Download Complex math basics

Complex math basics
material from Advanced Engineering Mathematics by E Kreyszig
and from
Liu Y; Tian Y; “Succinct formulas for decomposition of complex refraction angle,”
IEEE Antennas and Propagation Society International Symposium, vol.3, pp 487- 490, 2003
Chris Allen (callen@eecs.ku.edu)
Course website URL
people.eecs.ku.edu/~callen/823/EECS823.htm
1
Outline
Complex numbers and analytic functions
Complex numbers
•
•
•
•
Real, imaginary form: z  x  jy
Complex plane
Arithmetic operations
Complex conjugate
Polar form of complex numbers, powers, and roots
• Multiplication and division in polar form
• Roots
Elementary complex functions
• Exponential functions
• Trigonometric functions, hyperbolic functions
• Logarithms and general powers
Example applications
Summary
2
Complex numbers
Complex numbers provide solutions to some equations not
satisfied by real numbers.
A complex number z is expressed by a pair of real numbers
x, y that may be written as an ordered pair
z = (x, y)
The real part of z is x; the imaginary part of z is y.
x = Re{z} y = Im{z}
The complex number system is an extension of the real
number system.
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Complex numbers
Arithmetic with complex numbers
Consider z1 = (x1, y1) and z2 = (x2, y2)
Addition
z1 + z2 = (x1, y1) + (x2, y2) = (x1+ x2 , y1 + y2)
Multiplication
z1 z2 = (x1, y1) (x2, y2) = (x1x2 - y1y2 , x1y2 + x2y1)
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Complex numbers
The imaginary unit, denoted by i or j, where i = (0, 1)
which has the property that j2 = -1 or j = [-1]1/2
Complex numbers can be expressed as a sum of the real
and imaginary components as
z = x + jy
Consider z1 = x1 + jy1 and z2 = x2 + jy2
Addition
z1 + z2 = (x1+ x2) + j(y1 + y2)
Multiplication
z1 z2 = (x1x2 - y1y2) + j(x1y2 + x2y1)
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Complex plane
The complex plane provides a geometrical representation
of the complex number space.
With purely real numbers on the horizontal x axis and
purely imaginary numbers on the vertical y axis, the plane
contains complex number space.
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Complex plane
Graphically presenting complex numbers in the complex
plane provides a means to visualize some complex values
and operations.
addition
subtraction
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Complex conjugate
If z = x + jy then the complex conjugate of z is z* = x – jy
Conjugates are useful since:
2
z z  x 2  y 2  z , a purely real number
z  z*
 x  Re{z}
2
z  z*
 y  Im{z}
2
Conjugates are useful in complex division
z1 z1 z*2 z1 z*2
  * 
z2 z2 z2 z2 2
Note that
z1 x1x 2  y1y 2
x 2 y1  x1y 2


j
z2
x 22  y 22
x 22  y 22
1
j
  j and  1
j
j
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Polar form of complex numbers
Complex numbers can also be represented in polar format,
that is, in terms of magnitude and angle.
j
Here z  r e  r cos   j r sin  (Euler ' s fomula )
x  r cos and y  r sin 
r z  x y
2
2
 y
and   arctan  
x
Note that
if z  r e j
then z*  r e  j
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Multiplication, division, and trig identities
Multiplication
z1z 2  r1 e j 1 r2 e j2  r1 r2 e j1  2 
j 1
Division
Trig identities
z1 r1 e
r1 j1  2 

 e
j 2
z 2 r2 e
r2
e j  e j
cos  
2
e j  e j
sin  
2j
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Polar form to express powers
Powers
The cube of z is
z3  r 3 e j3   r 3 cos 3   j r 3 sin 3 
If we let r = e, where  is real (z = eej) then
z 3  e 3   j    e 3   j 3   e 3  e j 3 
The general power of z is
z n  r n e j n   r n cos n   j r n sin n 
or (for r = e)
z n  en  j   e n e j n 
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Polar form to express roots
Roots
Given w = zn where n = 1, 2, 3, …,
if w  0 there are n solutions
Each solution called an nth root of w can be written as
zn w
Note that w has the form w = r ej
and z has the form of z = R ej so zn = Rn ejn
where Rn = r and n =  + 2k (where k = 0, 1, …, n -1)
The kth general solution has the form
zk  r e
n
j
2 k 
n
2k  
2k  
 r (cos
 j sin
)
n
n
n
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Polar form to express roots
Example
Solve the equation zn = 1, that is w = 1, r = 1,  = 0
zk  e
j
2 k  
n
2k  
2k  
 cos
 j sin
n
n
For n = 3, solutions lie on the unit circle at angles 0, 2/3, 4/3
z3 = 1
z0 = ej0/3 = 1
z1 = ej2/3 = -0.5 + j0.866
z13 = ej2 = 1
z2 = ej4/3 = -0.5 – j0.866
z23 = ej4 = 1
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Polar form to express roots
Example
Solve the equation zn = 1, for n = 2,
solutions lie on the unit circle at angles 0, 
z0 = +1
z1 = -1
Solve the equation zn = 1, for n = 4,
solutions lie on the unit circle at angles 0, /2, , 3/2
z0 = +1
z1 = j
z2 = -1
z3 = -j
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Polar form to express roots
Example
Solve the equation zn = 1, for n = 5,
solutions lie on the unit circle at angles 0, 2/5, 4/5, 6/5, 8/5
z5 = 1
z0 = ej0/5 = 1
z1 = ej2/5 = 0.309 + j0.951
z2 = ej4/5 = -0.809 + j0.588
z3 = ej6/5 = -0.809 + j0.588
z4 = ej8/5 = 0.309 – j0.951
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Complex exponential functions
The complex exponential function ez can be expressed in
terms of its real and imaginary components
ez  e x  j y  e x cos y  j sin y
The product of two complex exponentials is
e z1 ez 2  e z1  z 2  e x1  x 2 cosy1  y 2   j sin y1  y 2 
Note that
e jy  cos y  j sin y  cos 2 y  sin 2 y  1
therefore |ez| = ex.
Also note that e  e
where n = 0, 1, 2, …
jy
j  y  2 n 
 cos y  jsin y
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Complex trigonometric functions
As previously seen for a real value x
cos x 

1 jx
e  e jx
2

and sin x 



1 jz
and sin z 
e  e  jz
2j


1 jx
e  e jx
2j
For a complex value z

1 jz
cos z  e  e  jz
2
Similarly Euler’s formula applies to complex values
e jz  cos z  j sin z
Focusing now on cos z we have
1 jx  jy   jx  jy  1 y j x
cos z  (e
e
)  (e e  e ye j x )
2
2
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Complex trigonometric functions
Focusing on cos z we have
1 jx  jy   jx  jy  1  y j x
cos z  (e
e
)  (e e  e y e  j x )
2
2
e y
ey
cos x  j sin x   cos x  j sin x 

2
2
1
j
 e y  e y cos x  e y  e y sin x
2
2




from calculus we know
about hyperbolic functions

1 y
e  ey
2
sinh y
tanh y 
cosh y
cosh y 

and
and

1 y
e  ey
2
cosh y
coth y 
sinh y
sinh y 

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Complex trigonometric functions
So we can say
cos z  cos x cosh y  j sin x sinh y
We can similarly show that
sin z  sin x cosh y  j cos x sinh y
cos z  cos 2 x  sinh 2 y
2
sin z  sin 2 x  sinh 2 y
2
Formulas for real trig functions hold for complex values
cos z1  z 2   cos z1 cos z 2  sin z1 sin z 2
sin z1  z 2   sin z1 cos z 2  sin z2 cos z1
cos 2 z  sin 2 z  1
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Complex trigonometric functions
Example
Solve for z such that cos z = 5
Solution
We know cos z  cos x cosh y  j sin x sinh y  5
Let x = 0 or ±2n (n = 0, 1, 2, …) such that z = jy or
cos jy  cosh y  5
acosh 5 = 2.2924
Therefore z = ±2n ± j 2.2924, n = 0, 1, 2, …
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Complex hyperbolic functions
Complex hyperbolic functions are defined as
cosh z 

1 z
e  e z
2

and
sinh z 

1 z
e  e z
2

Therefore we know
cosh jz  cos z
sinh jz  j sin z
and
cos jz  cosh z
sin jz  j sinh z
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Complex logarithmic functions
The natural logarithm of z = x + jy is denoted by ln z
and is defined as the inverse of the exponential function
w  ln z (for z  0) or e w  z
Recalling that z = re j we know that
ln z  ln r  j 
y
where r  z ,   arctan  
x
However note that the complex
natural logarithm is multivalued
ln z  ln r  j   2  n 
where n  0, 1, 2, ...
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Complex logarithmic functions
Examples (n = 0, 1, 2, …)
ln 1 = 0, ±2j, ±4j, …
ln 4 = 1.386294 ± 2jn
ln -1 = ±j, ±3j, ±5j, …
ln -4 = 1.386294 ± (2n + 1)j
ln j = j/2, -3j/2, 5j/2, …
ln 4j = 1.386294 + j/2 ± 2jn
ln -4j = 1.386294  j/2 ± 2jn
ln (3-4j) = ln 5 + j arctan(-4/3) = 1.609438  j0.927295 ± 2jn
Note
Formulas for natural logarithms hold for complex values
ln (z1 z2) = ln z1 + ln z2
ln(z1/z2) = ln z1 – ln z2
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General powers of complex numbers
General powers of a complex number z = x + jy is defined as
z c  ec ln z for c complex , z  0
If c = n = 1, 2, … then zn is single-valued
If c = 1/n where n = 2, 3, … then
z c  n z  e 1 n  ln z
since the exponent is multivalued with multiples of 2j/n
If c is real and irrational or complex, then zc is infinitely manyvalued.
Also, for any complex number a
a z  e z ln a
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General powers of complex numbers
Example
1  j2 j  exp 2  jln 1  j

1


 exp 2  jln 2  j  2nj
4



 1

1

 2e  4 2 n sin  ln 2   j cos ln 2 

2

 2
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Example application 1
from Liu Y; Tian Y; “Succinct formulas for decomposition of complex refraction angle”
IEEE Antennas and Propagation Society International Symposium, vol.3, pp 487- 490, 2003.
Refraction angle at an air/lossy medium interface
A plane wave propagating through air
is incident on a dissipative half space
with incidence angle 1.
From the refraction law we know
k 1x = k 2x = k 1 sin 1 = k 2 sin 2
We also know that
k 1z = k 1 cos 1 and k 2z =k 2 cos 2
where k1  
k2  k0
 o  o  k o and n1 
r 2 r 2
and n 2 
 r1  r1  1
r 2 r 2
Because k2 is complex, 2 must also be complex
k 2  k2  j k2 and 2  2  j 2
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Example application 1
Refraction angle at an air/lossy medium interface
k2 and 2 are both complex
k 2  k2  j k2 and 2  2  j2
The exponential part of the refraction field is
e  jk2 x x  jk2 z z 
e
Im  k 2 sin 2  x  Im  k 2 cos  2 z  jRe  k 2 sin  2  x  Re  k 2 cos  2 z 
The constant-phase plane results when
Rek 2 sin 2  x  Rek 2 cos 2  z  constant
The constant-amplitude plane results when
Imk 2 sin 2  x  Imk 2 cos 2  z  constant
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Example application 1
Refraction angle at an air/lossy medium interface
The aspect angle for the constant-phase plane is
 Re k 2 sin 2  
  tan 1 



Re
k
cos

2
2 

and the angle for the constant-amplitude plane is
 Imk 2 sin 2  
  tan 



Im
k
cos

2
2 

1
Since k 1x = k 2x = k 1 sin 1 = k 2 sin 2 are real, we know


1
  tan 
 and   0
 Re k 2 cot 2 
1
Thus the complex refraction angle results in a separation
of the planes of constant-phase and constant-amplitude.
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Example application 1
Refraction angle at an air/lossy medium interface
To get the phase velocity in medium 2 requires
analysis of the exponential part of the refraction
field
vp 
 sin 
k1 sin 1


ko
n12
 
sin 1  Re
2
n 22
 n12 sin 2 1

2

c
n eff
where neff dependent on 1 and n1.
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Example application 2
from Liu Y; Tian Y; “Succinct formulas for decomposition of complex refraction angle”
IEEE Antennas and Propagation Society International Symposium, vol.3, pp 487- 490, 2003.
Analysis of total internal reflection
First consider the case where both
regions 1 and 2 are lossless, i.e.,
n1 and n2 are real.
Letting N = n1 / n2 we have:
N sin 1  sin 2
If N > 1 (i.e., n1 > n2 ) and 1  sin-1(1/N) [the condition for total internal
reflection], then sin 2 is real and greater than unity.
Therefore the refraction angle becomes complex, 2  2  j2 .
Since sin 2  sin 2 cosh 2  j cos 2 sinh 2

1

we know 2  , 2  cosh N sin 1 
2
The refraction presents a surface wave propagating in the x direction.
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Example application 2
Analysis of total internal reflection
Now consider the case where region 1
is dissipative and region 2 is lossless.
Here k1 is complex, k2 is real, and sin 1 is complex.
From the previous example we know that
  2  Re 2  and   tan 1  cot 2   2 

2
Before addressing the value of 2 we know that the constant-phase plane is
perpendicular to the constant-amplitude plane because region 2 is lossless.
To find
2 we let N = Nr + jNi where Nr and Ni are real.

2
2
1

N
sin
1 
1 
  2  cos 


N
2

2
sin 1  1 
2
2
4 N i2 sin 2 1





31
Example application 2
Analysis of total internal reflection
2
2
2
Defining N  N r  Ni when 1 tends
to /2 we get the limited value for  as

2
1

N


1
   2  cos 
2


 N  1
2
2

4 N i2
2





Figure 2 shows the refraction angle
various non-dissipative (Ni = 0) and
dissipative (Ni > 0) as a function
of incidence angle.
Note that for Ni > 0, the abrupt slope
change at the critical angle becomes
smooth and that the maximum  values
are less than 90°.
Fig. 2. The influence of medium loss to critical
angle and refraction angle (Nr = 3.0).
32
Summary
z  x  jy  r e j  r cos   j sin 
r  z  x 2  y2 and   arctan y x 
ez  e x  j y  e x cos y  j sin y

sin z  e

 2 j  sin x cosh y  j cos x sinh y
cos z  e jz  e jz 2  cos x cosh y  j sin x sinh y
jz

sinh z  e
 e jz

 2   j sin jz
cosh z  ez  e z 2  cos jz
z
 ez
ln z  ln r  j   2  n  where n  0, 1, 2, ...
z c  ec ln z for c complex , z  0
33