X2 t01 09 de moivres theorem
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X2 t01 09 de moivres theorem X2 t01 09 de moivres theorem Presentation Transcript

  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n  r n cos n  i sin n 
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5  r n cos n  i sin n 
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5  r n cos n  i sin n  z  12   1 2  2   1 arg z  tan    1  1   4
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5 z  12   1 2        2cis    r n cos n  i sin n   4  5  2   1 arg z  tan    1  1   4
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5  z  12   1 2        2cis    r n cos n  i sin n   4  5  2   1 arg z  tan   5  1    5  2  cis    4   4 1
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5 z  12   1 2        2cis    r n cos n  i sin n   4  5  2   1 arg z  tan   5  1    5    2  cis    4   4 3   4 2cis    4  1
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5 z  12   1 2        2cis    r n cos n  i sin n   4  5  2   1 arg z  tan   5  1    5    2  cis    4   4 3   4 2cis    4  1 1  i  5  cos 3  i sin 3   4 2  4 4  
  • De Moivre’s Theorem  cos  i sin    cos n  i sin n n for all integers n this extends to; r cos  i sin   n e.g . 1  i  5 z  12   1 2        2cis    r n cos n  i sin n   4  5  2   1 arg z  tan   5  1    5    2  cis    4   4 3   4 2cis    4  1 1  i  5  cos 3  i sin 3   4 2  4 4   1 1   4 2   i  2 2    4  4i
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis   n   n k  0,1,, n  1
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis   n   n e.g .i  z 2  4i k  0,1,, n  1
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n e.g .i  z 2  4i  2 z  4cis 2 k  0,1,, n  1
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n e.g .i  z 2  4i  2 z  4cis 2  2k     2 z  2cis  2      k  0,1 k  0,1,, n  1
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n e.g .i  z 2  4i  2 z  4cis 2  2k     2 z  2cis  2      5  z  2cis ,2cis 4 4 k  0,1 k  0,1,, n  1
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n e.g .i  z 2  4i  2 z  4cis 2  2k     2  k  0,1 z  2cis  2      5  z  2cis ,2cis 4 4  1  1 i ,2  1  1 i  z  2    2   2 2   2 k  0,1,, n  1 z  2  2i, 2  2i
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n e.g .i  z 2  4i  OR 2 z  4cis 2  2k     2  k  0,1 z  2cis  2      5  z  2cis ,2cis 4 4  1  1 i ,2  1  1 i  z  2    2   2 2   2 k  0,1,, n  1 y x z  2  2i, 2  2i
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n e.g .i  z 2  4i  OR 2 z  4cis 2  2k     2  k  0,1 z  2cis  2      5  z  2cis ,2cis 4 4  1  1 i ,2  1  1 i  z  2    2   2 2   2 k  0,1,, n  1 y 2cis  x z  2  2i, 2  2i 4
  • Finding Roots If z n  x  iy z n  rcis  2k    z  rcis    n  n k  0,1,, n  1 e.g .i  z 2  4i  OR 2 y z  4cis  2 2cis 4  2k     2  k  0,1 z  2cis  2  x     3 2cis  5  z  2cis ,2cis 4 4 4  1  1 i ,2  1  1 i  z  2 z  2  2i, 2  2i    2   2 2   2
  •  ii  x 4  16  0
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4   k  0,1, 2,3
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4    k  0,1, 2,3 3 x  2cis 0, 2cis , 2cis , 2cis 2 2
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4   k  0,1, 2,3  3 x  2cis 0, 2cis , 2cis , 2cis 2 2 x  2, 2i, 2, 2i
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4   k  0,1, 2,3  3 x  2cis 0, 2cis , 2cis , 2cis 2 2 x  2, 2i, 2, 2i OR y x
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4   k  0,1, 2,3  3 x  2cis 0, 2cis , 2cis , 2cis 2 2 x  2, 2i, 2, 2i OR y 2cis 0 x
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4   k  0,1, 2,3  3 x  2cis 0, 2cis , 2cis , 2cis 2 2 x  2, 2i, 2, 2i OR y 2cis  2 2cis 2cis 0 x 2cis   2
  •  ii  x 4  16  0 x 4  16 x 4  16cis 0  2 k  x  2cis   4   k  0,1, 2,3  3 x  2cis 0, 2cis , 2cis , 2cis 2 2 x  2, 2i, 2, 2i OR y 2cis Patel: Exercise 4E; 1 to 4 ac  2 2cis 2cis 0 x 2cis  Cambridge: Exercise 7A; 1, 2, 3 abef, 5, 6, 7, 9 to 14, 16 to 18  2