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Chapter 2 Complex Numbers
February 10 Complex numbers
2.1 Introduction
2.2 Real and imaginary parts of a complex number
2.3 The complex plane
2.4 Terminology and notation
Solution to a quadratic equation:
 b  b 2  4ac
2
az  bz  c  0  z1, 2 
.
2a
What happens if b  4ac  0 ?
2
Answer : z1, 2
 b  i 4ac  b 2

2a
Especially z 2  1  z1, 2  i.
i 2  1 is the most fundamenta l equation in complex numbers.
Example p46.
A complex number has a real part and an imaginary part.
Re( 3  2i )  3, Im(3  2i )  2.
1
Representation of a complex number on the complex plane:
A complex number x + iy can be specified or represented by the following equivalent
methods on the complex plane:
1) The original rectangular form x + iy.
2) A point with the coordinates (x, y).
3) A vector that starts from the origin and ends at the point (x, y).
4) The polar form reiq that satisfies
x  iy  reiq  r (cos q  i sin q ).
Example p48.
Modulus (magnitude, absolute value) of a complex number:
Angle (argument, phase) of a complex number:
z  x  iy  re iq  r (cos q  i sin q )
z  x2  y2  r
arg( z )  q  arctan
y
( if z is in the 2nd or 3rd quadrants. )
x
Example p50.
2
Complex conjugate
z  x  iy  re iq  r (cos q  i sin q )
z *  x  iy  re iq  rcos( q )  i sin( q )  r (cos q  i sin q )
The complex conjugate pair z=x + iy and z*=x − iy are symmetric with respect to
the x-axis on the complex plane.
Problems 4.3,9,18.
3
Read: Chapter 2: 1-4
Homework:
2.4.1,3,5,15,18.
Due: February 19
4
February 15, 17 Complex algebra
2.1 Complex algebra
A. Simplifying to x + iy form
z1  x1  iy1 , z 2  x2  iy 2 .
z1  z 2   x1  iy1   x2  iy 2   x1  x2   i  y1  y2 
z1  z 2  x1  iy1   x2  iy 2    x1  x2   i  y1  y2 
z1 z 2  x1  iy1  x2  iy 2    x1 x2  y1 y2   i x1 y2  x2 y1 
cz  cx  iy   cx  icy ,  z  x  iy    x  iy
z1 x1  iy1 ( x1  iy1 )( x2  iy 2 ) ( x1 x2  y1 y2 )  i x2 y1  x1 y2 



z 2 x2  iy 2 ( x2  iy 2 )( x2  iy 2 )
x22  y22
Examples p51.1-4; Problems 5.3,7.
B. Complex conjugate of a complex expression
*
*
z1
*
*
*
*
* *  z1 
*
*
*
( z1  z 2 )  z1  z 2 , ( z1  z2 )  z1  z2 , ( z1 z2 )  z1 z 2 ,    *
z2
 z2 
The complex conjugate of any expression is just to change all i’s into –i.
Example p53.1
5
C. Finding the absolute value of z
z  x  iy  re iq  r (cos q  i sin q )
z  r  x 2  y 2  zz *
z1 z 2  z1  z 2
z
z1
1 1
 1 ,

z2
z2
z
z
Example p53.2; Problems 5.26,28,33.
D. Complex equation
z1  x1  iy1 , z2  x2  iy 2 .
z1  z2 if and only if x1  x2 and y1  y2 .
Example p54; Problems 5.36,43.
E. Graphs
Complex equations or inequalities have geometrical meanings.
In particular , | z  (a  ib ) | r is a circle centered at (a, b) with a radius of r.
Example p55.1-4; Problems 5.52,53,59.
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F. Physical applications
The position of a moving particle is represented by a vector. This vector also represents
a complex number. Addition and subtraction of complex numbers is analogous to the
addition and subtraction of vectors. Therefore the position, speed and acceleration of a
particle can be well represented by complex numbers.
However, because the multiplication of complex numbers is not in analogy with the
multiplication of the vectors, physical principles involving multiplication of vectors
can not be represented by complex algebra. Example: W=F·s.
Example p56.
7
Read: Chapter 2: 5
Homework:
2.5.2,7,23,26,33,36,47,59,68.
Due: February 26
8
February 19 Complex infinite series
2.6 Complex infinite series
Convergence of a complex infinite series:

 (a
n 1
n
 ibn ) is said to converge to S a  iS b if and only if

a
n 1

n
 S a and  bn  Sb .
n 1
Theorem: An absolutely convergent complex series converges.
Proof :

a
n 1

n
 ibn   an2  bn2 converges. Because an  an2  bn2 , bn  an2  bn2 ,
n 1

so  an and
n 1

b
n
n 1
converge according to the comparison test.


n 1
n 1
Then both  an and  bn converge, since they are absolutely convergent .

This means that
 (a
n 1
n
 ibn ) converges.
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Theorem: Ratio test for a complex series an:
If lim
n 


1) If   1,  an converges,
n 1



  , then 2) If   1,  an diverges.
n 1



3) If   1, further te st is needed.
an 1
an
Proof :
1) If   1, take  so that     1. Then for all n  N ,
an 1
 .
an
Form bn   n  N a N for n  N , then an  bn .
Since
b
n 1
n



converges,
a
n 1
n
must converge, and thus
a
n 1
n
converges,
a
2) If   1, take  so that     1. Then for all n  N , n 1    1, lim an  0.
n 
an

a
n 1
n
must diverge.
Example p57.1,2; Problems 6.2,6,7.
10
2.7 Complex power series; disk of convergence
Disk of convergence: An area on which the series is convergent.
Radius of convergence: The radius of the disk of convergence.
Example p58.7.2a-c; Problems 7.5,8,15.
Disk of convergence for the quotient of two power series:


Suppose  an z  f ( z ) converges within z  r1 , and  bn z n  g ( z ) converges within z  r2 .
n
n 0
n 0
Suppose z1 is the smallest solution (in magnitude) to g ( z )  0.
f ( z) 
Then the quotient series
  cn z n coverges at least with in r  min( r1 , r2 , z1 ).
g ( z ) n 0
Example p59.
11
Read: Chapter 2: 6-7
Homework:
2.6.3,5,6,13;
2.7.8,11,15.
Due: February 26
12
February 22 Elementary functions
2.8 Elementary functions of complex numbers
Elementary functions: powers and roots; trigonometric and inverse trigonometric
functions; logarithmic and exponential functions; and the combination of them.
Functions of a complex variable can be defined using their corresponding infinite series.
Examples :
(1) n z 2 n 1
z3 z5 z7
sin z  
 z    
3! 5! 7!
n  0 ( 2n  1)!


zn
z2 z3 z4
e    1 z    
2! 3! 4!
n  0 n!
z
2.9 Euler’s formula
(iq ) 2 (iq ) 3 (iq ) 4 (iq ) 5
e  1  iq 




2!
3!
4!
5!
 q2 q4
 

q3 q5
 1  
   iq      cos q  i sin q
2! 4!
3! 5!

 

iq
z  x  iy  r (cos q  i sin q )  re iq
Examples p.62.
13
Multiplication and division of complex numbers using Euler’s formula:
z1  z2  r1eiq1  r2eiq 2  r1r2ei (q1 q 2 ) .
z1 r1eiq1 r1 i (q1 q 2 )

 e
.
z2 r2eiq 2 r2
Example p.63; Problems 9.13,22,38.
2.10 Powers and roots of complex numbers
Power of a complex number: z n  reiq   r n (cos nq  i sin nq ).
n
Example p.64.1.
Roots of a complex number:
Fundamental theorem of algebra:
n
Any polynomial P( z )   am z m (n  0, an  0) has n roots.
m 0
z  re iq  re i (q  2 k ) , k  0,1,2, , n  1.

z1/ n  re

i (q  2 k ) 1 / n
 q 2 k 
i 

1/ n  n n 
r e
  q 2k 
 q 2k
 r 1/ n cos 
  i sin  
n 
n
n
 n

, k  0,1,2, , n  1.

14
Notes about roots of a complex number:
1)There are altogether n values for n z .
1/ n
i
q
2)The first root is r e n .
3)All roots are evenly distributed on the circle with a radius of
incremental phase change of
2 / n.
r1/.n Each root has a
Example p.65.2-4; Problem 10.18.
15
Read: Chapter 2: 8-10
Homework:
2.9.7,13,23,38;
2.10.2,18,21,27.
Due: March 4
16
February 24 Exponential and trigonometric functions
2.11 The exponential and trigonometric functions
z  x  iy  re iq  r (cos q  i sin q )
e z  e x iy  e x eiy  e x (cos y  i sin y )

e iq  e  i q
eiz  e iz
cos q 
 cos z 
eiq  cos q  i sin q  
2
2



e iq  e  iq
eiz  e iz
e iq  cos q  i sin q  
sin q 
 sin z 

2i
2i
Notes on trigonometric functions:
1)sinz and cosz are generally complex numbers. They can be more than 1 even if they are
real.
2)The trigonometric identities (such as sin 2 z  cos 2 z  )1 and the derivative rules (such
(sin z )'  cos
z hold.
as
) still
Examples p.68.1-4; Problems 11.6,9.
17
2.12 Hyperbolic functions
e z  e z
sinh z 
2
e z  ez
cosh z 
2
sinh z
tanh z 
, etc.
cosh z
sin iy  i sinh y

cos iy  cosh y
Example p.70; Problems 12.1,9,15.
18
Read: Chapter 2: 11-12
Homework:
2.11.6,8,10,11;
2.12.1,3,11,32,36.
Due: March 4
19
February 26 Complex roots and powers
2.13 Logarithms
z  x  iy  re iq  r (cos q  i sin q )
ln z  ln( re iq )  ln[ re i (q  2 n ) ]  Lnr  ln ei (q  2 n )  Lnr  i(q  2n ), n  0,1,2,
Notes on logarithms:
1)We use Lnr to represent the ordinary real logarithm of r.
2)Because a complex number can have an infinite number of phases, it will have an
infinite number of logarithms, differing by multiples of 2i.
3)The logarithm with the imaginary part 0q <2 is called the principle value.
Examples p.72. 1-2; Problems 14.3,6,7.
20
2.14 Complex roots and powers
a  ra eiq a
b  bx  ib y


a b  eb ln a  exp (bx  ib y ) ln ra ei (q a  2 n )
 exp (bx  ib y )Lnra  i (q a  2n )





 exp bx Lnra  by (q a  2n )  exp i by Lnra  bx (q a  2n ) , n  0,1,2, 
Notes on roots and powers:
1) There are many amplitudes as well as many phases for ab.
2) For the amplitude in most cases only the principle value (0 q a<2 and n=0) is
needed.
3) by=0, bx=m or 1/m gives us the real powers and real roots of a complex number.
Examples p.73. 1-3; Problems 14.8,12,14.
21
2.15 Inverse trigonometric and hyperbolic functions
eiz  e iz
z  arccos w, if w  cos z 
.
2
Example p.75. 1; Problems 15.3.
22
Read: Chapter 2: 13-15
Homework:
2.14.3,4,8,14,17;
2.15.1,3,15.
Due: March 4
23