Chris Allen ([email protected]) Course website URL people.eecs.ku/~callen/823/EECS823.htm
description
Transcript of Chris Allen ([email protected]) Course website URL people.eecs.ku/~callen/823/EECS823.htm
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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 ([email protected])
Course website URL people.eecs.ku.edu/~callen/823/EECS823.htm
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OutlineComplex numbers and analytic functions
Complex numbers• Real, imaginary form:
• 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
jyxz
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Complex numbersComplex 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 numbersArithmetic 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 numbersThe 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 planeThe 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 planeGraphically presenting complex numbers in the complex plane provides a means to visualize some complex values and operations.
subtraction
addition
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Complex conjugateIf z = x + jy then the complex conjugate of z is z* = x – jy
Conjugates are useful since:
Conjugates are useful in complex division
22
22
211222
22
2121
2
1
22
*21
*2
*2
2
1
2
1
yx
yxyxj
yx
yyxx
z
z
z
zz
z
z
z
z
z
z
}zIm{y2
*zz
}zRe{x2
*zz
numberrealpurelya,zyxzz222
1j
jandj
j
1
thatNote
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Polar form of complex numbersComplex numbers can also be represented in polar format, that is, in terms of magnitude and angle.
Here
sinryandcosrx
)fomulas'Euler(sinrjcosrerz j
x
yarctanandyxzr 22
j
j
er*zthen
erzif
thatNote
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Multiplication, division, and trig identities
Multiplication
Division
Trig identities
2121 j21
j2
j121 errererzz
21
2
1j
2
1j
2
j1
2
1 er
r
er
er
z
z
j2
eesin
2
eecos
jj
jj
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Polar form to express powersPowers
The cube of z is
If we let r = e, where is real (z = eej) then
The general power of z is
or (for r = e)
3sinrj3cosrerz 333j33
nsinrjncosrerz nnnjnn
3j33j3j33 eeeez
njnjnn eeez
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Polar form to express rootsRoots
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
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
n wz
)n
k2sinj
n
k2(cosrerz nn
k2j
nk
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Polar form to express rootsExample
Solve the equation zn = 1, that is w = 1, r = 1, = 0
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
n
k2sinj
n
k2cosez n
k2j
k
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Polar form to express rootsExample
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 rootsExample
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 functionsThe complex exponential function ez can be expressed in terms of its real and imaginary components
The product of two complex exponentials is
Note that
therefore |ez| = ex.
Also note that
where n = 0, 1, 2, …
ysinjycoseee xyjxz
2121xxzzzz yysinjyycoseeee 212121
1ysinycosysinjycose 22jy
ysinjycosee n2yjyj
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Complex trigonometric functionsAs previously seen for a real value x
For a complex value z
Similarly Euler’s formula applies to complex values
Focusing now on cos z we have
zjzjzjzj eej2
1zsinandee
2
1zcos
zsinjzcose zj
xjxjxjxj eej2
1xsinandee
2
1xcos
)eeee(2
1)ee(
2
1zcos xjyxjyjyxjjyxj
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Complex trigonometric functionsFocusing on cos z we have
from calculus we know
about hyperbolic functions
xsinee2
jxcosee
2
1
xsinjxcos2
exsinjxcos
2
e
)eeee(2
1)ee(
2
1zcos
yyyy
yy
xjyxjyjyxjjyxj
ysinh
ycoshycothand
ycosh
ysinhytanh
ee2
1ysinhandee
2
1ycosh yyyy
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Complex trigonometric functionsSo we can say
We can similarly show that
Formulas for real trig functions hold for complex values
ysinhxcosjycoshxsinzsin
ysinhxsinzsin 222
ysinhxcoszcos 222
212121 zsinzsinzcoszcoszzcos
122121 zcoszsinzcoszsinzzsin
1zsinzcos 22
ysinhxsinjycoshxcoszcos
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Complex trigonometric functionsExample
Solve for z such that cos z = 5
Solution
We know
Let x = 0 or ±2n (n = 0, 1, 2, …) such that z = jy or
acosh 5 = 2.2924
Therefore z = ±2n ± j 2.2924, n = 0, 1, 2, …
5ysinhxsinjycoshxcoszcos
5ycoshjycos
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Complex hyperbolic functionsComplex hyperbolic functions are defined as
Therefore we know
and
zzzz ee2
1zsinhandee
2
1zcosh
zsinjjzsinh
zcosjzcosh
zsinhjjzsin
zcoshjzcos
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Complex logarithmic functionsThe natural logarithm of z = x + jy is denoted by ln z
and is defined as the inverse of the exponential function
Recalling that z = re j we know that
However note that the complex
natural logarithm is multivalued
zeor)0zfor(zlnw w
x
yarctan,zrwhere
jrlnzln
...,2,1,0nwhere
n2jrlnzln
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Complex logarithmic functionsExamples (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 numbersGeneral powers of a complex number z = x + jy is defined as
If c = n = 1, 2, … then zn is single-valued
If c = 1/n where n = 2, 3, … then
since the exponent is multivalued with multiples of 2j/n
If c is real and irrational or complex, then zc is infinitely many-valued.
Also, for any complex number a
0z,complexcforez zlncc
zlnn1nc ezz
alnzz ea
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General powers of complex numbersExample
2ln2
1cosj2ln
2
1sine2
jn2j4
12lnj2exp
j1lnj2expj1
n24
j2
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Example application 1from 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 interfaceA plane wave propagating through airis incident on a dissipative half spacewith 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
Because k2 is complex, 2 must also be complex
2r2r22r2r02
1r1r1ooo1
nandkk
1nandkk
222222 jandkjkk
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Example application 1
Refraction angle at an air/lossy medium interfacek2 and 2 are both complex
The exponential part of the refraction field is
The constant-phase plane results when
The constant-amplitude plane results when
222222 jandkjkk
zcoskRexsinkRejzcoskImxsinkIm
zjkxjk
22222222
z2x2
e
e
constantzcoskRexsinkRe 2222
constantzcoskImxsinkIm 2222
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Example application 1
Refraction angle at an air/lossy medium interfaceThe aspect angle for the constant-phase plane is
and the angle for the constant-amplitude plane is
Since k 1x = k 2x = k 1 sin 1 = k 2 sin 2 are real, we know
Thus the complex refraction angle results in a separation of the planes of constant-phase and constant-amplitude.
22
221
coskRe
sinkRetan
22
221
coskIm
sinkImtan
0andcotkRe
1tan
22
1
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Example application 1
Refraction angle at an air/lossy medium interfaceTo get the phase velocity in medium 2 requires
analysis of the exponential part of the refraction
field
where neff dependent on 1 and n1.
eff2
122
1221
221o
11p
n
c
sinnnResinnk
sink
sinv
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Example application 2from 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 reflectionFirst consider the case where both
regions 1 and 2 are lossless, i.e.,
n1 and n2 are real.
Letting N = n1 / n2 we have:
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, .
Since
we know
The refraction presents a surface wave propagating in the x direction.
21 sinsinN
222 j
22222 sinhcosjcoshsinsin
11
22 sinNcosh,2
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Example application 2
Analysis of total internal reflectionNow 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
Before addressing the value of we know that the constant-phase plane is perpendicular to the constant-amplitude plane because region 2 is lossless.
To find we let N = Nr + jNi where Nr and Ni are real.
2
cottanandRe 221
22
2
2
2
sinN41sinNsinN1cos 1
22i
2
122
122
12
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Example application 2
Analysis of total internal reflectionDefining when 1 tends
to /2 we get the limited value for as
Figure 2 shows the refraction anglevarious 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 valuesare less than 90°.
2
N41NN1cos
2i
222
12
2
2i
2r
2NNN
Fig. 2. The influence of medium loss to critical angle and refraction angle (Nr = 3.0).
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Summary sinjcosrerjyxz j
xyarctanandyxzr 22
ysinjycoseee xyjxz
ysinhxsinjycoshxcos2eezcos zjzj
jzcos2eezcosh zz
ysinhxcosjycoshxsinj2eezsin zjzj
jzsinj2eezsinh zz
0z,complexcforez zlncc
...,2,1,0nwheren2jrlnzln