Mathematics Extension 2 Complex Numbers

ONLINE: MATHEMATICS EXTENSION 2
Topic 2
COMPLEX NUMBERS
2.2 ARITHMETIC OPERATIONS WITH COMPLEX
VARIABLES
You need to gain the ability to add, subtract, multiply and divide complex numbers.
The addition, subtraction, multiplication or division of two complex variables z1 and z2
z1  x1  i y1
z2  x2  i y2
x1 , y1, x2 , y2
real numbers
produces other complex variables
Addition
z3  z1  z2  x1  i y1  x2  i y2
z3   x1  x2   i  y1  y2 
add real parts
add imaginary parts
The vector representing z3  z1  z2 corresponds to the diagonal of a parallelogram with
the vectors z1 and z2 as adjacent sides.
From the parallelogram, it is obvious that
z1  z2  z1  z2
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
1
Subtraction
z4  z1  z2  x1  i y1   x2  i y2 
z4   x1  x2   i  y1  y2 
subtract real parts
subtract imaginary parts
For subtraction using the Argand diagram simply add the vectors z1 and -z2
z4  z1  z2  z1    z2 
Multiplication
z5  z1 z2   x1  i y1  x2  i y2    x1 x2  y1 y2   i  x1 y2  x2 y1 
Division
z6 
z1 z1 z2

z2 z2 z2
Multiply both the numerator and denominator by the complex conjugate of the
denominator so that you are now dividing by the real number z2 z2 .
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
2
Multiplication or division of two complex numbers is accomplished most easily when
they are in exponential form.
z5  z1 z2   R1 ei 1  R2 ei 2   R1 R2 e
i 1  2 
 R1 R2  1   2 
 R1 R2 cos 1   2   i sin 1   2  
z1  R1  i 1 2   R1 
 e
    1   2 
z2  R2 
 R2 
 R1 R2 cos 1   2   i sin 1   2  
z6 
A complex number can be shown as a vector on an Argard diagram.
z1  x1  i y1  R1 e i 1
R 1 x12  y12
tan 1 
y1
x1
We will consider the result
of multiplying the complex
number z1 by the complex
number e i 
z  e i  R1 e i 1  R1 e
i  1 
This means that the vector
for z1 is rotated
anticlockwise about the
origin through an angle  to
produce the vector for the
new vector z.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
3
z  e i  z1  R1 e i 1  R1 e
Let z1  3 e
i  1 
 
 
 4   3 cos   i sin    2.1213 1  i
 
4
4 

i 

i
  6 z  3 e
 
6
4

  3 e i 5 12
 0.7765  i  2.8978
 

 
 cos    i sin   3 e

2
2 

   2.1213  i 2.1213
 i 3 e



i  
i  
   2 z  e 2 3e 4
z
z
i 
i 
4
4
Multiplication by i produces an anticlockwise
rotation by ( / 2) rad.

 
 cos    i sin     3 e

    2.1213  i 2.1213   z
  3 e

 
 
 
z
z

z  e i  3e
 4
i 
i 
i 
4
4
1
    3 e   

 3e  
z  e
 
  3e  
z  e
 
  3 2 z  e
i 2 
i 
i 3
i 
2
i 
2


i 
2
4
4
4



i  
z  cos    i sin    3 e 4
2
2 


z  i 3 e


 4   2.1213  i 2.1213


i 
Clockwise rotation of z1 through ( / 2) rad.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
4
Example
Consider the two vectors z1  2  2 i
Find
z2  3  i
z  z1  z2


z  z1  z2  2  3   2  1 i  3.7321  3.0000 i
Find the magnitudes and arguments of z1, z2 and z
z  xi y
Know
z  R  x2  y2
2
Arg ( z )  1  a tan    0.7854 rad  45o
2
z1  R1  22  22  2.8284
z2  R2 
 y
Arg ( z )    a tan  
x
 1 
o
Arg ( z )  2  a tan 
  0.5236 rad  30
3


3  12  2.0000
 3 
o
Arg ( z )    a tan 
  0.6771 rad  38.7940
 3.7321 
z  R  3.73212  32  4.7883
Find z  z1  z2
z1  2 x  2 i

3xi


 z2  

z  z1  z2  z1    z2   2  3   2  1 i
 0.28284  1.0000 i

z  z1  z2 
 x1  x2 
2
  y1  y2 
2
hence the
magnitude of the complex number z1  z2 is equal to the distance between the
two points z1(x1, y1) and z2 (x2, y2) on the Argand diagram
 x1  x2 

2
  y1  y2  .
2
 y2  y1 
 y2  y1 
 but 
 is the slope of the line joining
 x2  x1 
 x2  x1 
the two points z1(x1, y1) and z2 (x2, y2) on the Argand diagram and so,
  Arg  z1  z2  is equal to the angle this line makes with the horizontal
  Arg  z1  z2   atan 
direction (line parallel to the real axis) as measured in an anticlockwise sense
with respect to the horizontal.
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
5
Find
z  z1 z2
z  xi y
Know
z1  x1  i y1
z 2  x2  i y 2
z  z1 z2   x1 x2  y1 y2   i  x1 y2  x2 y1 
z  x2  y2
 y
Arg ( z )    a tan  
x
  3  21  i 21   3 2  1.4641  5.4641 i
z   2
z  R  1.46412  5.46412  5.6569
 5.4641 
o
Arg ( z )    a tan 
  1.3090 rad  75
 1.4641 
Know
z  R e i    R1 e i 1  R2 e i 2   R1 R2 e
z   2.8284  2)  e
z  R  5.6569
i  0.78540.5236
 5.6569 e
i 1 2 
i 1.3090
Arg ( z )    1.3090 rad  75o
physics.usyd.edu.au/teach_res/hsp/math/math.htm
75
complex_002
o
 45o  30o 
6
Find
z
z1
z2
z  xi y
Know
z
z1  x1  i y1
z1 z2 z1 z2  x1 x2  y1 y2   i   x1 y2  x2 y1 


2
z2 z2
x2 2  y 2 2
z2
 y
Arg ( z )    a tan  
x
z  x2  y2
x1  2
z
2
z 2  x2  i y 2
y1  2 x2  3
 
3  2  i 2  2 3
4
y2  1
z2  2
  1.3660  0.3660
z  R  1.36602  0.36602  1.4142
 0.3660 
o
Arg ( z )    a tan 
  0.2618 rad  15
 1.3660 
Know
z  R e i    R1 e i 1  R2 e i 2   R1 R2 e
R1  2.8284
R2 
z
i 1 2 
1  0.7854 rad = 45o
3  12  2.0000
2  0.5236 rad  30o
z1  R1  i 1 2   2.8284  i 0.78540.5236
  e

 1.4142 e i 0.2618
e
z2  R2 
 2 
z  R  1.4142
Arg ( z )    0.2618 rad  15o
physics.usyd.edu.au/teach_res/hsp/math/math.htm
15
complex_002
o
 45o  30o 
7
Modulus and Complex conjugate relationships

arg  z1 z2   arg  z1   arg  z2 
z1 z2  z1 z2
Proof
z1  R1 e i 1
z1  R1
z2  R2
z1 z2  R1 R2  z1 z2
arg  z1   1 arg  z2    2

z1 z2   R1 e i 1   R2 e i 2   R1 R2 e i 1 2 
z2  R2 e i 2
arg  z1 z2   1   2  arg  z1   arg  z2 
z 
arg  1   arg  z1   arg  z2 
 z2 
z
z1
 1
z2
z2
Proof
z1  R1 e i 1
z1  R1
z
z1
R
 1  1
z2 R2 z2
z2  R2
arg  z1   1 arg  z2    2

zn  z
z1 R1 e i 1 R1 i 1  2 


e
z2 R2 e i 2 R2
z2  R2 e i 2
z 
arg  1   1   2  arg  z1   arg  z2 
 z2 
arg  z n   n arg  z 
n
Proof
z  Re i 
z n  R ne i n
z  Rn  z
n

arg  z n   n   n arg  z 
n
z1*  z2*   z1  z2 
*
 z1 z2 
*
 z1* z2*
* used for complex conjugate
Proof
z1  z2   x1  x2   i  y1  y2 
z1  x1  i y1
z 2  x2  i y 2
z1*  x1  i y1
z 2 *  x2  i y 2
z1*  z2*   z1  z2 
z1  R1 e i 1
 z1 z2 
*
*
  x1  x2   i  y1  y2 
*
z1*  R1 e  i 1
 R1R2 e
 z1  z2 
 i 1  2 
z2  R2 e i 2
z2*  R2 e  i 1 2
  R1 e  i 1   R2 e  i 1 2   z1* z2*
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
8
QUADRATIC FUNCTION
A quadratic function has the general form
a x2  b x  c  0
a0
and its graph is a parabola.
If there are real values of x for which the quadratic function a x 2  b x  c  0 then the
curve will intersect the X axis at the values of x given by the formula
x
b  b 2  4 a c
2a
Two real roots when the parabola
crosses the X axis twice
x1 = 0.5806
x2 = -1.9139
When the parabola touches the X
axis at one point, there is only one
real root
x1 = x2 = -0.6667
The are no real roots when the
parabola does not cut the X axis.
The two roots are now imaginary
x1 = - 0.6667 + i (0.8165)
x2 = -1.9139 - i (0.8165)
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
9
Example
0ABC is a square on an Argand diagram. The point A is given by the ordered pair (has
the ordered


3, 1 . Find the complex numbers for the points B and C.
zA  xA  i yA  3  i
z A  RA  x A2  y A2 
2
3  12  2
 1 
o
Arg ( z A )   A  a tan 
  0.5236 rad  30
 3
The angle between 0A and OC must be 90o. Multiplication by i produces an
anticlockwise rotation of 90o (/2 rad). Therefore, we can determine the coordinates of
the point C from
zC  i z A  1  3 i

Hence the order pair for C is 1, 3

or
 1, 1.7321 .
2
zC  RC  xC 2  yC 2  12  3  2
 3
Arg ( zC )  C    a tan 
 2.0944 rad  120 o
 1 


0B must correspond to the diagonal of the square whose side length is z  RA  2 .
Hence,
zB  RB  22  22  2.8284
The angle between the side 0A and the diagonal OB is 45o (/4 rad). Therefore,
 B   A  45o  30o  45o  1.3090 rad
Hence the complex number zB is
zB  2.8284 e
i 1.3090
xB  RB cos    2.8284  cos 1.3090   0.7320
y B  RB sin    2.8284  sin 1.3090   2.7320
Hence the order pair for B is  0.7320, 2.7320 .
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
10
The square is now rotated about O through an angle of 60o ( / 3 rad) in a clockwise
direction. Find the ordered pairs for the vertices of the square A1, B1 and C1.
know
multiplication of a complex number z by the complex number e
 
i  
 3
corresponds to a rotation of the vector representing z on an Argand
diagram through an angle ( / 3) rad in a clockwise direction.
z1  e
 
i  
 3
ze
 
i  
 3
R e i R e
 
i   
3

The new vertices of the square A1, B1 and C1 are determined by the rotation of the
vectors representing zA, zB and zC by ( / 3) rad
z A1  RA e
zB1  RB e
zC1  RC e


i  A  
3

2e


i  B  
3



i  C  
3



i  0.5236  
3



i  1.3090  
3

 2.8284 e
2e
2e


i  2.0944  
3

i  0.5236 
 2.8284 e i 0.2618
 2 e i 1.0472 
 A1  0.5236 rad  30o  B1  0.2618 rad  15o C1  1.0472 rad  60o
know
The rectangular form of the complex numbers is found from
x  R cos 
y  R sin 
x A1  2 cos  0.5236  1.7320
y A1  2 sin  0.5236  1.0000
xB1  2.8284 cos  0.2618  2.7320
xC1  2 cos 1.0472   1.0000
y B1  2.8284 sin  0.2618  0.7320
yC1  2 sin 1.0472   1.7321
The order pairs are A1 (1.7320, -1.0000), B1 (2.7320, 0.7320), C1 (1.0000, 1.7321).
rotation by
60o clockwise 
physics.usyd.edu.au/teach_res/hsp/math/math.htm
complex_002
11

Download Report