0-2 Operations with Complex Numbers
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1 Simplify. 1. i i 2 + i i 3 + i 20 1 i esolutions Manual - Powered by Cognero Page 1
2 4. i i 77 i 6. i 4 + i i 5 + i 9 2i esolutions Manual - Powered by Cognero Page 2
3 8. i 18 1 Simplify. 9. (3 + 2i) + ( 4 + 6i) 1 + 8i 10. (7 4i) + (2 3i) 9 7i 11. (0.5 + i) (2 i) i esolutions Manual - Powered by Cognero Page 3
4 12. ( 3 i) (4 5i) 7 + 4i 13. ( i) ( 1 6.3i) i 14. (2 + 3i) + ( 6 + i) 4 + 4i 15. ( 2 + 4i) + (5 4i) 3 esolutions Manual - Powered by Cognero Page 4
5 16. (5 + 7i) ( 5 + i) i 17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I as a variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 3j ohms. Add these complex numbers to find the total impedance in the circuit j ohms Simplify. 18. ( 2 i) i esolutions Manual - Powered by Cognero Page 5
6 19. (1 + 4i) i 20. (5 + 2i) i 21. (3 + i) i esolutions Manual - Powered by Cognero Page 6
7 22. (2 + i)(4 + 3i) i 23. (3 + 5i)(3 5i) (5 + 3i)(2 + 6i) i esolutions Manual - Powered by Cognero Page 7
8 25. (6 + 7i)(6 7i) 85 Simplify i esolutions Manual - Powered by Cognero Page 8
9 27. + i 28. i esolutions Manual - Powered by Cognero Page 9
10 29. + i i esolutions Manual - Powered by Cognero Page 10
11 31. i esolutions Manual - Powered by Cognero Page 11
12 32. esolutions Manual - Powered by Cognero Page 12
13 33. esolutions Manual - Powered by Cognero Page 13
14 ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I Z. Find the voltage (in volts) in each of the following circuits given the current and impedance. 34. I = 1 + 3j amps, Z = 7 5j ohms Substitute the given values for I and Z into the equation E = I Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is j volts j volts 35. I = 2 7j amps, Z = 4 3j ohms Substitute the given values for I and Z into the equation E = I Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is 13 34j volts j volts esolutions Manual - Powered by Cognero Page 14
15 36. I = 5 4j amps, Z = 3 + 2j ohms Substitute the given values for I and Z into the equation E = I Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is 23 2j volts. 23 2j volts 37. I = j amps, Z = 6 j ohms Substitute the given values for I and Z into the equation E = I Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is j volts j volts esolutions Manual - Powered by Cognero Page 15
16 Solve each equation x = 0 i 39. 4x = 0 4i esolutions Manual - Powered by Cognero Page 16
17 40. 2x = x = x = 0 i esolutions Manual - Powered by Cognero Page 17
18 43. 3x = 0 13i 44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance, and the reactance due to inductance, and can be written as a complex number R + (X L X C )j. The values (in ohms) for R,, and in the first and second parts of a particular series circuit are shown. a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms. a. Substitute the given values for R, X C, and into the equation Z = R + (X L X C )j. The impedance in the first part of the circuit is 10 j ohms and in the second of the circuit is 3 + 0j ohms. b. (10 j ) + (3 + 0j ) = 13 j ohms c. The reciprocal of a number x is, so the reciprocal of is. Simplify this fraction by rationalizing the denominator. esolutions Manual - Powered by Cognero Page 18
19 a. 10 j ohms; 3 + 0j ohms b. 13 j ohms c. siemens Find the values of x and y to make each equation true x + 2iy = i To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 2, x + 3iy = 5 6i To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 1, 2 esolutions Manual - Powered by Cognero Page 19
20 47. x iy = 3 + 4i To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 3, x + 3iy = 10 9i To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 2, x + 3iy = i To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 6, 4 esolutions Manual - Powered by Cognero Page 20
21 50. 4x iy = 8 + 7i To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 2, 7 Simplify. 51. (2 i)(3 + 2i)(1 4i) 12 31i 52. ( 1 3i)(2 + 2i)(1 2i) 12 16i esolutions Manual - Powered by Cognero Page 21
22 53. (2 + i)(1 + 2i)(3 4i) i 54. ( 5 i)(6i + 1)(7 i) i esolutions Manual - Powered by Cognero Page 22
0-2 Operations with Complex Numbers
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