CHAPTER 45 COMPLEX NUMBERS

CHAPTER 45 COMPLEX NUMBERS EXERCISE 87 Page 50. Solve the quadratic equation: x + 5 0 Since x + 5 0 then x 5 x 5 ( )(5) 5 j 5 from which, x ± j5. Solve the quadratic equation: x x + 0 Since x x + 0 then [ ] ± ( ) 4()() ± 4 ± ( )(4) ± ( ) (4) ± j (4) x () ± j 4 ± j ± j ± j 3. Solve the quadratic equation: x 4x + 5 0 Since x 4x + 5 0 then [ ] 4 ± ( 4) 4()(5) 4± 4 4 ± ( )(4) 4 ± ( ) (4) 4 ± j (4) x () 4 ± j 4 4 ± j ± j ± j 4. Solve the quadratic equation: x 6x + 0 0 Since x 6x + 0 0 then [ ] 6 ± ( 6) 4()(0) 6± 4 6 ± ( )(4) 6 ± ( ) (4) 6 ± j (4) x () 757 04, John Bird

6 ± j 4 6 ± j 3 ± j 5. Solve the quadratic equation: x x + 0 Since x x + 0 then [ ] ± ( ) 4()() ± 4 ± ( )(4) ± ( ) (4) ± j (4) x () 4 4 4 4 ± j 4 ± j 0.5 ± j0.5 4 4 4 4 6. Solve the quadratic equation: x 4x + 8 0 Since x 4x + 8 0 then [ ] 4 ± ( 4) 4()(8) 4 ± 6 4 ± ( )(6) 4 ± ( ) (6) 4 ± j (6) x () 4 ± j 6 4 ± j 4 ± j 7. Solve the quadratic equation: 5x 0x + 0 Since 5x 0x + 0 then [ ] 0 ± ( 0) 4(5)() 0 ± 00 0 ± ( )(00) 0 ± ( ) (00) x (5) 50 50 50 0 ± j (00) 50 0 00 0 0 ± j ± j 0. ± j0. 50 50 50 50 8. Solve the quadratic equation: x + 3x + 4 0 Since x + 3x+ 4 0 then [ ] 3 ± 3 4()(4) 3 ± 3 3 ± ( )(3) 3 ± ( ) (3) 3 ± j (3) x () 4 4 4 4 758 04, John Bird

3 ± j 3 or ( 0.750 ± j.99) 4 4 9. Solve the quadratic equation: 4t 5t + 7 0 Since 4t 5t + 7 0 then [ ] 5 ± ( 5) 4(4)(7) 5 ± 87 5 ± ( )(87) 5 ± ( ) (87) 5 ± j (87) t (4) 8 8 8 8 5 ± j 87 or (0.65 ± j.66) 8 8 0. Evaluate (a) j 8 (b) j 7 (c) 4 j 3 (a) 8 j ( j ) 4 ( ) 4 3 3 (b) 7 6 ( ) ( ) j j j j j j j Hence, j j j j j7 j j j( j) j ( ) j (c) ( ) 6 j3 j j j j j ( ) 6 j Hence, 4 j 3 ( j) j j j j j( j) j 759 04, John Bird

EXERCISE 88 Page 53. Evaluate (a) (3 + j) + (5 j) and (b) ( + j6) (3 j) and show the results on an Argand diagram. (a) (3 + j) + (5 j) (3 + 5) + j( ) 8 + j (b) ( + j6) (3 j) + j6 3 + j ( 3) + j(6 + ) 5 + j8 (8 + j) and ( 5 + j8) are shown on the Argand diagram below.. Write down the complex conjugates of (a) 3 + j4, (b) j. (a) The complex conjugate of 3 + j4 is: 3 j4 (b) The complex conjugate of j is: + j 3. If z + j and w 3 j evaluate (a) z + w (b) w z (c) 3z w (d) 5z + w (e) j(w 3z) (f) jw jz (a) z + w ( + j) + (3 j) + j + 3 j 5 (b) w z (3 j) ( + j) 3 j j j (c) 3z w 3( + j) (3 j) 6 + j3 6 + j j5 (d) 5z + w 5( + j) + (3 j) 0 + 5j + 6 j 6 + j3 (e) j(w 3z) j[(6 j) (6 +j3)] j[6 j 6 j3] j( j5) j 5 ( )5 5 (f) jw jz j(3 j) j( + j) j6 j j j j6 ( ) j ( ) j6 + j + 3 + j4 4. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) + 3 (b) + 4 760 04, John Bird

(a) + 3 + j + 4 j3 ( + j3) + j + 4 j3 + j3 ( + 4 + ) + j( 3 3) 7 j4 (b) + 4 (4 j3) ( + j) + ( 5 j) 4 j3 j 5 j (4 5) + j( 3 ) j6 5. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) (b) 3 4 (a) ( + j)(4 j3) 4 j3 + j8 j 6 4 j3 + j8 + 6 0 + j5 (b) 3 4 ( + j3)( 5 j) 0 + j j5 j 3 0 + j j5 + 3 3 j3 6. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) 3 + 4 (b) 3 (a) 3 4 + ( + j)( + j3) + ( 5 j) + j3 j4 + j 6 5 j + j3 j4 6 5 j 3 j (b) 3 ( + j)(4 j3)( + j3) (4 j3 + j8 j 6)( + j3) ( 0 + j5)( + j3) 0 + j30 j0 + j 5 0 + j30 j0 5 35 + j0 7. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) (b) + 3 4 (a) j j j j j j j j j j j j + (+ )(4+ 3) 4+ 3+ 8+ 6 4+ 3+ 8 6 + 4 3 (4 3)(4 + 3) 4 + 3 5 5 + j 5 5 (b) + 3 + j + + j + j + j + j j + j + j + j j j j j + j + 4 ( ) ( 3) 5 ( 5)(9 ) 9 45 0 9 43 (4 3) ( 5 ) 9 (9 )(9 ) 9 85 9 43 + j 85 85 76 04, John Bird

8. Evaluate in a + jb form, given + j, 4 j3, 3 + j3 and 4 5 j (a) 3 + 3 (b) + + 3 4 (a) 3 + 3 (+ j)( + j3) + j3 j4+ j6 8 j (+ j) + ( + j3) + j5 + j5 ( 8 j)( j5) 8 + j40 + j+ j5 3 + j4 ( + j5)( j5) + 5 6 3 + j 4 6 6 (b) + + (4 j3) + + j 5 j 3 4 (+ j)( 5 + j) + ( + j3) 4 j3 + 5 + + j3 4 j3 + + + 6 5 j j0 j + j3 4 j3 + 7 j9 6 + j3 7 j 9 6 6 5 7 j 9 45 j 9 6 6 6 6 6 9. Evaluate (a) + j j (b) + j (a) j ( j)( j) j j+ j j j ( j)( j) + + + 0 j j (b) ()( j) j j + + + j ( j)( j) j 0. Show that 5 + j j5 3+ j4 j 57 + j4 + j ( + j)(3 j4) 3 j4 + j6 j8 + j + j 3 + j4 3 + 4 5 5 5 5 j5 ( j5)( j) j j5 5+ j j j( j) j 5+ j 76 04, John Bird

L.H.S. 5 + j j5 5 5 j (5 j) + + 5 + j 3+ j4 j 5 5 5 5 5 5 50 5 4 48 j j + 5 5 5 5 5 4 5 48 + j 57 + j4 R.H.S. 5 5 763 04, John Bird

EXERCISE 89 Page 54. Solve: ( + j)(3 j) a + jb ( + j)(3 j) a + jb Hence, 6 j4 + j3 j a + jb 8 j a + jb Thus, a 8 and b. Solve: + j j j(x + jy) + j j ( x + jy) j hence, ( + j)( + j) j( x + jy) ( j)( + j) Hence, x 3 + j+ j+ j + jx + + j3 jx y j y 3 + j y + jx and y 3. Solve: ( j3) ( a+ b) Squaring both sides gives: ( ) ( j3) ( a + jb) j3 a + jb ( j3)( j3) a + jb 4 j6 j6 + j 9 a + jb 5 j a + jb Hence, a 5 and b 4. Solve: (x jy) (y jx) + j 764 04, John Bird

(x jy) (y jx) + j Hence, (x y) + j( y + x) + j x y () and x y () () () gives: y Substituting in () gives: x from which, x 3 5. If R + jωl + /jωc, express in (a + jb) form when R 0, L 5, C 0.04 and ω 4 R + jωl + jω C 0 + j(4)(5) + j (4)(0.04) 0 + j0 + 6.5 j 0 + j0 + 6.5( j ) j( j) 0 + j0 6.5 0 + j0 j6.5 j 0 + j3.75 765 04, John Bird

EXERCISE 90 Page 57. Determine the modulus and argument of (a) + j4 (b) 5 j (c) j( j) (a) + j4 lies in the first quadrant as shown below Modulus, r 4 + 4.47 Argument, θ tan 4 63.43 (b) 5 j lies in the third quadrant as shown below Modulus, r 5 + 5.385 α tan.80 5 Hence, argument, θ (80.80 ) 58.0 (c) j( j) j j j + or + j + j lies in the first quadrant as shown below. Modulus, r +.36 Argument, θ tan 63.43 766 04, John Bird

. Express in polar form, leaving answers in surd form: (a) + j3 (b) 4 (c) 6 + j (a) + j3 From the diagram below, r + 3 3 and θ 3 tan 56.3 or 56 9' Hence, + j3 3 56.3 in polar form (b) 4 4 + j0 and is shown in the diagram below, where r 4 and θ 80 Hence, 4 4 80 in polar form (c) 6 + j From the diagram below, r 6 + 37 and α tan 9.46 6 thus θ 80 9.46 70.54 Thus, 6 + j 37 70.54 3. Express in polar form, leaving answers in surd form: (a) j3 (b) ( + j) 3 (c) j 3 ( j) 767 04, John Bird

(a) j3 From the diagram below, r 3 and θ 90 Hence, j3 3 90 in polar form (b) ( + j) 3 ( + j)( + j)( + j) (4 j j + j )( + j) (3 j4)( + j) 6 + j3 + j8 j 4 + j From the diagram below, r + 5 and α tan 79.70 and θ 80 79.70 00.30 Hence, ( ) 3 j + + j 5 00.30 in polar form (c) j3( j) (j)( j )( j) j( j) j + j j From the diagram below, r + and α tan 45 and θ 80 45 35 Hence, j 3 ( j) j 35 4. Convert into (a + jb) form giving answers correct to 4 significant figures: (a) 5 30 (b) 3 60 (c) 7 45 768 04, John Bird

(a) 5 30 5 cos 30 + j 5 sin 30 4.330 + j.500 (b) 3 60 3 cos 60 + j 3 sin 60.500 + j.598 (c) 7 45 7 cos 45 + j 7 sin 45 4.950 + j4.950 5. Convert into (a + jb) form giving answers correct to 4 significant figures: (a) 6 5 (b) 4 π (c) 3.5 0 (a) 6 5 6 cos 5 + j 6 sin 5 3.44 + j4.95 (b) 4 π 4 cos π + j sin π 4.000 + j0 (Note that π is radians) (c) 3.5 0 3.5 cos( 0 ) + j 3.5 sin( 0 ).750 j3.03 6. Evaluate in polar form: (a) 3 0 5 45 (b).4 65 4.4 (a) 3 0 5 45 3 5 (0 + 45 ) 45 65 (b).4 65 4.4.4 4.4 (65 + ) 0.56 44 7. Evaluate in polar form: (a) 6.4 7 5 (b) 5 30 4 80 0 40 (a) 6.4 7 5 6.4 7 6.4 7 5 5 3. 4 (b) 5 30 4 80 0 40 5 30 4 80 5 4 (30 + 80 40 ) 0 40 0 50 π π 8. Evaluate in polar form: (a) 4 + 3 (b) 0 + 5. 58.6 40 6 8 π π (a) 4 + 3 4cos π j4sin π 3cos π j3sin π + + + 6 8 6 6 8 8 (3.464 + j) + (.77 + j.48) 6.36 + j3.48 From the diagram below, r 6.36 + 3.48 6.986 769 04, John Bird

and θ 3.48 tan 6.79 or 0.467 rad 6.36 π π Hence, 4 + 3 6.986 6.79 or 6.986 0.467 rad 6 8 (b) 0 + 5. 58.6 40 ( cos 0 + j sin 0 ) + (5. cos 58 + j5. sin 58 ) (.6 cos( 40 ) + j.6 sin( 40 )) ( + j.73) + (.756 + j4.40) (.6 j.08) + j.73 +.756 + j4.40.6 + j.08 0.530 + j7.70 From the diagram below, r 0.530 + 7.70 7.90 and θ 7.70 tan 85.77 0.530 Hence, 0 + 5. 58.6 40 7.90 85.77 770 04, John Bird

EXERCISE 9 Page 59. Determine the resistance R and series inductance L (or capacitance C) for each of the following impedances assuming the frequency to be 50 Hz: (a) (3 + j8) Ω (b) ( j3) Ω (c) j4 Ω (d) 8 60 Ω (a) If (3 + j8) Ω then resistance, R 3 Ω and inductive reactance, X L 8 Ω (since the j X L πfl 8 hence, inductance, L 8 8 0.055 H or 5.5 mh π f π(50) term is positive) (b) If ( j3) Ω then resistance, R Ω and capacitive reactance, X C 3 Ω (since the j X C term is negative) π fc 3 hence, capacitance, C.06 0 3 or 06 0 6 π f (3) π(50)(3) 06 µf (c) If j4 Ω (0 + j4) Ω then resistance, R 0 Ω and X L 4 Ω πfl 4 hence, inductance, L 4 π (50) 0.04456 H or 44.56 mh (d) If 8 60 Ω 8 cos( 60 ) + j8 sin( 60 ) (4 j6.98) Ω Hence, resistance, R 4 Ω and X C 6.98 Ω π fc 6.98 and capacitance, C 459.4 0 6 459.4 µf π (50)(6.98). Two impedances, (3 + j6) Ω and (4 j3) Ω are connected in series to a supply voltage of 0 V. Determine the magnitude of the current and its phase angle relative to the voltage In a series circuit, total impedance, TOTAL + (3 + j6) + (4 j3) (7 + j3) Ω Since voltage V 0 0 V, then current, I V 3 7 + 3 tan 7 7.66 3.0 Ω the current is 5.76 A and is lagging the voltage by 3.0 0 0 5.76 3.0 A 7.66 3.0 77 04, John Bird

3. If the two impedances in Problem are connected in parallel, determine the current flowing and its phase relative to the 0 V supply voltage. In a parallel circuit shown below, the total impedance T is given by: 3 j6 4+ j3 3 6 4 3 + + + j + + j j j T 3 + 6 4 3 3 + 6 4 + 3 45 45 5 5 T admittance, YT 0.667 j0.0333 0.7 3.37 siemen V Current, I VYT (0 0 )(0.7 3.37 ) 7.5 3.37 A T the current is 7.5 A and is lagging the voltage by 3.37 4. A series circuit consists of a Ω resistor, a coil of inductance 0.0 H and a capacitance of 60 µf. Calculate the current flowing and its phase relative to the supply voltage of 40 V, 50 Hz. Determine also the power factor of the circuit. R Ω, inductive reactance, X L πfl π(50)(0.0) 3.46 Ω and capacitive reactance, X C π fc π 50 60 0 9.894 Ω 6 ( )( ) Hence, impedance, R + j( XL XC) + j(3.46 9.894) ( + j.5) Ω 6.64 43.83 Ω Current flowing, I V 40 0 6.64 43.83 4.4 43.83 A Phase angle 43.83 lagging ( I lags V by 43.83 ) Power factor cos ϕ cos 43.83 0.7 77 04, John Bird

5. For the circuit shown, determine the current I flowing and its phase relative to the applied voltage. 30 + j0 40 j50 + + + + + + j j T 3 30 0 40 + 50 5 30 + 0 40 + 50 5 T 30 0 40 50 + j + j + 300 300 400 400 5 admittance, Y T 0.0783 + j0.0039 0.079.5 S V Current, I VYT (00 0 )(0.079.5 ) 4.6.5 A T the current is 4.6 A and is leading the voltage by.5 6. Determine, using complex numbers, the magnitude and direction of the resultant of the coplanar forces given below, which are acting at a point. Force A, 5 N acting horizontally, Force B, 9 N acting at an angle of 35 to force A, Force C, N acting at an angle of 40 to Force A. Resultant force F + F + F 5 0 + 9 35 + 40 A B C (5 + j0) + ( 6.364 + j6.364) + ( 6 j0.39) 5 + j0 6.364 + j6.364 6 j0.39) 7.364 j4.08 8.394 5.3 or 8.394 08.68 N Hence, the magnitude of the force that has the same effect as the three forces acting separately is: 8.39 N and its direction is 08.68 to the horizontal ( from Force A) 773 04, John Bird

7. A delta-connected impedance A is given by: A + + 3 3 Determine A in both Cartesian and polar form given (0 + j0) Ω, (0 j0) Ω and 3 (0 + j0) Ω A + 3 + 3 (0 + j0)(0 j0) + (0 j0)(0 + j0) + (0 + j0)(0 + j0) (0 j0) j00 j00 j00 + 00 + j00 00 j00 00 j00 j0 j0 j0 j0 j00 + 0 (0 + j 0) Ω 0 From the diagram below, r 0 + 0.36 and θ 0 tan 63.43 0 Hence, (0 + j0) Ω.36 63.43 Ω A 8. In the hydrogen atom, the angular momentum p of the de Broglie wave is given by jh pψ (± jmψ). Determine an expression for p. π jh π If pψ ( ± jmψ) jh ± jmψ jh h π Ψ π π then p ( ± jm ) ( j )( ± m) h π ± mh ± π ( m) 774 04, John Bird

9. An aircraft P flying at a constant height has a velocity of (400 + j300)km/h. Another aircraft Q at the same height has a velocity of (00 j600) km/h. Determine (a) the velocity of P relative to Q, and (b) the velocity of Q relative to P. Express the answers in polar form, correct to the nearest km/h. (a) The velocity of P relative to Q vp vq (400 + j300) (00 j600) 400 + j300 00 + j600 00 + j900 9 77.47 the velocity of P relative to Q is 9 km/h at 77.47 (b) The velocity of Q relative to P v Q v P (00 j600) (400 + j300) 00 j600 400 j300 00 j900 9 0.53 the velocity of Q relative to P is 9 km/h at 0.53 0. Three vectors are represented by P, 30, Q, 3 90 and R, 4 60. Determine in polar form the vectors represented by (a) P + Q + R, (b) P Q R. (a) P + Q + R 30 + 3 90 + 4 60 (.73 + j) + (0 + j3) + ( j3.464) (3.73 + j0.536) 3.770 8.7 (b) P Q R 30 3 90 4 60 (.73 + j) (0 + j3) ( j3.464) ( 0.68 + j.464) From the diagram below, r.488 and α.464 tan 79.63 0.68 and θ 80 79.63 00.37 775 04, John Bird

Hence, P Q R.488 00.37. In a Schering bridge circuit, x ( RX jx C X ) R where 4 4 X C π fc, jx C,. At balance: ( )( ) ( )( ) X 3 4 3. ( R3 )( jx C3 ) ( R3 jx C3 ) and Show that at balance R X C3R4 and C C X C R R 3 4 Since ( X )( 3) ( )( 4) then Thus, ( R3 )( jx C3 ) ( RX jx CX ) jx C R R3 jx C3 ( R jx ) X CX ( )( 4 ) ( R3 jx C3)( jx C)( R4) ( R3 )( jx C3 ) jr X R j X 3X R ( R jx ) + X C X 3 C 4 C C 4 ( R3)( jx C3) ( R3)( jx C3) XCR4 XCR4 ( RX jx CX ) X R ( j) C 3 C 3 ( 3 ) XCR X R + X jr 4 C 4 3 ( R jx ) X C X XCR X R j X R C 3 4 C 4 3 Equating the real parts gives: R X R X C R π fc fc X C3 π fc π fc 4 4 π 3 3 R 4 CR 3 4 RX C XC R4 Equating the imaginary parts gives: X CX R3 R4 π fc R4 π fc R π fc R X 3 3 from which, C X CR R 3 4 776 04, John Bird

. An amplifier has a transfer function T given by T 500 + jω 5 0 4 ( ) where ω is the angular frequency. The gain of the amplifier is given by the modulus of T and the phase is given by the argument of T. If ω 000 rad/s, determine the gain and the phase (in degrees). When ω 000 rad/s, transfer function T 500 500 500 ( ) + jω 5 0 4 + j(000)(5 0 4) + j Hence, T 500 (500)( j) 500 j500 500 j500 500 500 j + j (+ j)( j) + 50 j50 353.6 45 Hence, the gain of the amplifier 353.6 and the phase is 45 3. The sending end current of a transmission line is given by I S VS tanh PL. Calculate the value of the sending current, in polar form, given V 00 V, 0 560 + j40 Ω, P 0.0 and L 0 S 0 V 00 00 tanh (560 + j40) (560 + j40) S Sending current, IS tanh PL tanh ( 0.0 0) 0 9.8 (9.8)(560 j40) (9.8)(560 j40) (560 + j40) (560 + j40)(560 j40) 560 + 40 ( ) (9.8) 700 36.87 0.75 36.87 A 490000 the sending end current, I S 75 36.87 ma 777 04, John Bird