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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 ([email protected]) Course website URL people.eecs.ku.edu/~callen/823/EECS823.htm 1 Outline Complex numbers and analytic functions Complex numbers • • • • Real, imaginary form: z x jy 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 2 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. 3 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) 4 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) 5 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. 6 Complex plane Graphically presenting complex numbers in the complex plane provides a means to visualize some complex values and operations. addition subtraction 7 Complex conjugate If z = x + jy then the complex conjugate of z is z* = x – jy Conjugates are useful since: 2 z z x 2 y 2 z , a purely real number z z* x Re{z} 2 z z* y Im{z} 2 Conjugates are useful in complex division z1 z1 z*2 z1 z*2 * z2 z2 z2 z2 2 Note that z1 x1x 2 y1y 2 x 2 y1 x1y 2 j z2 x 22 y 22 x 22 y 22 1 j j and 1 j j 8 Polar form of complex numbers Complex numbers can also be represented in polar format, that is, in terms of magnitude and angle. j Here z r e r cos j r sin (Euler ' s fomula ) x r cos and y r sin r z x y 2 2 y and arctan x Note that if z r e j then z* r e j 9 Multiplication, division, and trig identities Multiplication z1z 2 r1 e j 1 r2 e j2 r1 r2 e j1 2 j 1 Division Trig identities z1 r1 e r1 j1 2 e j 2 z 2 r2 e r2 e j e j cos 2 e j e j sin 2j 10 Polar form to express powers Powers The cube of z is z3 r 3 e j3 r 3 cos 3 j r 3 sin 3 If we let r = e, where is real (z = eej) then z 3 e 3 j e 3 j 3 e 3 e j 3 The general power of z is z n r n e j n r n cos n j r n sin n or (for r = e) z n en j e n e j n 11 Polar form to express roots 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 zn w 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 zk r e n j 2 k n 2k 2k r (cos j sin ) n n n 12 Polar form to express roots Example Solve the equation zn = 1, that is w = 1, r = 1, = 0 zk e j 2 k n 2k 2k cos j sin n n 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 13 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 14 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 15 Complex exponential functions The complex exponential function ez can be expressed in terms of its real and imaginary components ez e x j y e x cos y j sin y The product of two complex exponentials is e z1 ez 2 e z1 z 2 e x1 x 2 cosy1 y 2 j sin y1 y 2 Note that e jy cos y j sin y cos 2 y sin 2 y 1 therefore |ez| = ex. Also note that e e where n = 0, 1, 2, … jy j y 2 n cos y jsin y 16 Complex trigonometric functions As previously seen for a real value x cos x 1 jx e e jx 2 and sin x 1 jz and sin z e e jz 2j 1 jx e e jx 2j For a complex value z 1 jz cos z e e jz 2 Similarly Euler’s formula applies to complex values e jz cos z j sin z Focusing now on cos z we have 1 jx jy jx jy 1 y j x cos z (e e ) (e e e ye j x ) 2 2 17 Complex trigonometric functions Focusing on cos z we have 1 jx jy jx jy 1 y j x cos z (e e ) (e e e y e j x ) 2 2 e y ey cos x j sin x cos x j sin x 2 2 1 j e y e y cos x e y e y sin x 2 2 from calculus we know about hyperbolic functions 1 y e ey 2 sinh y tanh y cosh y cosh y and and 1 y e ey 2 cosh y coth y sinh y sinh y 18 Complex trigonometric functions So we can say cos z cos x cosh y j sin x sinh y We can similarly show that sin z sin x cosh y j cos x sinh y cos z cos 2 x sinh 2 y 2 sin z sin 2 x sinh 2 y 2 Formulas for real trig functions hold for complex values cos z1 z 2 cos z1 cos z 2 sin z1 sin z 2 sin z1 z 2 sin z1 cos z 2 sin z2 cos z1 cos 2 z sin 2 z 1 19 Complex trigonometric functions Example Solve for z such that cos z = 5 Solution We know cos z cos x cosh y j sin x sinh y 5 Let x = 0 or ±2n (n = 0, 1, 2, …) such that z = jy or cos jy cosh y 5 acosh 5 = 2.2924 Therefore z = ±2n ± j 2.2924, n = 0, 1, 2, … 20 Complex hyperbolic functions Complex hyperbolic functions are defined as cosh z 1 z e e z 2 and sinh z 1 z e e z 2 Therefore we know cosh jz cos z sinh jz j sin z and cos jz cosh z sin jz j sinh z 21 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 w ln z (for z 0) or e w z Recalling that z = re j we know that ln z ln r j y where r z , arctan x However note that the complex natural logarithm is multivalued ln z ln r j 2 n where n 0, 1, 2, ... 22 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 23 General powers of complex numbers General powers of a complex number z = x + jy is defined as z c ec ln z for c complex , z 0 If c = n = 1, 2, … then zn is single-valued If c = 1/n where n = 2, 3, … then z c n z e 1 n ln z since the exponent is multivalued with multiples of 2j/n If c is real and irrational or complex, then zc is infinitely manyvalued. Also, for any complex number a a z e z ln a 24 General powers of complex numbers Example 1 j2 j exp 2 jln 1 j 1 exp 2 jln 2 j 2nj 4 1 1 2e 4 2 n sin ln 2 j cos ln 2 2 2 25 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 k1 k2 k0 o o k o and n1 r 2 r 2 and n 2 r1 r1 1 r 2 r 2 Because k2 is complex, 2 must also be complex k 2 k2 j k2 and 2 2 j 2 26 Example application 1 Refraction angle at an air/lossy medium interface k2 and 2 are both complex k 2 k2 j k2 and 2 2 j2 The exponential part of the refraction field is e jk2 x x jk2 z z e Im k 2 sin 2 x Im k 2 cos 2 z jRe k 2 sin 2 x Re k 2 cos 2 z The constant-phase plane results when Rek 2 sin 2 x Rek 2 cos 2 z constant The constant-amplitude plane results when Imk 2 sin 2 x Imk 2 cos 2 z constant 27 Example application 1 Refraction angle at an air/lossy medium interface The aspect angle for the constant-phase plane is Re k 2 sin 2 tan 1 Re k cos 2 2 and the angle for the constant-amplitude plane is Imk 2 sin 2 tan Im k cos 2 2 1 Since k 1x = k 2x = k 1 sin 1 = k 2 sin 2 are real, we know 1 tan and 0 Re k 2 cot 2 1 Thus the complex refraction angle results in a separation of the planes of constant-phase and constant-amplitude. 28 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 vp sin k1 sin 1 ko n12 sin 1 Re 2 n 22 n12 sin 2 1 2 c n eff 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: N sin 1 sin 2 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, 2 2 j2 . Since sin 2 sin 2 cosh 2 j cos 2 sinh 2 1 we know 2 , 2 cosh N sin 1 2 The refraction presents a surface wave propagating in the x direction. 30 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 2 Re 2 and tan 1 cot 2 2 2 Before addressing the value of 2 we know that the constant-phase plane is perpendicular to the constant-amplitude plane because region 2 is lossless. To find 2 we let N = Nr + jNi where Nr and Ni are real. 2 2 1 N sin 1 1 2 cos N 2 2 sin 1 1 2 2 4 N i2 sin 2 1 31 Example application 2 Analysis of total internal reflection 2 2 2 Defining N N r Ni when 1 tends to /2 we get the limited value for as 2 1 N 1 2 cos 2 N 1 2 2 4 N i2 2 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 z x jy r e j r cos j sin r z x 2 y2 and arctan y x ez e x j y e x cos y j sin y sin z e 2 j sin x cosh y j cos x sinh y cos z e jz e jz 2 cos x cosh y j sin x sinh y jz sinh z e e jz 2 j sin jz cosh z ez e z 2 cos jz z ez ln z ln r j 2 n where n 0, 1, 2, ... z c ec ln z for c complex , z 0 33