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 z 1 = (x 1, y 1 ) and z 2 = (x 2, y 2 ) Addition z 1 + z 2 = (x 1, y 1 ) + (x 2, y 2 ) = (x 1 + x 2, y 1 + y 2 ) Multiplication z 1 z 2 = (x 1, y 1 ) (x 2, y 2 ) = (x 1 x 2 - y 1 y 2, x 1 y 2 + x 2 y 1 )

5 Complex numbers The imaginary unit, denoted by i or j, where i = (0, 1) which has the property that j 2 = -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 z 1 = x 1 + jy 1 and z 2 = x 2 + jy 2 Addition z 1 + z 2 = (x 1 + x 2 ) + j(y 1 + y 2 ) Multiplication z 1 z 2 = (x 1 x 2 - y 1 y 2 ) + j(x 1 y 2 + x 2 y 1 )

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. subtraction addition

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 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  e j  ) then The general power of z is or (for r = e  )

12 Polar form to express roots Roots Given w = z n where n = 1, 2, 3, …, if w  0 there are n solutions Each solution called an n th root of w can be written as Note that w has the form w = r e j  and z has the form of z = R e j  so z n = R n e jn  where R n = r and n  =  + 2k  (where k = 0, 1, …, n -1 ) The k th general solution has the form

13 Polar form to express roots Example Solve the equation z n = 1, that is w = 1, r = 1,  = 0 For n = 3, solutions lie on the unit circle at angles 0, 2  /3, 4  /3 z 3 = 1 z 0 = e j0/3 = 1 z 1 = e j2  /3 = j0.866 z 1 3 = e j2  = 1 z 2 = e j4  /3 = -0.5 – j0.866 z 2 3 = e j4  = 1

14 Polar form to express roots Example Solve the equation z n = 1, for n = 2, solutions lie on the unit circle at angles 0,  z 0 = +1 z 1 = -1 Solve the equation z n = 1, for n = 4, solutions lie on the unit circle at angles 0,  /2, , 3  /2 z 0 = +1 z 1 = j z 2 = -1 z 3 = -j

15 Polar form to express roots Example Solve the equation z n = 1, for n = 5, solutions lie on the unit circle at angles 0, 2  /5, 4  /5, 6  /5, 8  /5 z 5 = 1 z 0 = e j0/5 = 1 z 1 = e j2  /5 = j0.951 z 2 = e j4  /5 = j0.588 z 3 = e j6  /5 = j0.588 z 4 = e j8  /5 = – j0.951

16 Complex exponential functions The complex exponential function e z can be expressed in terms of its real and imaginary components The product of two complex exponentials is Note that therefore |e z | = e x. 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 H owever 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 (z 1 z 2 ) = ln z 1 + ln z 2 ln(z 1 /z 2 ) = ln z 1 – ln z 2

24 General powers of complex numbers General powers of a complex number z = x + jy is defined as If c = n = 1, 2, … then z n 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 z c 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 , 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 k 2 is complex,  2 must also be complex

27 Example application 1 Refraction angle at an air/lossy medium interface k 2 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 n eff dependent on  1 and n 1.

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 , Analysis of total internal reflection First consider the case where both regions 1 and 2 are lossless, i.e., n 1 and n 2 are real. Letting N = n 1 / n 2 we have: If N > 1 ( i.e., n 1 > n 2 ) 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 k 1 is complex, k 2 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 = N r + jN i where N r and N i are real.

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 (N i = 0) and dissipative (N i > 0) as a function of incidence angle. Note that for N i > 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 (N r = 3.0).

33 Summary