ENGIN 211, Engineering Math. Complex Numbers

Transcription

1 ENGIN 211, Engineering Math Complex Numbers 1

2 Imaginary Number and the Symbol J Consider the solutions for this quadratic equation: x = 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

3 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

4 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 = j4 + j6 3 + (j6)( j4) = 6 + j8 + j18 j 2 24 = 6 + j = 18 + j26 4

5 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 = = 25 5

6 Division of Complex Numbers Division by a real number 3+j4 2 Division by another complex number = j2 3 + j4 3 + j4 2 + j3 = 2 j3 2 j3 2 + j j(8 + 9) = = j1.31 Division by itself = 6 + j j4 3 + j4 = 3 + j4 3 j4 3 + j4 3 j4 = = 1 6

7 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

8 Complex Plane (Argand Diagram) Imaginary axis z = a + jb b (a, b) a Real axis Graphical addition of two complex numbers

9 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 = = 10, θ = tan = 36o 52 9

10 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

11 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

12 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 = o, z 5 = 2 5 e j5 120o = o = o z 120 o 120 o z 5 12

13 Roots Roots of a complex number is somewhat complicated and requires careful attention in manipulation. Example, w = 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 = o because z 5 = 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 = o is still not recovered. 13

14 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 = 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 = o When do we stop? 14

15 Roots (Cont d) Let s plot these roots in the complex plane. We can see that these five roots are o o o 32e j120o 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

16 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