Download Complex Numbers and Functions

Many engineering problems can be treated and solved by using complex numbers and complex
functions. We will look at complex numbers, complex functions and complex differentiation.
Part I Complex Numbers
In algebra we discovered that many equations are not satisfied by any real numbers. Examples
are:
x 2  2
x 2  2 x  40  0
or
We must introduce the concept of complex numbers.
Definition: A complex number is an ordered pair z  ( x, y ) of real numbers x and y. We call x
the real part of z and y the imaginary part, and we write
Re z  x ,
Example 1: Re(4,6)  4
and
Im z  y .
Im(4,6)  6
Two complex numbers are equal where z1  ( x1, y1 ) and z2  ( x2 , y2 ) :
z1  z2 if and only if x1  x2 and y1  y2
Addition and Subtraction of Complex Numbers: We define for two complex numbers, the
sum and difference of z1  ( x1, y1 ) and z2  ( x2 , y2 ) :
z1  z2  ( x1  x2 , y1  y2 ) and z1  z2  ( x1  x2 , y1  y2 ) .
Multiplication of two complex numbers is defined as follows:
z1z2  ( x1x2  y1 y2 , x1 y2  x2 y1 )
Example 2: Let z1  (3, 4) and z2  (5, 6) then
z1  z2  (3  5,4  (6))  (8, 2)
and
z1  z2  (3  5,4  (6))  (2,10)
and
z1z2  (3  5  4  (6),3  (6)  4  5)  (39,2) .
We need to represent complex numbers in a manner that will make addition and multiplication
easier to do.
1
Complex numbers represented as z  x  iy
A complex number whose imaginary part is 0 is of the form ( x,0) and we have
( x1,0)  ( x2 ,0)  ( x1  x2 ,0) and ( x1,0)  ( x2 ,0)  ( x1  x2 ,0)
and
( x1,0)  ( x2 ,0)  ( x1x2 ,0)
which looks like real addition, subtraction and multiplication. So we identify ( x,0) with the real
number x and therefore we can consider the real numbers as a subset of the complex numbers.
We let the letter i  (0,1) and we call i a purely imaginary number.
Now consider i 2  i  i  (0,1)  (0,1)  (1,0) and so we can consider the complex number
i 2  1 = the real number 1 . We also get yi  y  (0,1)  (0, y )
And so we have:
( x, y )  ( x,0)  (0, y )  x  iy
Now we can write addition and multiplication as follows:
z1  z2  ( x1  x2 , y1  y2 )  x1  x2  i( y1  y2 )
and
z1z2  ( x1x2  y1 y2 , x1 y2  x2 y1 ) = x1x2  y1 y2  i( x1 y2  x2 y1 ) .
Example 3: Let z1 = (2,3)= 2 +3i and z 2 = (5,-4) = 5 -4i , then
z1  z2  (2  3i)  (5  4i)  7  i
and
z1  z2  (2  3i )  (5  4i )  10  15i  8i  12i 2  22  7i
The Complex Plane
The geometric representation of complex numbers is to represent the complex number ( x, y ) as
the point ( x, y ) .
y-axis
x  iy
2
1
1
2
x-axis
(2, 3)  2  3i
2
So the real number ( x,0) is the point on the horizontal x-axis, the purely imaginary number
yi  (0, y ) is on the vertical y-axis. For the complex number ( x, y ) , x is the real part and y is
the imaginary part.
Example 4. Locate 2-3i on the graph above.
How do we divide complex numbers? Let’s introduce the conjugate of a complex number then
go to division.
Given the complex number z  x  iy , define the conjugate z  x  iy  x  iy
We can divide by using the following:
z1 x1  iy1 x1  iy1 x2  iy2 x1 x2  y1 y2  i ( x2 y1  x1 y2 )



z2 x2  iy2 x2  iy2 x2  iy2
x22  y22
Example 5.
2  3i (2  3i)(3  4i) 6  12i 2  8i  9i
6
17


  i
2
3  4i (3  4i)(3  4i)
9  16i
25 25
Problem Set I
Find
1. i 2
2. i 3
3. i 4
4. i 21
5. i 1
6. i 2
7. i 3
8. i 4
9. i 24
Let z1  5  6i and z2  3  2i and z3  1  3i , find
10. z1  z2  z3
13. z1 z2
Graph the following:
11. 2 z2  3z1
z
14. 1
z3
16. 2  i
17. i
19. Find the solutions of
x 2  2 .
20. Find the solution of
x 2  2 x  40  0 .
12. z3  z 2
i
15.
z2
18. 1  3i and its conjugate.
Complex Numbers in Polar Form
It is possible to express complex numbers in polar form. If the point z  ( x, y )  x  iy is
represented by polar coordinates r ,  , then we can write x  r cos , y  r sin  and
3
z  r cos  ir sin   rei . r is the modulus or absolute value of z, | z | r  x 2  y 2 , and  is
z
2
 y
the argument of z,   arctan   . The values of r and  determine z uniquely, but the
x
converse is not true. The modulus r is determined uniquely by z, but  is only determined up to a
multiple of 2. There are infinitely many values of  which satisfy the equations
x  r cos , y  r sin  , but any two of them differ by some multiple of 2. Each of these angles
 is called an argument of z, but, by convention, one of them is called the principal argument.
If z is a non-zero complex number, then the unique real number , which satisfies
x  z cos , y  z sin  ,      
is called the principal argument of z, denoted by   arg( z ) .
Definition
Note: The distance from the origin to the point ( x, y ) is z , the modulus of z; the argument of z
is the angle   arctan
y
. Geometrically,  is the directed angle measured from the positive xx
axis to the line segment from the origin to the point ( x, y ) . When z  0 , the angle  is
undefined.
The polar form of a complex number allows one to multiply and divide complex number
more easily than in the Cartesian form. For instance, if z1  r1ei1 and z2  r2ei2 then
z1 z2  r1r2ei (1 2 ) ,
z1 r1 i (1 2 )
 e
. These formulae follow directly from DeMoivre’s formula.
z2 r2
zz
1
y
2
 
1
z
2
1

1

2
x
4
Example 6. For z  1  i , we get r  12  12  2 and   arctan
principal value of  is

9
, but
would work also.
4
4
y

 arctan1  . The
x
4
Multiplication and Division in Polar Form
Let z1  r1 cos1  ir1 sin 1  r1 (cos1  i sin 1 ) and z2  r2 (cos2  i sin 2 ) then we have
z1z2  r1r2 (cos(1  2 )  i sin(1  2 ))
and
z1 r1
 (cos(1   2 )  i sin(1   2 ))
z2 r2






z2  3  i  2  cos  i sin 
Example 7: z1  1  i  2  cos  i sin  and
4
4
6
6



 


5
5 


 i sin
Then z1z2  2  cos  i sin  2  cos  i sin  = 2 2  cos

4
4 
6
6
12
12 


  10 5

Since  
4 6
24 12
And
z1
=
z2



2  cos  i sin 
2

 
4
4

=
cos  i sin 




2 
12
12 
2  cos  i sin 
6
6

We can use z 2  z  z  r  r (cos(   )  i sin(   ) = r 2 (cos 2  i sin 2 )
And so:
DeMoivre's Theorem:
z n  r n (cos n  i sin n )
where n is an positive integer.
5
Let r  1 to get:
(cos  i sin  )n  cos n  i sin n .
Example 8: Compute (1  i)6
1  i 
6
 

 
  2  cos  i sin  
4
4 
 
6
6


 2  cos 6   i sin 6  
4
4

3
3 

 8  cos
 i sin

2
2 

 8i
Roots of Complex Numbers:
Consider z  r (cos  i sin  ) = wn  R n (cos n  i sin n )
(Equation 1)
where w  R(cos   i sin  ) . Then R  n r , and so   n or  
However n    2 also satisfies Equation 1 and so  



n
.
2
. And
n
n
4
 6
n    4 implies   
. However n    6 implies   
.
n n
n n

And continuing n    k implies  
We get
n

n

k
. for k any integer up to n.
n

   k 2 ) 
   k 2 )  
z  n x  cos 
  i sin 
,
n
n





Example 9:
k=0, 1, 2, 3,
, (n-1).
Find the square roots of i.
 
 
Since i  cos    i sin   , we let
2
2



1
1
1  
 1  
w1  1  cos     i sin      cos  i sin 
i
is one square root of i . The
4
4
2
2
2 2
 2 2 

second square root of i is :

5
5
1
1




w1  1  cos      i sin       cos
 i sin

i
.
4
4
2
2
4

4


6
Example 10. Find the sixth root of 1.
There will be six roots:

3 1
 
  
z1  6 1  cos    i sin    
 i
6
 6  2 2


  2
z2  6 1  cos  
6 6


  2  
  i sin  
  i

 6 6 

  4 
  4
z3  6 1  cos  
  i sin  
6 6 
6 6


  6
z4  6 1  cos  
6 6

3 1

 i
  
2 2

3 1

  6  
 i
  i sin  
  
2 2

 6 6 

  8 
  8  
z5  6 1  cos  
  i sin  
   i
6 6 
 6 6 


3 1
  10 
  10  
z6  6 1  cos  
 i
  i sin  
  
6 
6 
2 2
6
6

7
Example 11: Compute
4
4i
3
3 

4  cos
 i sin  = 2(cos   i sin  )
2
2 

(3  k  2 )
3 7 11 15
,
,
,
Where  
. So  
24
8 8 8
8
Solution:
4
4
4i =
4

 3 
 3  
4i  2  cos    i sin     2(0.3827  0.9239i)
 8 
 8 


 7 
 7  
= 2  cos 
  i sin 
   2(0.9239  0.3827i)
 8 
 8 


 11 
 11  
 2  cos 
  i sin 
    2(0.3827  0.9239i)
 8 
 8 


 15 
 15  
 2  cos 
  i sin 
   2(0.9239  0.3827i) .
 8 
 8 

Problem Set II
Write in polar form:
1. 3i
2. 2  2i
2  3i
3.
Write in rectangular form:
3
3 

 i sin
4. 32  cos

4
4 

5. 3(cos   i sin  )
7. Find (2  2i) 4
8. Find
9. Find 4 16  16i
10. Find

3 i
5

5
32
8
6.



2  cos  i sin 
4
4

Part II Functions, Neighborhoods and Limits
We consider the concept of a function of a complex number. For y  f ( x)  x 2 , where x and y
are real numbers, we know about limits, continuity and derivatives. Let the complex number
w  f ( z )  z 2 , where z  x  iy and w  ( x  iy)2  x 2  y 2  2 xyi . Here x and y are independent
variables and w  u  iv where u  x 2  y 2 and v  2 xy . This is an example of a complex
valued function, w, of a complex variable z. In general, let
w  u  iv
where u  u ( x, y ) and v  v( x, y ).
w
z
Distance in the complex plane.
In the complex plane let z1  x1  iy1 and z2  x2  iy2 then
| z1  z2 | ( x1  x2 )2  ( y1  y2 )2
is the distance between the complex numbers z1 and z2 Note that | z | x 2  y 2 is the distance
from the origin to z.
Definition: A neighborhood N ( z0 , r ) of the point z0 in the complex plane is the set of points z
where r  0 and {z | z  z0 | r}
Definition: A function w  f ( z ) is said to have a limit L as z approaches z0 ( and written
lim f ( z)  L ) iff f ( z ) is defined in a neighborhood N ( z0 , r ) , r  0 except at z0 and the
z z0
values of f are close to L for all z close to z0 . Mathematically: for every   0 , we can find
  0 such that for all z  z0 in the neighborhood N ( z0 , r ) , so that if 0 | z  z0 |  , then
| f ( z )  L |  .
Notice that this definition of a limit is similar to the definition in calculus. The difference here is
that z can approach z0 from any direction in the complex plane.
Definition: A function w  f ( z ) is continuous at z  z0 if f ( z0 ) is defined and lim f ( z)  L .
z z0
By definition a complex function that is continuous at z0 is defined in a neighborhood of z0 .
f ( z ) is continuous in a domain D if it is continuous at each point in the domain D.
9
The derivative of a function f ( z ) at a point z is defined as:
f '( z )  lim
h0
f ( z  h)  f ( z )
h
provided that limit exists.
Example 12: Let f ( z )  z 2 . To compute f '( z ) we consider
( z  h) 2  z 2
z 2  2 zh  h 2  z 2
2 zh  h 2
 lim
 lim
 lim(2 z  h)  2 z
h0
h0
h0
h0
h
h
h
lim
The rules for derivatives are the same for calculus:
(cf ) '  cf ' c=constant
( f  g ) '  f ' g '
( fg ) '  f ' g  fg '
 f 
f ' g  fg '
 ' 
g2
g
Definition: A function f ( z ) is defined to be analytic in the domain D if f ( z ) is defined and has
a derivative at all points of D.
The Exponential Function e z
The definition of the exponential function e z  e xiy  e xeiy is given in terms of the real functions
e x , sin y , and cos y.
e z  e xiy  e xeiy  e x (cos y  i sin y )
To show why this is true, consider the power series expansion: e x = 1  x 
x 2 x3
xn
     
2! 3!
n!
and substitute iy for x, to get:
(iy)2 (iy)3
(iy) n
e  1  iy 

  
  = 1  iy 
2!
3!
n!
 y2 y4 y6
 
y3 y5
eiy = 1  
    + i  y   
2! 4! 6!
3! 5!

 
iy
10
y 2 iy 3
(iy )n

  
 
2! 3!
n!

y7
   = cos y  i sin y .
7!

The formula
e z  e xiy  e x (cos y  i sin y) is known as Euler’s formula.
Theorem:
e z is never zero.
Proof: e z  e  z  e0  1 , therefore, e z cannot be zero.
If y is real, then eiy  1 .
Theorem:
2
Proof: eiy  cos 2 y  sin 2 y  1 and eiy  0 , which implies eiy  1 .
Theorem:
e z  1 if and only if z  2n i , where n is an integer.
Proof: If z  2n i , e z  cos 2n  i sin 2n  1 . Conversely, suppose e z  1 , then
e x cos y  1, e x sin y  0 . Since e x  0 , this implies sin y  0 ; and y  k where k is an integer.
Substituting this into e x cos y  1 implies e x (1)k  1 . Hence k is even, since e x  0 . Therefore,
e x  1 and x  0 .
Theorem:
e z1  e z2 if and only if z1  z2  2n i , where n is an integer.
Proof: e z1  e z2 if and only if e z1  z2  1 . Then use the previous theorem.
Definition of sin z and cos z
To get the definitions of sin z and cos z we substitute  y for y in Euler’s Formula to get:
 y2 y4 y6
 

y3 y5 y 7
e  iy = 1  
    + i   y       = cos y  i sin y .
2! 4! 6!
3! 5! 7!

 

Adding eiy and e  iy together and dividing by 2, we get:
eiy  eiy
 cos y .
2
Subtracting e
 iy
eiy  eiy
from e , and dividing by 2i , we get: sin y 
.
2i
We now define sin z 
iy
eiz  eiz
2i
and
cos z 
11
eiz  eiz
.
2
Substitute ix for y in cos y 
eiy  eiy
e x  e x
to get cos(ix) 
 cosh x .
2
2
Similarly we get sin(ix)  i sinh x .
Facts:
1. For z  x  0i , to get e z  e xi 0  e x
2. The derivative of e z is e z .
3. e z1 z2  e z1 e z2
4. For z  iy we get eiy  cos y  i sin y .
5. In polar form z  x  iy  r (cos  i sin  )  rei .
6. It now follows that e z 2 i  e z e 2 i  e z 1  e z .
7. e i  1
8. cosh z 
e z  e z
2
sinh z 
9. sin 2 z  cos 2 z  1
e z  e z
2
sin( z )   sin z
cos( z )  cos z
Logarithmic Functions
If z  e w then we write w  ln z , called the natural logarithm of z. Thus the natural logarithm
function is the inverse of the exponential function and can be defined by
w  ln z  ln r  i(  2  k ) where k  0, 1, 2, 3, 
where z  rei  rei 2 i .
So we have that w  ln z is multiple-valued, and in fact is infinitely many-valued. If we let
k  0 then w  ln z  ln r  i(  2  k )  ln r  i , where 0    2 called the principal value or
principal branch of ln z .
To find the values of ln(1  i ) , let z  e w , where z  rei  r (cos  i sin  ) and so
w  u  iv which implies z  euiv  eu (cos v  i sin v) .
12
We equate the real and imaginary parts:
(2) eu cos v  r sin 
(1) eu cos v  r cos 
Squaring (1) and (2) and adding, we find e 2u  r 2
or eu  r . Thus, from (1) and (2),
r cos v  r cos
r sin v  r sin
and so v    2k .
Hence we have
w  u  iv  ln r  i (  2k )
Example 13: Compute ln(1  i ).
Since 1  i  2e 1  i  2e
7 i
 2 k i
4
, we have
7 i
 7 i
 1
ln(1  i )  ln 2  
 2k i   ln 2 
 2 k i
4
 4
 2
1
7 i
The principle value is ln 2 
(let k  0 ).
2
4
Problem Set III.
Sketch the point sets defined by the following
1. | z  1| 2
2. | z  2 | 3
3. | z  1  i | 2
Show that
4. sin z  sin( z )
5. cosh z  cosh( z ) 6. (e z )2  e2 z
Evaluate each of the following
7. e 2 i
8. ei 1

9. sin(3   i ) 10. e 2
i
11. cos  i
Find all possible values of ln z :
12. ln 2i
13. ln(9)
14. ln(1  2i )
15. ln(i 3 )
16. ln(1  i )
17. ln(1)
18. cos(2  3 i)
19. sec(1  i ) 
1
cos(1  i )
20 Show that e z  1 if and only if z  2n i where n is any integer.
13