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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 (callen@eecs.ku.edu)

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