Números Complexos e Operações

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Part 6 Complex Numbers 34 Complex numbers 34.1 Cartesian complex numbers Since x² + 4 = 0 then x² = -4 and x = √-4 i.e., (i) If the quadratic equation solved using the quadratic formula then: = ±j2, (since j = X = (Note that ±j2 may also be written as ±2j). 2(1) Problem 2. Solve the quadratic equation: = = 2 2 = = 2 Using the quadratic formula, It is not possible to evaluate in real 2(2) terms. However, if an operator j is defined as j = then the solution may be = ± 4 = 4 expressed as x = (ii) -1+j2 and -1-j2 are known as complex numbers. Both solutions are of the form = 4 a + jb, 'a' being termed the real part and jb the imaginary part. A complex number of the form a + jb is called a Cartesian Hence x = 4 3 +j 4 or 0.750 complex number. correct to 3 decimal places. (iii) In pure mathematics the symbol i is used to indicate (i being the first letter of the word imaginary). However i is the symbol (Note, a graph of y = + 3x + 5 does not cross of electric current in engineering, and to the x-axis and hence + 3x + 5 = 0 has no real avoid possible confusion the next letter in roots). the alphabet, j, is used to represent √-1 Problem 3. Evaluate Problem 1. Solve the quadratic equation: x²+4=0 (a) j³ (b) (c)292 ENGINEERING MATHEMATICS (a) Imaginary axis (b) B j4 j3 j2 A j =jx1=j -3 -2 -1 0 2 3 Real axis Hence -j2 -j3 D -j4 j5 Now try the following exercise Exercise 121 Further problems on the Figure 34.1 introduction to Cartesian complex numbers also shows the Argand points B, C and D represent- ing the complex numbers (-2+ (-3- j5) and In Problems 1 to 3, solve the quadratic equa- j3) respectively. tions. 1. 34.3 Addition and subtraction of 2. complex numbers 4 3 ± j 4 or 0.750 ± ] Two complex numbers are added/subtracted by adding/subtracting separately the two real parts and 3. + 7 = the two imaginary parts. For example, if = jb and Z₂ = 5 8 8 or ± j1.166 then = c) + + 4. Evaluate (a) (b) 1 (c) 4 and [(a) 1 (b) j (c) j2] Thus, for example, 34.2 The Argand diagram =5-j1 and A complex number may be represented pictorially on rectangular or Cartesian axes. The horizontal (or =-1+j7 x) axis is used to represent the real axis and the ver- The addition and subtraction of complex numbers tical (or y) axis is used to represent the imaginary may be achieved graphically as shown in the Argand axis. Such a diagram is called an Argand diagram. diagram of Fig. 34.2. (2+j3) is represented by vector In Fig. 34.1, the point A represents the complex OP and (3-j4) by vector 0Q. In Fig. 34.2(a), by vec- number (3+j2) and is obtained by plotting the co- tor addition, (i.e. the diagonal of the parallelogram), ordinates (3, j2) as in graphical work. Figure 34.1 OP 0Q = OR. R is the point (5,-j1).COMPLEX NUMBERS 293 Imaginary axis Problem 4. Given = + j4 and Z₂=3-j determine (a) + Z₂, (b) Z₂, (c) and show the j3) j3 results on an Argand diagram j2 (a) + Z₂ = (2 + j4) + (3 j) 0 = (2 + 3) + j(4 1) = 5+j3 1 2 3 4 \ 5 Real axis -j (b) Z₁ - Z₂ = (2 + j4) - (3 j) -j2 = (2 3) + j(4 - (-1)) = -j3 (c) Z₂ = j) - (2 + j4) -j4 Q(3 j4) = (3 2) + 4) = 1 j5 (a) Each result is shown in the Argand diagram of Fig. 34.3. Imaginary axis S(-1+j7) Imaginary j7 axis j5 j4 (5+j3) Q' j3 j3) j2 -1 0 1 2 3 4 5 Real axis -3 -2 -1 0 1 2 3 Real axis -j -j2 -j2 -j3 -j3 -j4 -j4 Q(3 j4) -j5 (b) Figure 34.3 Figure 34.2 Hence 34.4 Multiplication and division of complex numbers In Fig. 34.2(b), vector 0Q is reversed (shown as since it is being subtracted. (Note 0Q = 3- j4 (i) Multiplication of complex numbers is and 0Q' = j4) = + j4). achieved by assuming all quantities involved OP 0Q = OP + 0Q' = os is found to be the are real and then using = 1 to simplify. Argand point (-1, j7). Hence Hence = ac + a(jd) + (jb)c + (jb)(jd)294 ENGINEERING MATHEMATICS = ac + jad + jbc + = (-2 + 15) + + since = = bd) + since Thus (b) Z₃ = -3-j4 1 j3 = -3 1 j3 j4 X -3+j4 -3+j4 = = (12 (-10)) + = 25 j13 = 25 9 25 13 (ii) The complex conjugate of a complex num- or ber is obtained by changing the sign of the imaginary part. Hence the complex conju- gate of a + jb is jb. The product of a (c) Z₁Z₂ = complex number and its complex conjugate is always a real number. For example, = 13 -1+j2 + from part (a), = [(a+jb)(a-jb) may be evaluated 'on sight' - = (iii) Division of complex numbers is achieved by multiplying both numerator and denom- = 9-j37 5 = 9 5 j 5 or 1.8-j7.4 inator by the complex conjugate of the denominator. (d) Z₁Z₂Z₃ = + since For example, = 13 + j11, from part (a) j5 = j5 X j4) j4 + j4 j4) = -39 j52 j33 = Problem 6. Evaluate: = 25 j23 = -14 25 j 25 23 (a) + 2 (b) j Problem 5. If = Z₂ = (a) and determine in jb form: =1+j+j-1=j2 (a) Z₁Z₂ (b) Z₃ Z₁ (c) Z₁Z₂ (d) Z₁Z₂Z₃ Hence + 2 = -4 2 = 2 1 (a) 1+j3 j3 1+j2 (b) j2 = j2 X 1+j2COMPLEX NUMBERS 295 = j5 = 5 (a) 26 3 + 41 (b) 45 26 j 9 8. Evaluate (a) 1-j 1+j (b) 1+j 1 (a) j (b) 2 1 Hence j = = = 9. Show that: = 57 + j24 since Now try the following exercise 34.5 Complex equations Exercise 122 Further problems on opera- If two complex numbers are equal, then their real tions involving Cartesian parts are equal and their imaginary parts are equal. complex numbers Hence if a+jb=c+jd, then a= and 1. Evaluate (a) (3 + j2) + (5 j) and (b) (-2 + j6) j2) and show the Problem 7. Solve the complex equations: results on an Argand diagram. (a) 2(x + jy) = 6 j3 [(a) 8+j 2. Write down the complex conjugates of (a) 3 + j4, (b) j (a) 2(x + jy) = j3 hence [(a) j4 (b) 2 + j] Equating the real parts gives: In Problems 3 to 7 evaluate in a + jb form given = j2, Z₂ = j3, Z₃ = and Equating the imaginary parts gives: 3. (a) (b) [(a) j4 (b) (b) 4. (a) Z₁Z₂ (b) Z₃Z₄ Hence 4 j7 = + jb [(a) 10+j5 (b) Equating real and imaginary terms gives: 5. (a) Z₁Z₃ + Z₄ (b) Z₁Z₂Z₃ Problem 8. Solve the equations: 6. (a) Z₂ (b) Z₂ Z₃ (a) -2 25 + 11 (b) -19 85 43 (a) j3) = + 7. (a) Z₁Z₃ (b) Hence296 ENGINEERING MATHEMATICS i.e. 34.6 The polar form of a complex Hence number and -5-j12=a+jb (i) Let a complex number Z be x+jy as shown in the Argand diagram of Fig. 34.4. Let Thus a = -5 and 12 distance OZ be r and the angle OZ makes with the positive real axis be (b) From trigonometry, Hence Hence Equating real and imaginary parts gives: = sin (1) Z = r(cos j sin is usually abbreviated and (2) to Z = which is known as the polar i.e. two stimulaneous equations to solve form of a complex number. Multiplying equation (1) by 2 gives: Imaginary (3) axis, Z Adding equations (2) and (3) gives: r jy i.e. x = -7 A Real axis From equation (1), y = 9, which may be X checked in equation (2) Figure 34.4 Now try the following exercise (ii) r is called the modulus (or magnitude) of Z and is written as mod Z or |Z|. Exercise 123 Further problems on com- r is determined using Pythagoras' theorem plex equations on triangle OAZ in Fig. 34.4, In Problems 1 to 4 solve the complex equa- tions. i.e. 1. (iii) is called the argument (or amplitude) of [a = 8, and is written as arg 2. 3 = By trigonometry on triangle OAZ, 3. x (iv) Whenever changing from Cartesian form 4. to polar form, or vice-versa, a sketch is invaluable for determining the quadrant in which the complex number occurs 5. If Z = R + + express Z in (a + jb) form when R = 10, L = 5, Problem 9. Determine the modulus and = 0.04 and w = 4 argument of the complex number and express Z in polar formCOMPLEX NUMBERS 297 Z = 2+ j3 lies in the first quadrant as shown in Imaginary Fig. 34.5. axis Modulus, = r = + = or 3.606, (-3+j4) j4 (3+j4) correct to 3 decimal places. j3 Argument, arg = = 3 r j2 r 2 j = 56.31° or 56°19' -3 1 2 3 Real axis In polar form, 2+ j3 is written as or -j r -j2 r Imaginary -j3 axis -4 (-3 j4) j4) j3 Figure 34.6 r Hence the argument = 180° + 53.13° = 233.13°, which is the same as -126.87° 0 2 Real axis Hence( -j4) = or 126.87° (By convention the principal value is nor- Figure 34.5 mally used, i.e. the numerically least value, such that -π