Download 0-2 Operations with Complex Numbers

0-2 Operations with Complex Numbers
Simplify.
1. i–10
SOLUTION: 2. i2 + i8
SOLUTION: 3. i3 + i20
SOLUTION: 4. i100
SOLUTION: 5. i77
SOLUTION: 6. i4 + i–12
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7. i5 + i9
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SOLUTION: 0-2 Operations with Complex Numbers
6. i4 + i–12
SOLUTION: 7. i5 + i9
SOLUTION: 8. i18
SOLUTION: Simplify.
9. (3 + 2i) + (–4 + 6i)
SOLUTION: 10. (7 – 4i) + (2 – 3i)
SOLUTION: 11. (0.5 + i) – (2 – i)
SOLUTION: 12. (–3 – i) – (4 – 5i)
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SOLUTION: 0-2 Operations with Complex Numbers
12. (–3 – i) – (4 – 5i)
SOLUTION: 13. (2 + 4.1i) – (–1 – 6.3i)
SOLUTION: 14. (2 + 3i) + (–6 + i)
SOLUTION: 15. (–2 + 4i) + (5 – 4i)
SOLUTION: 16. (5 + 7i) – (–5 + i)
SOLUTION: 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.
SOLUTION: Simplify.
18. (–2 – i)2
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in the circuit.
SOLUTION: 0-2 Operations with Complex Numbers
Simplify.
18. (–2 – i)2
SOLUTION: 19. (1 + 4i)2
SOLUTION: 20. (5 + 2i)2
SOLUTION: 21. (3 + i)2
SOLUTION: 22. (2 + i)(4 + 3i)
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0-2 Operations with Complex Numbers
22. (2 + i)(4 + 3i)
SOLUTION: 23. (3 + 5i)(3 – 5i)
SOLUTION: 24. (5 + 3i)(2 + 6i)
SOLUTION: 25. (6 + 7i)(6 – 7i)
SOLUTION: Simplify.
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0-2 Operations with Complex Numbers
Simplify.
26. SOLUTION: 27. SOLUTION: 28. SOLUTION: eSolutions Manual - Powered by Cognero
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0-2 Operations with Complex Numbers
28. SOLUTION: 29. SOLUTION: 30. SOLUTION: eSolutions Manual - Powered by Cognero
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0-2 Operations with Complex Numbers
30. SOLUTION: 31. SOLUTION: 32. SOLUTION: eSolutions Manual - Powered by Cognero
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0-2 Operations with Complex Numbers
32. SOLUTION: 33. SOLUTION: eSolutions Manual - Powered by Cognero
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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.
0-2 Operations with Complex Numbers
33. SOLUTION: 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
SOLUTION: 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 22 + 16j volts.
35. I = 2 – 7j amps, Z = 4 – 3j ohms
SOLUTION: Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.
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0-2 Therefore,
Operations
with ofComplex
the voltage
the circuit isNumbers
22 + 16j volts.
35. I = 2 – 7j amps, Z = 4 – 3j ohms
SOLUTION: 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.
36. I = 5 – 4j amps, Z = 3 + 2j ohms
SOLUTION: 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.
37. I = 3 + 10j amps, Z = 6 – j ohms
SOLUTION: 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 28 + 57j volts.
Solve each equation.
38. 5x2 + 5 = 0
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0-2 Operations with Complex Numbers
Therefore, the voltage of the circuit is 28 + 57j volts.
Solve each equation.
38. 5x2 + 5 = 0
SOLUTION: 39. 4x2 + 64 = 0
SOLUTION: 40. 2x2 + 12 = 0
SOLUTION: 41. 6x2 + 72 = 0
SOLUTION: 2
42. 8x Manual
+ 120 =
0
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0-2 Operations with Complex Numbers
42. 8x2 + 120 = 0
SOLUTION: 43. 3x2 + 507 = 0
SOLUTION: 44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance
and the reactance due to inductance
ohms) for R,
, and
,
, and can be written as a complex number R + (X L – X C)j . The values (in
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.
SOLUTION: a. Substitute the given values for R, X C, and
The impedance in the first part of the circuit is
b.
ohms
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c. The reciprocal of a number x is
denominator.
into the equation Z = R + (X L – X C)j .
ohms and in the second of the circuit is , so the reciprocal of
is ohms.
. Simplify this fraction by rationalizingPage
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0-2 Operations with Complex Numbers
44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance
and the reactance due to inductance
ohms) for R,
, and
,
, and can be written as a complex number R + (X L – X C)j . The values (in
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.
SOLUTION: a. Substitute the given values for R, X C, and
The impedance in the first part of the circuit is
b.
ohms
c. The reciprocal of a number x is
into the equation Z = R + (X L – X C)j .
ohms and in the second of the circuit is , so the reciprocal of
is ohms.
. Simplify this fraction by rationalizing the
denominator.
Find the values of x and y to make each equation true.
45. 3x + 2iy = 6 + 10i
SOLUTION: 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.
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0-2 Operations with Complex Numbers
Find the values of x and y to make each equation true.
45. 3x + 2iy = 6 + 10i
SOLUTION: 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.
46. 5x + 3iy = 5 – 6i
SOLUTION: 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.
47. x – iy = 3 + 4i
SOLUTION: 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.
48. –5x + 3iy = 10 – 9i
SOLUTION: 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.
49. 2x + 3iy = 12 + 12i
SOLUTION: 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.
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50. 4x – iy = 8 + 7i
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equation
and then solve for x and y respectively.
0-2 Operations with Complex Numbers
49. 2x + 3iy = 12 + 12i
SOLUTION: 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.
50. 4x – iy = 8 + 7i
SOLUTION: 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.
Simplify.
51. (2 – i)(3 + 2i)(1 – 4i)
SOLUTION: 52. (–1 – 3i)(2 + 2i)(1 – 2i)
SOLUTION: 53. (2 + i)(1 + 2i)(3 – 4i)
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SOLUTION: Page 16
0-2 Operations with Complex Numbers
53. (2 + i)(1 + 2i)(3 – 4i)
SOLUTION: 54. (−5 – i)(6i + 1)(7 – i)
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