Complex Math Basics from Advanced Engineering Mathematics

1. Complex math basics material from Advanced Engineering Mathematics by E Kreyszigand fromLiu 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

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 thenthe complex conjugate of zis 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 • Multiplication • Division • Trig identities

11. Polar form to express powers • 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 • Roots • Given w = zn where n = 1, 2, 3, …, if w  0there 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 = Rnejn • 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 = -0.5 + 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 • z0 = +1 • 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 = 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

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 = 2.2924 • Therefore z = ±2n ± j 2.2924, 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 = rej 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 = 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

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 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 interface • A 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

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 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 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 1is 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 anglevarious non-dissipative (Ni = 0) and dissipative (Ni > 0) as a functionof incidence angle. Note that for Ni > 0, the abrupt slopechange at the critical angle becomes smooth and that the maximum  valuesare less than 90°. Fig. 2. The influence of medium loss to critical angle and refraction angle (Nr = 3.0). 32

33. Summary