ENGIN 211, Engineering Math Complex Numbers 1
Imaginary Number and the Symbol J Consider the solutions for this quadratic equation: x 2 + 1 = 0 x = ± 1 1 is called the imaginary number, and we use the symbol j to represent it: j = 1. Thus, the solutions Obviously, x = ±j j 2 = 1 This notation allowed us to deal with a large number of quadratic equations of the form: ax 2 + bx + c = 0 with the solution In case b 2 4ac < 0, so we can rewrite x = b± b2 4ac 2a x = b ± j 4ac b2 2a 2
Complex Numbers Example: x 2 + 2x + 3 = 0 We have two roots x 1 = 1 + j 2 and x 2 = 1 j 2 Both roots consisting of a real number and an imaginary number are complex number. Re(x 1 ) = 1 real part of the complex number x 1 Im(x 1 ) = 2 imaginary part of the complex number x 1 Re(x 2 ) = 1 real part of the complex number x 2 Im x 2 = 2 imaginary part of the complex number x 2 Note: Both Re(x) and Im(x) are real numbers themselves. 3
Operations of Complex Numbers Addition: 2 + j6 + 3 j4 = 1 + j2 Subtraction: 2 + j6 3 j4 = 5 + j10 Powers of j, j 2 = 1, j 3 = j 2 j = j, j 4 = j 2 j 2 = 1 1 = 1, Multiplication: 2 + j6 3 j4 = 2 3 + 2 j4 + j6 3 + (j6)( j4) = 6 + j8 + j18 j 2 24 = 6 + j26 1 24 = 18 + j26 4
Complex Conjugates Complex conjugate: a + jb and a jb are complex conjugate pairs a + jb a jb = a 2 jb 2 = a 2 + b 2 always a real number Example: 3 + j4 3 j4 = 3 2 + 4 2 = 25 5
Division of Complex Numbers Division by a real number 3+j4 2 Division by another complex number = 1.5 + j2 3 + j4 3 + j4 2 + j3 = 2 j3 2 j3 2 + j3 6 12 + j(8 + 9) = 4 + 9 = 0.46 + j1.31 Division by itself = 6 + j17 13 3 + j4 3 + j4 = 3 + j4 3 j4 3 + j4 3 j4 = 25 25 = 1 6
Equal Complex Numbers If two complex number are equal, their corresponding real and imaginary parts must be equal: Implies: Because we can write a + jb = c + jd a = c, and b = d a c = j(d b) And a real number never equals an imaginary number. Example: Then x 1 = 6, and x 2 = 8. x = x 1 + jx 2 = 6 + j8 7
Complex Plane (Argand Diagram) Imaginary axis z = a + jb b (a, b) a Real axis Graphical addition of two complex numbers 3 2 1 2 3 5 8
Polar Form Imaginary axis z = a + jb r b θ a Conversion between polar and rectangular forms r: modulus of the complex number z θ: argument of the complex number z Real axis r = a 2 + b 2, θ = tan 1 b a a = r cos θ, b = r sin θ, π θ π Example: z = 6 j8 can be converted into polar form, r = 6 2 + 8 2 = 10, θ = tan 1 8 6 = 36o 52 9
Exponential Form Maclaurin Series Let x = jθ, e x = 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + e jθ = 1 + jθ + jθ 2 Thus, r cos θ + j sin θ 2! + jθ 3 3! + jθ 4 4! + jθ 5 5! + = 1 θ2 2! + θ4 θ3 + j θ 4! 3! + θ5 5! = cos θ + j sin θ Euler s identity = re jθ exponential form Three ways: z = a + jb = r cos θ + j sin θ = re jθ 10
Which Form Is Better? It depends on what you want to do: For addition and subtraction, it is better to work with the rectangular form z = a + jb For multiplication and division, it is better to work with exponential form, z = re jθ, e.g., 4e j60o 2e j30o = 2e j 60o 30 o = 2e j30o To convert between rectangular and exponential form, we work thru polar form, z = r cos θ + j sin θ 3 + j4 = 5 cos 36 o 52 + j sin 36 o 52 = 5e j36o 52 11
Powers Consider: z = re jθ z 2 = re jθ 2 = r 2 e j2θ = r 2 cos 2θ + j sin 2θ, z n = re jθ n = r n e jnθ = r n cos nθ + j sin nθ DeMoivre s theorem If nθ > π, we need to add or subtract multiples of 2π so that the equivalent angle φ stays within φ π. Example: z = 2e j120o = 2 120 o, z 5 = 2 5 e j5 120o = 32 600 o = 32 120 o z 120 o 120 o z 5 12
Roots Roots of a complex number is somewhat complicated and requires careful attention in manipulation. Example, w = 32 120 o = 32e j120o, let s find w 1/5 =? 1) We can take do the following 32e j120o 1/5 = 32 1/5 e j120o /5 = 2e j24o = 2 24 o But we certainly expect to recover z = 2e j120o = 2 120 o because z 5 = 32 120 o from the previous example of powers, so what happened? 2) Of course, we can say 120 o is also 360 o 120 o = 240 o, so now 32e j120o 1/5 = 32e j240 o 1/5 = 32 1/5 e j240o /5 = 2e j48o = 2 48 o But the solution z = 2e j120o = 2 120 o is still not recovered. 13
Roots (Cont d) 3) Well, we can keep adding 720 o 120 o = 600 o, so now 32e j120o 1/5 = 32e j600 o 1/5 = 32 1/5 e j600o /5 = 2e j120o = 2 120 o This is it, but how do we know we have gotten them all? 4) We can also subtract 360 o, so 360 o 120 o = 480 o, then 32e j120o 1/5 = 32e j480 o 1/5 = 32 1/5 e j480o /5 = 2e j96o = 2 96 o 5) Of course, we can also obtain 32e j120o 1/5 = 32e j840 o 1/5 = 32 1/5 e j840o /5 = 2e j168o = 2 168 o When do we stop? 14
Roots (Cont d) Let s plot these roots in the complex plane. We can see that these five roots are 2 120 o 2 3 2 168 o 4 2 96 o 32e j120o 1 5 2 48 o 2 24 o equally separated by 360o 5 = 72 o. It is generally true that the n-th order roots have n number of values that are equally separated by 360o n. So in practice, we only need to find one as we did in Step 1), and the remaining n 1 roots will be generated similar to the illustration in the complex plane plot. The one nearest to the positive real axis is called the principal root. 15
Summary Key points: Recognize j = 1 for imaginary number Three forms of the complex number: rectangular, polar, and exponential Conversions between the three forms Complex conjugate pairs Addition, subtraction, multiplication (powers), and division Roots of n-th order (equally spaced by 360 o /n on a circle) 16