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X2 T01 08 factorising complex expressions (2011)

N
Nigel Simmons

The document discusses factorizing complex expressions. It states that if a polynomial's coefficients are real, its roots will appear in complex conjugate pairs. It notes that every polynomial of degree n can be factorized as a mixture of quadratic and linear factors over real numbers or as n linear factors over complex numbers. Odd degree polynomials must have a real root. Examples of factorizing polynomials over both real and complex numbers are provided.

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Factorising Complex Expressions
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in complex conjugate pairs.
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in complex conjugate pairs. Every polynomial of degree n can be;
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in complex conjugate pairs. Every polynomial of degree n can be; • factorised as a mixture of quadratic and linear factors over the real field
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root e.g . i  x 2  2 x  2
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root e.g . i  x 2  2 x  2   x  12  1
Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root e.g . i  x 2  2 x  2   x  12  1   x  1  i  x  1  i 
ii  z 4  z 2  12  0
ii  z 4  z 2  12  0 z 2  3z 2  4   0
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  Factorise 2 x 3  3x 2  8 x  5
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  Factorise 2 x 3  3x 2  8 x  5 as it is a cubic it must have a real factor
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0  2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0  2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5   2 x  1  x  1  4 2 
ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0  2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5   2 x  1  x  1  4 2   2 x  1 x  1  2i  x  1  2i 
iv  z 5  z 4  z 3  z 2  z  1  0
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6 
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1   2 2 z  cis , cis  , cis , cis  , cis 3 3 3 3
iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1   2 2 z  cis , cis  , cis , cis  , cis 3 3 3 3 1 3 1 3 1 3 1 3 z  i,  i,   i,   i, 1 2 2 2 2 2 2 2 2
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9  
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9 
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0 sum of roots 
v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0 sum of roots     2   3   4   5   6   7   8  1
 2 4 1 c) Hence show that cos cos cos  9 9 9 8
 2 4 1 c) Hence show that cos cos cos  9 9 9 8  A B   A B  sin A  sin B  2sin   cos    2   2 
 2 4 1 c) Hence show that cos cos cos  9 9 9 8  A B   A B  (2 sine half sum sin A  sin B  2sin   cos    2   2  cos half diff)
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 
 2 4 1 c) Hence show that cos cos cos  9 9 9 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)
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs 2 4 6 8 2cos  2cos  2cos  2cos  1 9 9 9 9
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs 2 4 6 8 2cos  2cos  2cos  2cos  1 9 9 9 9 2 4 6 8 1 cos  cos  cos  cos  9 9 9 9 2
 2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs 2 4 6 8 2cos  2cos  2cos  2cos  1 9 9 9 9 2 4 6 8 1 cos  cos  cos  cos  9 9 9 9 2 3  7  1 2cos cos  2cos cos   9 9 9 9 2
3  7  1 2cos cos  2cos cos   9 9 9 9 2
3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4
3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4
3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8
3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8 5 4 But cos   cos 9 9
3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8 5 4 But cos   cos 9 9  2 4 1  cos cos cos  9 9 9 8
3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8 5 4 But cos   cos 9 9  2 4 1  cos cos cos  9 9 9 8  2 4 1 cos cos cos  9 9 9 8
OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9 
OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9   2 2  2 4    z  1  z  2cos z  1 z  2cos z  1  9  9   8   z 2  z  1  z 2  2cos  9 z  1 
OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9   2 2  2 4    z  1  z  2cos z  1 z  2cos z  1  9  9   8   z 2  z  1  z 2  2cos  9 z  1  Let z  i
OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9   2 2  2 4    z  1  z  2cos z  1 z  2cos z  1  9  9   8   z 2  z  1  z 2  2cos  9 z  1  Let z  i  2  4   8  i 9  1   i  1  2cos i  2cos i   i   2cos i  9  9   9 
 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9 
 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9 
 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9
 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9 2 4   1  8cos cos   cos  9 9  9
 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9 2 4   1  8cos cos   cos  9 9  9  2 4 1 cos cos cos  9 9 9 8
 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9 2 4   1  8cos cos   cos  9 9  9  2 4 1 cos cos cos  9 9 9 8 Exercise 4J; 1 to 4, 7ac

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X2 T01 08 factorising complex expressions (2011)

  • 1. Factorising Complex Expressions
  • 2. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in complex conjugate pairs.
  • 3. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in complex conjugate pairs. Every polynomial of degree n can be;
  • 4. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in complex conjugate pairs. Every polynomial of degree n can be; • factorised as a mixture of quadratic and linear factors over the real field
  • 5. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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
  • 6. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root
  • 7. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root e.g . i  x 2  2 x  2
  • 8. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root e.g . i  x 2  2 x  2   x  12  1
  • 9. Factorising Complex Expressions If a polynomial’s coefficients are all real then the roots will appear in 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 root e.g . i  x 2  2 x  2   x  12  1   x  1  i  x  1  i 
  • 10. ii  z 4  z 2  12  0
  • 11. ii  z 4  z 2  12  0 z 2  3z 2  4   0
  • 12. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2
  • 13. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers
  • 14. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0
  • 15. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  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  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i
  • 17. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  Factorise 2 x 3  3x 2  8 x  5
  • 18. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0
  • 20. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0  2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5
  • 21. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0  2 x 3  3 x 2  8 x  5  2 x  1x 2  2 x  5   2 x  1  x  1  4 2 
  • 22. ii  z 4  z 2  12  0 z  3z  4  0 2 2 z  3 z  3z  4  0 2 factorised over Real numbers z  3 z  3 z  2i  z  2i   0 factorised over Complex numbers z   3 or z  2i iii  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    5 1 1 1 1          2  2  2  2 1 3    45 4 4 0  2 x 3  3 x 2  8 x  5  2 x  1x 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  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
  • 25. iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And 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  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And 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  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1
  • 28. iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1
  • 29. iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1   2 2 z  cis , cis  , cis , cis  , cis 3 3 3 3
  • 30. iv  z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1   2 2 z  cis , cis  , cis , cis  , cis 3 3 3 3 1 3 1 3 1 3 1 3 z  i,  i,   i,   i, 1 2 2 2 2 2 2 2 2
  • 31. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer
  • 32. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1
  • 33. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9  
  • 34. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9 
  • 35. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k
  • 36. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1
  • 37. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0
  • 38. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0
  • 39. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0 sum of roots 
  • 40. v 1996 HSC 2 2 Let   cos  i sin 9 9 a) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8  9   2 k z  cis     9  z  k b) Prove that    2   3   4   5   6   7   8  1 z9 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 1 c) Hence show that cos cos cos  9 9 9 8
  • 42.  2 4 1 c) Hence show that cos cos cos  9 9 9 8  A B   A B  sin A  sin B  2sin   cos    2   2 
  • 43.  2 4 1 c) Hence show that cos cos cos  9 9 9 8  A B   A B  (2 sine half sum sin A  sin B  2sin   cos    2   2  cos half diff)
  • 44.  2 4 1 c) Hence show that cos cos cos  9 9 9 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 1 c) Hence show that cos cos cos  9 9 9 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 1 c) Hence show that cos cos cos  9 9 9 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 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)
  • 48.  2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1
  • 49.  2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos 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 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs 2 4 6 8 2cos  2cos  2cos  2cos  1 9 9 9 9
  • 51.  2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs 2 4 6 8 2cos  2cos  2cos  2cos  1 9 9 9 9 2 4 6 8 1 cos  cos  cos  cos  9 9 9 9 2
  • 52.  2 4 1 c) Hence show that cos cos cos  9 9 9 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 sum cos A  cos B  2sin   sin    2   2  sine half diff)    2   3   4   5   6   7   8  1 roots appear in conjugate pairs 2 4 6 8 2cos  2cos  2cos  2cos  1 9 9 9 9 2 4 6 8 1 cos  cos  cos  cos  9 9 9 9 2 3  7  1 2cos cos  2cos cos   9 9 9 9 2
  • 53. 3  7  1 2cos cos  2cos cos   9 9 9 9 2
  • 54. 3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4
  • 55. 3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4
  • 56. 3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8
  • 57. 3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8 5 4 But cos   cos 9 9
  • 58. 3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8 5 4 But cos   cos 9 9  2 4 1  cos cos cos  9 9 9 8
  • 59. 3  7  1 2cos cos  2cos cos   9 9 9 9 2  3 7  1 cos  cos  cos  9 9 9  4  5 2  1 cos  2cos cos  9 9 9  4  2 5 1 cos cos cos  9 9 9 8 5 4 But cos   cos 9 9  2 4 1  cos cos cos  9 9 9 8  2 4 1 cos cos cos  9 9 9 8
  • 60. OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
  • 61. OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9 
  • 62. OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9   2 2  2 4    z  1  z  2cos z  1 z  2cos z  1  9  9   8   z 2  z  1  z 2  2cos  9 z  1 
  • 63. OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9   2 2  2 4    z  1  z  2cos z  1 z  2cos z  1  9  9   8   z 2  z  1  z 2  2cos  9 z  1  Let z  i
  • 64. OR z9 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5   2  4    z  1  z 2  2cos z  1 z 2  2cos z  1  9  9   2 6  2 8   z  2cos z  1 z  2cos z  1  9  9   2 2  2 4    z  1  z  2cos z  1 z  2cos z  1  9  9   8   z 2  z  1  z 2  2cos  9 z  1  Let z  i  2  4   8  i 9  1   i  1  2cos i  2cos i   i   2cos i  9  9   9 
  • 65. 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9 
  • 66. 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9 
  • 67. 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9
  • 68. 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9 2 4   1  8cos cos   cos  9 9  9
  • 69. 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9 2 4   1  8cos cos   cos  9 9  9  2 4 1 cos cos cos  9 9 9 8
  • 70. 2  4   8  i  1   i  1  2cos 9 i  2cos i   i   2cos i  9  9   9   2  4  8  i  1  i  i  1  2cos 4  2cos  2cos   9  9  9  2 4 8 1  8cos cos cos 9 9 9 2 4   1  8cos cos   cos  9 9  9  2 4 1 cos cos cos  9 9 9 8 Exercise 4J; 1 to 4, 7ac