# x2 t01 08 factorising complex expressions (2011)

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1. Factorising ComplexExpressions 2. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs. 3. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs.Every polynomial of degree n can be; 4. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs.Every polynomial of degree n can be; factorised as a mixture of quadratic and linear factors over thereal field 5. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs.Every polynomial of degree n can be; factorised as a mixture of quadratic and linear factors over thereal field factorised to n linear factors over the complex field 6. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs.Every polynomial of degree n can be; factorised as a mixture of quadratic and linear factors over thereal field factorised to n linear factors over the complex fieldNOTE: odd ordered polynomials must have a real root 7. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs. Every polynomial of degree n can be; factorised as a mixture of quadratic and linear factors over the real field factorised to n linear factors over the complex field NOTE: odd ordered polynomials must have a real roote.g . i x 2 2 x 2 8. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs. Every polynomial of degree n can be; factorised as a mixture of quadratic and linear factors over the real field factorised to n linear factors over the complex field NOTE: odd ordered polynomials must have a real roote.g . i x 2 2 x 2 x 12 1 9. Factorising Complex ExpressionsIf a polynomials coefficients are all real then the roots will appearin complex conjugate pairs. Every polynomial of degree n can be; factorised as a mixture of quadratic and linear factors over the real field factorised to n linear factors over the complex field NOTE: odd ordered polynomials must have a real roote.g . i x 2 2 x 2 x 12 1 x 1 i x 1 i 10. ii z 4 z 2 12 0 11. ii z 4 z 2 12 0z 2 3z 2 4 0 12. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 13. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers 14. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 15. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbers 16. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbersz 3 or z 2i 17. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbers z 3 or z 2iiii Factorise 2 x 3 3x 2 8 x 5 18. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbersz 3 or z 2iiii Factorise 2 x 3 3x 2 8 x 5 as it is a cubic it must have a real factor 19. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbersz 3 or z 2iiii Factorise 2 x 3 3x 2 8 x 5 as it is a cubic it must have a real factor 3 2 P 2 3 8 511 11 2 2 2 2 1 3 45 4 40 20. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbersz 3 or z 2iiii Factorise 2 x 3 3x 2 8 x 5 as it is a cubic it must have a real factor 3 2 P 2 3 8 51 1 1 1 2 2 2 2 1 3 45 4 40 2 x 3 3 x 2 8 x 5 2 x 1x 2 2 x 5 21. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbersz 3 or z 2iiii Factorise 2 x 3 3x 2 8 x 5 as it is a cubic it must have a real factor 3 2 P 2 3 8 51 1 1 1 2 2 2 2 1 3 45 4 40 2 x 3 3 x 2 8 x 5 2 x 1x 2 2 x 5 2 x 1 x 1 4 2 22. ii z 4 z 2 12 0z 3z 4 022z 3 z 3z 4 0 2 factorised over Real numbers z 3 z 3 z 2i z 2i 0 factorised over Complex numbersz 3 or z 2iiii Factorise 2 x 3 3x 2 8 x 5 as it is a cubic it must have a real factor 3 2 P 2 3 8 51 1 1 1 2 2 2 2 1 3 45 4 40 2 x 3 3 x 2 8 x 5 2 x 1x 2 2 x 5 2 x 1 x 1 4 2 2 x 1 x 1 2i x 1 2i 23. iv z 5 z 4 z 3 z 2 z 1 0 24. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1 25. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1And z 6 1 0 has solutions; 2 k z cis k 0, 1, 2,3 6 26. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1And z 6 1 0 has solutions; 2 k z cis k 0, 1, 2,3 6 z5 z4 z3 z2 1 0 27. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1And z 6 1 0 has solutions; 2 k z cis k 0, 1, 2,3 6 z5 z4 z3 z2 1 0 z 1z 5 z 4 z 3 z 2 10 z 1 28. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1And z 6 1 0 has solutions; 2 k z cis k 0, 1, 2,3 6 z5 z4 z3 z2 1 0 z 1z 5 z 4 z 3 z 2 10 z 1 z 6 1 0, z 1 29. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1And z 6 1 0 has solutions; 2 k z cis k 0, 1, 2,3 6 z5 z4 z3 z2 1 0 z 1z 5 z 4 z 3 z 2 10 z 1z 6 1 0, z 1 22z cis , cis , cis, cis , cis 3 33 3 30. iv z 5 z 4 z 3 z 2 z 1 0Now z 6 1 z 1z 5 z 4 z 3 z 2 z 1And z 6 1 0 has solutions; 2 k z cis k 0, 1, 2,3 6 z5 z4 z3 z2 1 0 z 1z 5 z 4 z 3 z 2 10 z 1z 6 1 0, z 1 22z cis , cis , cis, cis , cis 3 33 313 1 3131 3z i, i, i, i, 12 2 2 2 2 2 2 2 31. v 1996 HSC22Let cos i sin 9 9a) Show that k is a solution of z 9 1 0, where k is an integer 32. v 1996 HSC22Let cos i sin 9 9a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 33. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 34. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 35. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 z k 36. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 z kb) Prove that 2 3 4 5 6 7 8 1 37. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 z kb) Prove that 2 3 4 5 6 7 8 1z9 1 0 38. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 z kb) Prove that 2 3 4 5 6 7 8 1z9 1 0 1 2 3 4 5 6 7 8 0 39. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 z kb) Prove that 2 3 4 5 6 7 8 1z9 1 0 1 2 3 4 5 6 7 8 0 sum of roots 40. v 1996 HSC 22Let cos i sin 99a) Show that k is a solution of z 9 1 0, where k is an integer z9 1 2k z cis k 0,1,2,3,4,5,6,7,8 9 2kz cis 9 z kb) Prove that 2 3 4 5 6 7 8 1z9 1 0 1 2 3 4 5 6 7 8 0 sum of roots 2 3 4 5 6 7 8 1 41. 2 4 1c) Hence show that cos cos cos 999 8 42. 2 4 1c) Hence show that cos cos cos 999 8 A B A B sin A sin B 2sin cos 2 2 43. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) 44. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) A B cos A B cos A cos B 2 cos 2 2 45. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) A B cos A B cos A cos B 2 cos(2 cos half sum 2 2 cos half diff) 46. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) A B cos A B cos A cos B 2 cos(2 cos half sum 2 2 cos half diff) A B A B cos A cos B 2sin sin 2 2 47. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) A B cos A B cos A cos B 2 cos(2 cos half sum 2 2 cos half diff) A B A B (minus 2 sine half sumcos A cos B 2sin sin 2 2 sine half diff) 48. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) A B cos A B cos A cos B 2 cos(2 cos half sum 2 2 cos half diff) A B A B (minus 2 sine half sumcos A cos B 2sin sin 2 2 sine half diff) 2 3 4 5 6 7 8 1 49. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum sin A sin B 2sin cos 2 2 cos half diff) A B cos A B cos A cos B 2 cos(2 cos half sum 2 2 cos half diff) A B A B (minus 2 sine half sumcos A cos B 2sin sin 2 2 sine half diff) 2 3 4 5 6 7 8 1 roots appear in conjugate pairs 50. 2 4 1c) Hence show that cos cos cos 999 8 A B A B (2 sine half sum

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