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

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

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.

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)

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)

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.

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

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

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

Multiplication, division, and trig identities

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)

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

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

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

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

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, …

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

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

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

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, …

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

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

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

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

General powers of complex numbers Example

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 Because k2 is complex, 2 must also be complex

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

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.

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

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: 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.

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

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

Summary