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X2 t01 04 mod arg form(2013)

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Mod-Arg Form
Mod-Arg Form y z = x + iy y O x x Modulus The modulus of a complex number is the length of the vector OZ
Mod-Arg Form y z = x + iy r z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x
Mod-Arg Form y z = x + iy r z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x z  x2  y2
Mod-Arg Form y z = x + iy r z   arg z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x z  x2  y2 Argument The argument of a complex number is the angle the vector OZ makes with the positive real (x) axis
Mod-Arg Form y z = x + iy r z   arg z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x z  x2  y2 Argument The argument of a complex number is the angle the vector OZ makes with the positive real (x) axis  y arg z  tan    x 1    arg z  
e.g . Find the modulus and argument of 4  4i
e.g . Find the modulus and argument of 4  4i 4  4i  4 2   4  2  32 4 2
e.g . Find the modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1
e.g . Find the modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1   4
e.g . Find the modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1   4 Every complex number can be written in terms of its modulus and argument
e.g . Find the modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1   4 Every complex number can be written in terms of its modulus and argument z  x  iy  r cos  ir sin   r cos  i sin  
e.g . Find the modulus and argument of 4  4i 4  4i  4   4  2 2  32 4 2  4 arg4  4i   tan    4   tan 1  1 1   4 Every complex number can be written in terms of its modulus and argument z  x  iy  r cos  ir sin   r cos  i sin   The mod-arg form of z is; z  r cos  i sin   z  rcis where; r  z   arg z
e.g. i  4  4i
 e.g. i  4  4i  4 2cis      4
 e.g. i  4  4i  4 2cis      4 ii  3  i
 e.g. i  4  4i  4 2cis      4 ii  3  i z  3  4 2 2  12
 e.g. i  4  4i  4 2cis      4 ii  3  i z  3  4 2 2 1 2 arg z  tan 1   6 1 3
 e.g. i  4  4i  4 2cis      4 ii  3  i z  3 2 1 2 arg z  tan 1   4 2  3  i  2cis  6  6 1 3
 e.g. i  4  4i  4 2cis      4 ii  3  i z  3 2 arg z  tan 1 1 2   4 2  3  i  2cis  ii  Convert 6cis  6  6 to Cartsian form  6 1 3
 e.g. i  4  4i  4 2cis      4 ii  3  i z  3 2 arg z  tan 1 1 2   4 2  3  i  2cis  ii  Convert 6cis 6cis  6  6   6 to Cartsian form  6(cos    i sin ) 6 6 6 1 3
 e.g. i  4  4i  4 2cis      4 ii  3  i z  3 2 arg z  tan 1 1 2   4 2  3  i  2cis  ii  Convert 6cis 6cis  6    6 to Cartsian form  6(cos    i sin ) 6 6 6 3 1  6(  i) 2 2  3 3  3i 6 1 3
Mod-Arg Relations
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 NOTE: Multiplication rotates z1 by arg z2
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2  NOTE: Multiplication rotates z1 by arg z2
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2 
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2 
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2   z1 z 2  r1r2  z1 z 2
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2   z1 z 2  r1r2 arg z1 z 2   1   2  z1 z 2  arg z1  arg z 2
Mod-Arg Relations 1 z1 z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2   z1 z 2  r1r2 arg z1 z 2   1   2  z1 z 2  arg z1  arg z 2 NOTE: it follows that; z1 z 2 z3  z n  z1 z 2 z3  z n arg z1 z 2 z3  z n   arg z1  arg z 2  arg z3    arg z n
z1 z1 2   z2 z2 z  arg 1   arg z1  arg z 2  z2 
z1 z1 2   z2 z2 z  arg 1   arg z1  arg z 2  z2  NOTE: it follows that; z1 z 2 z1 z 2  z3 z 4 z3 z 4  z1 z 2  arg   z z   arg z1  arg z 2  arg z3  arg z 4  3 4
z1 z1 2   z2 z2 z  arg 1   arg z1  arg z 2  z2  NOTE: it follows that; z1 z 2 z1 z 2  z3 z 4 z3 z 4  z1 z 2  arg   z z   arg z1  arg z 2  arg z3  arg z 4  3 4 3 z  z n n argz n   n arg z
e.g. Find the modulus and argument of z  5  i  2  i  3  2i
e.g. Find the modulus and argument of z  2 2 52  12  2    1 z 32  2 2 5  i  2  i  3  2i
e.g. Find the modulus and argument of z  2 2 52  12  2    1 z 32  2 2 26 5  13  10 5  i  2  i  3  2i
5  i  2  i  e.g. Find the modulus and argument of z  3  2i 2 2 2 2 5  1  2    1 1 2 1 z 2 2 arg z  tan 1    tan 1    tan 1         3 2 5   2  3 26 5  13  10
5  i  2  i  e.g. Find the modulus and argument of z  3  2i 2 2 2 2 5  1  2    1 1 2 1 z 2 2 arg z  tan 1    tan 1    tan 1         3 2 5   2  3 26 5   1119   153 26  33 41 13  175 48  10
5  i  2  i  e.g. Find the modulus and argument of z  3  2i 2 2 2 2 5  1  2    1 1 2 1 z 2 2 arg z  tan 1    tan 1    tan 1         3 2 5   2  3 26 5   1119   153 26  33 41 13  175 48  10 Patel: Exercise 4B; evens Patel: Exercise 4C; 1 to 10 evens Cambridge: Exercise 1D; 9, 11, 13, 14, 16, 20

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X2 t01 04 mod arg form(2013)

  • 1.
  • 2.
    Mod-Arg Form y z =x + iy y O x x Modulus The modulus of a complex number is the length of the vector OZ
  • 3.
    Mod-Arg Form y z =x + iy r z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x
  • 4.
    Mod-Arg Form y z =x + iy r z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x z  x2  y2
  • 5.
    Mod-Arg Form y z =x + iy r z   arg z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x z  x2  y2 Argument The argument of a complex number is the angle the vector OZ makes with the positive real (x) axis
  • 6.
    Mod-Arg Form y z =x + iy r z   arg z O x Modulus The modulus of a complex number is the length of the vector OZ r 2  x2  y2 r  x2  y2 y x z  x2  y2 Argument The argument of a complex number is the angle the vector OZ makes with the positive real (x) axis  y arg z  tan    x 1    arg z  
  • 7.
    e.g . Findthe modulus and argument of 4  4i
  • 8.
    e.g . Findthe modulus and argument of 4  4i 4  4i  4 2   4  2  32 4 2
  • 9.
    e.g . Findthe modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1
  • 10.
    e.g . Findthe modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1   4
  • 11.
    e.g . Findthe modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1   4 Every complex number can be written in terms of its modulus and argument
  • 12.
    e.g . Findthe modulus and argument of 4  4i 4  4i  4   4  2  32 4 2 2  4 arg4  4i   tan    4   tan 1  1 1   4 Every complex number can be written in terms of its modulus and argument z  x  iy  r cos  ir sin   r cos  i sin  
  • 13.
    e.g . Findthe modulus and argument of 4  4i 4  4i  4   4  2 2  32 4 2  4 arg4  4i   tan    4   tan 1  1 1   4 Every complex number can be written in terms of its modulus and argument z  x  iy  r cos  ir sin   r cos  i sin   The mod-arg form of z is; z  r cos  i sin   z  rcis where; r  z   arg z
  • 14.
    e.g. i 4  4i
  • 15.
     e.g. i 4  4i  4 2cis      4
  • 16.
     e.g. i 4  4i  4 2cis      4 ii  3  i
  • 17.
     e.g. i 4  4i  4 2cis      4 ii  3  i z  3  4 2 2  12
  • 18.
     e.g. i 4  4i  4 2cis      4 ii  3  i z  3  4 2 2 1 2 arg z  tan 1   6 1 3
  • 19.
     e.g. i 4  4i  4 2cis      4 ii  3  i z  3 2 1 2 arg z  tan 1   4 2  3  i  2cis  6  6 1 3
  • 20.
     e.g. i 4  4i  4 2cis      4 ii  3  i z  3 2 arg z  tan 1 1 2   4 2  3  i  2cis  ii  Convert 6cis  6  6 to Cartsian form  6 1 3
  • 21.
     e.g. i 4  4i  4 2cis      4 ii  3  i z  3 2 arg z  tan 1 1 2   4 2  3  i  2cis  ii  Convert 6cis 6cis  6  6   6 to Cartsian form  6(cos    i sin ) 6 6 6 1 3
  • 22.
     e.g. i 4  4i  4 2cis      4 ii  3  i z  3 2 arg z  tan 1 1 2   4 2  3  i  2cis  ii  Convert 6cis 6cis  6    6 to Cartsian form  6(cos    i sin ) 6 6 6 3 1  6(  i) 2 2  3 3  3i 6 1 3
  • 23.
  • 24.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2
  • 25.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 NOTE: Multiplication rotates z1 by arg z2
  • 26.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2  NOTE: Multiplication rotates z1 by arg z2
  • 27.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2 
  • 28.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2 
  • 29.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2   z1 z 2  r1r2  z1 z 2
  • 30.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2   z1 z 2  r1r2 arg z1 z 2   1   2  z1 z 2  arg z1  arg z 2
  • 31.
    Mod-Arg Relations 1 z1z2  z1 z2 arg z1 z 2   arg z1  arg z 2 Proof: let z1  r1cis1 and z 2  r2 cis 2 NOTE: Multiplication rotates z1 by arg z2 z1 z 2  r1 cos1  i sin 1   r2 cos 2  i sin  2   r1r2 cos1 cos 2  i sin 1 cos 2  i cos1 sin  2  sin 1 sin  2   r1r2 cos1 cos 2  sin 1 sin  2   isin 1 cos 2  cos1 sin  2   r1r2 cos1   2   i sin 1   2   z1 z 2  r1r2 arg z1 z 2   1   2  z1 z 2  arg z1  arg z 2 NOTE: it follows that; z1 z 2 z3  z n  z1 z 2 z3  z n arg z1 z 2 z3  z n   arg z1  arg z 2  arg z3    arg z n
  • 32.
    z1 z1 2   z2z2 z  arg 1   arg z1  arg z 2  z2 
  • 33.
    z1 z1 2   z2z2 z  arg 1   arg z1  arg z 2  z2  NOTE: it follows that; z1 z 2 z1 z 2  z3 z 4 z3 z 4  z1 z 2  arg   z z   arg z1  arg z 2  arg z3  arg z 4  3 4
  • 34.
    z1 z1 2   z2z2 z  arg 1   arg z1  arg z 2  z2  NOTE: it follows that; z1 z 2 z1 z 2  z3 z 4 z3 z 4  z1 z 2  arg   z z   arg z1  arg z 2  arg z3  arg z 4  3 4 3 z  z n n argz n   n arg z
  • 35.
    e.g. Find themodulus and argument of z  5  i  2  i  3  2i
  • 36.
    e.g. Find themodulus and argument of z  2 2 52  12  2    1 z 32  2 2 5  i  2  i  3  2i
  • 37.
    e.g. Find themodulus and argument of z  2 2 52  12  2    1 z 32  2 2 26 5  13  10 5  i  2  i  3  2i
  • 38.
    5  i 2  i  e.g. Find the modulus and argument of z  3  2i 2 2 2 2 5  1  2    1 1 2 1 z 2 2 arg z  tan 1    tan 1    tan 1         3 2 5   2  3 26 5  13  10
  • 39.
    5  i 2  i  e.g. Find the modulus and argument of z  3  2i 2 2 2 2 5  1  2    1 1 2 1 z 2 2 arg z  tan 1    tan 1    tan 1         3 2 5   2  3 26 5   1119   153 26  33 41 13  175 48  10
  • 40.
    5  i 2  i  e.g. Find the modulus and argument of z  3  2i 2 2 2 2 5  1  2    1 1 2 1 z 2 2 arg z  tan 1    tan 1    tan 1         3 2 5   2  3 26 5   1119   153 26  33 41 13  175 48  10 Patel: Exercise 4B; evens Patel: Exercise 4C; 1 to 10 evens Cambridge: Exercise 1D; 9, 11, 13, 14, 16, 20