃﻭ‬Z = − i ‫ﺇﻤﺎ‬ Z 2 + 2( −2 + i ) Z + 11 + 2i = 0 ‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬∆′ = − 8 − 6i : ‫∆ ﻭﻤﻨﻪ‬′ = ( −2 + i )2 − 11 − 2i

‫ﻨﺒﺤﺙ ﻋﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﺘﺭﺒﻴﻌﻴﻴﻥ ﻟﻠﻌﺩﺩ ‪. ∆′‬‬ ‫ﻨﻔﺭﺽ ‪ δ = x + iy‬ﺠﺫﺭ ﺘﺭﺒﻴﻌﻲ ﻟﻠﻌﺩﺩ ‪ ∆′‬ﺃﻱ ‪δ2 = ∆′‬‬ ‫‪x2 − y2 = − 8‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪xy = − 3‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪‬‬ ‫‪x2‬‬ ‫‪+‬‬ ‫‪y2‬‬ ‫=‬ ‫‪10‬‬ ‫‪‬‬ ‫ﻨﺤل ﺍﻟﺠﻤﻠﺔ ﻓﻨﺠﺩ ‪δ2 = − 1 + 3i ، δ1 = 1 − 3i :‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬ ‫‪Z2 = 2 − i + 1 − 3i ، Z1 = 2 − i − 1 + 3i‬‬ ‫ﺇﺫﻥ ‪Z0 = − i ، Z2 = 3 − 4i ، Z1 = 1 + 2i :‬‬ ‫‪ -2‬ﺇﺜﺒﺎﺕ ﻭﺠﻭﺩ ﺍﻟﺘﺸﺎﺒﻪ ‪Z′ = aZ + b :‬‬ ‫ﻟﺩﻴﻨﺎ ‪ S ( A) = B :‬ﻭﻤﻨﻪ ‪1 + 2i = a ( −i ) + b :‬‬ ‫‪ S (C ) = D‬ﻭﻤﻨﻪ ‪−5 − 4i = a ( 3 − 4i ) + b :‬‬ ‫ﺒﺎﻟﻁﺭﺡ ﻨﺠﺩ ‪6 + 6i = a ( −3 + 3i ) :‬‬‫‪a‬‬ ‫=‬ ‫×‪2‬‬ ‫‪−2i‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪a‬‬ ‫=‬ ‫)‪6(1+ i‬‬ ‫×‬ ‫‪−1‬‬ ‫‪−‬‬ ‫‪i‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫)‪3(−1 + i‬‬ ‫‪−1‬‬ ‫‪−‬‬ ‫‪i‬‬ ‫ﻭﻋﻠﻴﻪ ‪ a = − 2i :‬ﺇﺫﻥ ‪1 + 2i = − 2i ( −i ) + b :‬‬ ‫ﻭﻤﻨﻪ ‪ 1 + 2i = − 2 + b :‬ﻭﻋﻠﻴﻪ ‪b = 3 + 2i :‬‬ ‫ﺇﺫﻥ ‪Z′ = − 2iZ + 3 + 2i :‬‬ ‫ﻨﺴﺒﺔ ﺍﻟﺘﺸﺎﺒﻪ ﻫﻲ ‪ −2i :‬ﺃﻱ ‪. 2‬‬ ‫‪( )−π‬‬ ‫‪−2i‬‬ ‫ﺯﺍﻭﻴﺔ ﺍﻟﺘﺸﺎﺒﻪ ﻫﻲ ﻋﻤﺩﺓ‬ ‫ﺃﻱ ‪. 2‬‬ ‫‪b‬‬ ‫‪Z0‬‬ ‫=‬ ‫‪1−a‬‬ ‫ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬ ‫‪M0‬‬ ‫ﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬

Z0 = 7 − 4i : ‫ﻭﻤﻨﻪ‬ Z0 = 3 + 2i × 1 − 2i : ‫ﺃﻱ‬ 5 1 + 2i 1 − 2i Z′ = 2e − i π Z + 3 + 2i : ‫ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻠﺘﺸﺎﺒﻪ‬-3 2 : SoS ‫ ﺤﺴﺎﺏ‬-4 M ( Z ) S→ M1 ( Z1 ) S→ M′( Z′) Z1 = 2e − i π Z + 3+ 2i ; Z′ = 2e − i π Z1 + 3 + 2i 2 2 Z′ = 2e − i π  2e − i π Z + 3 + 2i  + 3 + 2i 2  2   4e Z− i π + π  e− i π ( )Z′ 2 2  2 2 2i = + 3 + 2i + 3+ Z′ = 4e−iπZ + (6 + 4i )( −i ) + 3 + 2i Z′ = 4e−iπ Z − 6i + 4 + 3 + 2i Z′ = 4e−iπ Z + 7 − 4i: ‫ ﺤﻴﺙ‬Z0 ‫ ﺫﺍﺕ ﺍﻟﻼﺤﻘﺔ‬M0 ‫ ﻭﻤﺭﻜﺯﻩ ﺍﻟﻨﻘﻁﺔ‬−π ‫ ﻭﺯﺍﻭﻴﺘﻪ‬4 ‫ ﻫﻭ ﺘﺸﺎﺒﻪ ﻨﺴﺒﺘﻪ‬SoS . Z0 = b = 7 − 4i 1−a 5 . 9‫ﺍﻟﺘﻤﺭﻴﻥ‬ Z′ = 3 Z + b : H ‫ ﻋﺒﺎﺭﺓ‬-1 2 ( )H A = A : ‫ ﻓﺈﻥ‬A ‫ﻭﺒﻤﺎ ﺃﻥ ﺍﻟﻤﺭﻜﺯ ﻫﻭ‬: ‫ﺃﻱ ﺃﻥ‬ b = 1 − i − 3 + 3 i : ‫ﻭﻋﻠﻴﻪ‬ 1− i = 3 (1 − i ) + b : ‫ﻭﻤﻨﻪ‬ 2 2 2 3 1 1 1 1 Z′ = 2 Z − 2 + 2 i : ‫ﺇﺫﻥ‬ b = − 2 + 2 i

Z′ = aZ + b : r ‫ ﻋﺒﺎﺭﺓ‬-a = 1 − 3 i : ‫ﺃﻱ ﺃﻥ‬ a = cos −π  + i sin −π  : ‫ﺤﻴﺙ‬ 2 2 3  3 b = 3i − 3ia : ‫ ﺃﻱ ﺃﻥ‬3i = 3ia + b : ‫ ﻭﻤﻨﻪ‬r ( B) = B : ‫ﻭﻟﺩﻴﻨﺎ‬ b = 3i  1 − 1 + 3 i  : ‫ﺃﻱ‬ b = 3i (1 − a ) : ‫ﻭﻋﻠﻴﻪ‬  2 2    b = 3 i − 33 : ‫ﻭﻤﻨﻪ‬ b = 3i  1 + 3 i  : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ 2 2  2 2    Z′ =  1 − 3 i  Z + 3 i − 3 3 : ‫ﻭﻋﻠﻴﻪ‬  2 2  2 2   : roH ‫ ﺤﺴﺎﺏ‬-2 M ( Z ) H→ M1 ( Z1 ) r→ M′( Z′)Z1 = 3 Z − 1 1 i ، Z′ =  1 − 3 i  Z1 + 3 i − 3 3 2 2+ 2  2 2  2 2  Z′ =  1 − 3 i  3 Z − 1 + 1 i  + 3 i − 3 3 : ‫ﻭﻤﻨﻪ‬  2 2   2 2 2  2 2  Z′ = 3 1 − 3 i  Z +  1 − 3 i  − 1 + 1 i  + 3 i − 33 2 2 2   2 2  2 2  2 2  ( )Z′=3 1− 3 Z − 1 + 1 i + 3 + 3 + 3 i − 3 3 4 4 4 4 4 2 2 ( )Z′ = 3 1− 3 Z − 1 − 3 + 7 i 4 4 4

‫ﻭﻤﻨﻪ ‪ roH‬ﻫﻭ ﺘﺸﺎﺒﻪ ‪( ).‬‬ ‫‪3‬‬ ‫‪1−‬‬ ‫‪3‬‬ ‫=‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪2‬‬‫= ‪(Z′ = x′ + iy′‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬ ‫‪ -1‬ﻜﺘﺎﺒﺔ ‪ Z ′‬ﺒﺩﻻﻟﺔ ‪: Z‬‬ ‫‪) ( )3x − y + i x + 3 y‬‬ ‫‪= x 3 − y + ix + i 3 y‬‬ ‫‪Z′ = 3 ( x + iy) + i2 y + ix‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪Z′ = 3Z + i ( x + iy‬‬ ‫‪Z′ = 3Z + iZ‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪( )Z′ = 3 + i Z‬‬ ‫ﻭﻫﻭ ﻤﻥ ﺍﻟﺸﻜل ‪Z′ = aZ :‬‬ ‫‪.‬‬ ‫‪π‬‬ ‫ﻫﻲ‬ ‫‪a‬‬ ‫ﻋﻤﺩﺓ‬ ‫‪،‬‬ ‫‪a =2‬‬ ‫‪6‬‬ ‫‪.‬‬ ‫‪O‬‬ ‫ﻭﻤﺭﻜﺯﻩ‬ ‫‪π‬‬ ‫ﻭﻤﻨﻪ ﻨﺴﺒﺔ ﺍﻟﺘﺸﺎﺒﻪ ﻫﻲ ‪ 2‬ﻭﺯﺍﻭﻴﺘﻪ‬ ‫‪6‬‬ ‫‪Z′‬‬ ‫=‬ ‫‪2e‬‬ ‫‪i‬‬ ‫‪π‬‬ ‫‪Z‬‬ ‫‪6‬‬ ‫ﺍﻟﻌﺒﺎﺭﺓ ﺍﻷﺴﻴﺔ ﻟﻠﺘﺸﺎﺒﻪ ﻫﻲ ‪:‬‬ ‫‪Z0 = 1 (2‬‬‫= ‪(Z1‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ M1 = f M0 :‬ﻓﺈﻥ ‪) ( )3 + i Z0 :‬‬ ‫ﻭﻤﻨﻪ ‪Z1 = 3 + i :‬‬‫ﻭﺒﻤﺎ ﺃﻥ ‪ M2 = f M1 :‬ﻓﺈﻥ ‪( ) ( )Z2 = 3 + i Z1 :‬‬ ‫‪( )2‬‬ ‫ﻭﻤﻨﻪ ‪Z2 = 3 + i :‬‬

‫ﻭﺒﻤﺎ ﺃﻥ ‪ M3 = f M2 :‬ﻓﺈﻥ ‪( ) ( )Z3 = 3 + i Z2 :‬‬ ‫‪( )3‬‬‫ﻭﻤﻨﻪ ‪Z3 = 3 + i :‬‬ ‫‪( )n‬‬‫ﻭﻨﻼﺤﻅ ﺃﻥ ‪Zn = 3 + i :‬‬‫‪( ) ( )n‬‬‫ﻟﻨﺒﺭﻫﻥ ﺒﺎﻟﺘﺭﺍﺠﻊ ﻋﻠﻰ ﺫﻟﻙ ‪p n : Zn = 3 + 1 :‬‬ ‫‪( )0‬‬‫* ﻤﻥ ﺃﺠل ‪Z0 = 3 + 1 = 1 : n = 0 :‬‬ ‫ﻭﻤﻨﻪ ‪ p 0‬ﺼﺤﻴﺤﺔ ‪( ).‬‬‫* ﻨﻔﺭﺽ ﺼﺤﺔ ‪ p n‬ﻭﻨﺒﺭﻫﻥ ﺼﺤﺔ ‪( ) ( ). p n + 1‬‬ ‫‪n‬‬ ‫‪3 +1‬‬‫= ‪( )p(n) : Zn‬‬‫‪( )p( n + 1) : Zn+1 = 3 + 1 n+1‬‬ ‫ﻟﺩﻴﻨﺎ ‪( )Mn+1 = f Mn :‬‬‫‪( )Zn+1 = 3 + i Zn‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﻭﻤﻨﻪ ‪3 + i n :‬‬‫‪( )( )Zn+1 = 3 + i‬‬ ‫‪3 + i n+1‬‬‫= ‪( )Zn+1‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﻭﻤﻨﻪ ‪ p n + 1‬ﺼﺤﻴﺤﺔ ‪( ).‬‬ ‫‪n‬‬‫ﺇﺫﻥ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ ‪ n‬ﻓﺈﻥ ‪3 + i :‬‬‫= ‪( ). Zn‬‬