1. 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 , 2003
Chris Allen
Course website URL people.eecs.ku.edu/~callen/823/EECS823.htm

2. Outline Complex 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

3. 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.

4. 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)

5. 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)

6. 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.

7. Complex plane
Graphically presenting complex numbers in the complex plane provides a means to visualize some complex values and operations.
addition
subtraction

8. Complex conjugate
If z = x + jy then the complex conjugate of z is z* = x – jy
Conjugates are useful since:
Conjugates are useful in complex division

9. Polar form of complex numbers
Complex numbers can also be represented in polar format, that is, in terms of magnitude and angle.
Here

10. Multiplication, division, and trig identities

11. Polar form to express powers
The cube of z is
If we let r = e, where  is real (z = eej) then
The general power of z is
or (for r = e)

12. Polar form to express 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
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

13. Polar form to express roots
Example
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 = j0.866
z13 = ej2 = 1
z2 = ej4/3 = -0.5 – j0.866
z23 = ej4 = 1

14. 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
z1 = j
z2 = -1
z3 = -j

15. 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 = j0.951
z2 = ej4/5 = j0.588
z3 = ej6/5 = j0.588
z4 = ej8/5 = – j0.951

16. Complex exponential functions
The 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, …

17. Complex trigonometric functions
As 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

18. Complex trigonometric functions
Focusing on cos z we have
from calculus we know
about hyperbolic functions

19. Complex trigonometric functions
So we can say
We can similarly show that
Formulas for real trig functions hold for complex values

20. Complex trigonometric functions
Example
Solve for z such that cos z = 5
Solution
We know
Let x = 0 or ±2n (n = 0, 1, 2, …) such that z = jy or
acosh 5 =
Therefore z = ±2n ± j , n = 0, 1, 2, …

21. Complex hyperbolic functions
Complex hyperbolic functions are defined as
Therefore we know
and

22. 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
Recalling that z = re j we know that
However note that the complex
natural logarithm is multivalued

23. Complex logarithmic functions
Examples (n = 0, 1, 2, …)
ln 1 = 0, ±2j, ±4j, … ln 4 = ± 2jn
ln -1 = ±j, ±3j, ±5j, … ln -4 = ± (2n + 1)j
ln j = j/2, -3j/2, 5j/2, … ln 4j = j/2 ± 2jn
ln -4j =  j/2 ± 2jn
ln (3-4j) = ln 5 + j arctan(-4/3) =  j ± 2jn
Note
Formulas for natural logarithms hold for complex values
ln (z1 z2) = ln z1 + ln z2 ln(z1/z2) = ln z1 – ln z2

24. General powers of complex numbers
General 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

25. General powers of complex numbers
Example

26. 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 , 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
Because k2 is complex, 2 must also be complex

27. Example application 1
Refraction angle at an air/lossy medium interface
k2 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

28. Example application 1
Refraction angle at an air/lossy medium interface
The 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.

29. 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
where neff dependent on 1 and n1. 29

30. 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 , 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:
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.

31. 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
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. 31

32. Example application 2 Analysis of total internal reflection
Defining when 1 tends to /2 we get the limited value for  as
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

33. Summary