The document discusses converting complex numbers between rectangular and polar forms, performing arithmetic operations on complex numbers in both rectangular and polar forms, and representing sinusoidal waveforms using phasor notation. Some key points: - Complex numbers can be represented in rectangular form (a + jb) or polar form (r∠θ) and converted between the two forms. - Arithmetic operations like addition, subtraction, multiplication, and division can be performed on complex numbers in both rectangular and polar forms. - Sinusoidal waveforms can be expressed using phasor notation as a magnitude and phase angle to denote the peak value and phase offset from a reference time or angle.

1. Conversion Between Forms
37.
a) 4 + 𝑗3
𝑎 = 4, 𝑏 = 3
𝑟 = √42 + 322
= 5.00
𝜃 = tan−1
3
4
= 36.87°
𝑠𝑜4 + 𝑗3 = 5.00∠36.87°
b) 2 + 𝑗2
𝑎 = 2, 𝑏 = 2
𝑟 = √22 + 222
= 2.83
𝜃 = tan−1
2
2
= 45.00°
𝑠𝑜 2 + 𝑗2 = 2.83∠45.00°
c. 4 + 𝑗12
𝑎 = 4, 𝑏 = 12
𝑟 = √42 + 1222
= 12.65
𝜃 = tan−1
12
4
= 71.57°
𝑠𝑜 4 + 𝑗12 = 12.65∠71.57°
38.
a) −8 − 𝑗16
𝑎 = −8, 𝑏 = −16
𝑟 = √(−8)2 + (−16)22
= 17.89
𝜃 = tan−1
−16
−8
= 243.43°
𝑠𝑜 − 8 − 𝑗16 = 17.89∠243.43°
b) +8 − 𝑗4
𝑎 = +8, 𝑏 = −4
𝑟 = √(8)2 + (−4)22
= 8.94
𝜃 = tan−1
−4
8
= 26.57°
𝑆𝑖𝑛𝑐𝑒 + 8 − 𝑗4 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡, 𝜃 = 360 − 26.57 = 333.43°

2. 𝑠𝑜 + 8 − 𝑗4 = 8.94∠333.43°
c) 0.02−𝑗0.003
𝑎 = 0.02, 𝑏 = −0.003
𝑟 = √(0.02)2 + (−0.003)22
= 0.0202
𝜃 = tan−1
−0.003
0.02
= 8.53°
𝑠𝑜 0.02 − 𝑗0.003 = 0.0202∠8.53°
39. Convert from polar to rectangular
a) 6∠40°
6.00(cos40°
+ 𝑗 sin 40°
)
= (6.00 cos40°
)+ (𝑗6.00sin 40°
)
= 4.60 + 𝑗3.86
b) 12∠120°
12.00(cos120°
+ 𝑗 sin 120°
)
= (12.00cos120°
)+ (𝑗12.00sin 120°
)
= −6.00 + 𝑗10.39
c) 2000∠−90°
2000(cos270°
+ 𝑗 sin 270°
)
= (2000cos270°
)+ (𝑗2000sin 270°
)
= 0.00 − 𝑗2000
40. Convert from polar to rectangular
a) 42∠0.15°
42.00(cos0.15°
+ 𝑗 sin 0.15°
)
= (42.00cos0.15°
)+ (𝑗42.00sin 0.15°
)
= 42 + 𝑗0.11
b) 2002∠−60°
2002(cos300°
+ 𝑗 sin 300°
)
= (2002cos300°
)+ (𝑗2002sin 300°
)
= 1001 + −𝑗1733.78
c) 0.006∠−120°
0.006(cos240°
+ 𝑗 sin 240°
)
= (0.006cos240°
)+ (𝑗0.006sin 240°
)
= −0.003 + −𝑗5.2 × 10−3
Section 14.9

3. 41.
a) (4.2 + 𝑗6.8) + (7.6 + 𝑗0.2)
𝑌𝑜𝑢 𝑎𝑑𝑑 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑡𝑠 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦 𝑡ℎ𝑒𝑛 𝑎𝑑𝑑 𝑡ℎ𝑒 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡𝑠
= (4.2 + 7.6) + 𝑗(6.8 + 0.2)
= 11.8 + 𝑗7.00
b) (142 + 𝑗7)+ (9.8 + 𝑗42)+ (0.1 + 𝑗0.9)
= (142 + 9.8 + 0.1) + 𝑗(7 + 42 + 0.9)
= 151.9 + 𝑗49.9
c) (4 × 10−6
+ 𝑗76)+ (7.2 × 10−7
− 𝑗5)
= (4 × 10−6
+ 7.2 × 10−7) + 𝑗(76 − 5)
= 4.72 × 10−6
+ 𝑗71
42.
a) (9.8 + 𝑗6.2) − (4.6 + 𝑗4.6)
= (9.8 − 4.6) + 𝑗(6.2 − 4.6)
= 5.2 + 𝑗1.6
b) (167 + 𝑗243)− (−42.3 − 𝑗68)
= (167 + 42.3) + 𝑗(243 + 68)
= 209.3 + 𝑗311
c) (−36 + 𝑗78)− (−4 − 𝑗6) + (10.8 − 𝑗72)
= (−36 + 4 + 10.8) + 𝑗(78 + 6 − 72)
= −21.2 + 𝑗12
43.
a) 6∠20°
+ 8∠80°
= 12.16∠54.72°
b) 42∠45°
+ 62∠60°
− 70∠120°
= 146.3∠79.88°
c) 20∠−120°
− 10∠−150°
+ 8∠−210°
+ 8∠+240°
= 38.89∠139.53°
44. Perform the following multiplication in rectangular form
a) (2 + 𝑗3)(6+ 𝑗8)
= (2 × 6 − 3 × 8) + 𝑗(6 × 3 + 2 × 8)
= −12 + 𝑗34
b) (7.8 + 𝑗1)(4+ 𝑗2)(7+ 𝑗6)

4. = 86.8 + 𝑗312.4
c) (400 − 𝑗200)(−0.01 − 𝑗0.5)(−1 + 𝑗3)
= 698 + 𝑗114
45. Perform the following multiplications in polar form
a) (2∠60°
)(4∠ − 40°
)
= (2 × 4)∠(60− 40)
8∠20°
b) (6.9∠8°
)(7.2∠ − 72°
= (6.9 × 7.2)∠(8− 72)
849.68∠−62°
c) (0.002∠120°
)(0.5∠200°
)(40∠80°
= (0.002 × 0.5 × 40)∠(120+ 200 + 80)
0.04∠400°
46. Perform the following divisions in polar
a)
42∠10°
7∠ 60°
=
42
7
∠10 − 60
6∠−50°
b)
0.006∠120°
30∠ 60°
=
0.006
30
∠120 − 60
2 × 10−4
∠60°
c)
4360∠−20°
40∠−210°
=
4360
40
∠ − 20 + 210
109∠190°
47. Perform the following divisions and leave the answer in rectangular form
a)
(8+𝑗8)
(2+𝑗2)
(8 + 𝑗8)
(2 + 𝑗2)
= 4

5. Section 14.11 Phasors
52. Express the following in phasor form
a) √2(160)sin( 𝜔𝑡 + 30°)
= 226.27∠30°
b) √2(25 × 10−3)sin(157𝑡 − 40°)
= 0.0354∠−40°
c) 100 sin( 𝜔𝑡 − 90°)
= 100∠270°
53. Express the following in phasor form
a) 20 sin(377𝑡 − 180°)
= 20∠−180°
b) 6 × 10−6
cos 𝜔𝑡
6 × 10−6
c) 3.6 × 10−6
cos(745𝑡 − 20°)
= 3.6 × 10−6
∠−20°
54. Express the following phasor currents and voltages as sine waves if the frequency is 60 Hz.
a) 𝐼 = 40𝐴∠20°
𝐼 = 40cos(2𝜋 × 60𝑡 + 20°)
𝐼 = 40cos(120𝜋𝑡 + 20°) 𝐴
b) 𝑉 = 120𝑉∠10°
𝑉 = 120cos(2𝜋 × 60𝑡 + 10°)
𝐼 = 120cos(120𝜋𝑡 + 10°) 𝑉
c) 𝐼 = 8 × 10−3
𝐴∠−110°
𝐼 = 40sin(2𝜋 × 60𝑡 − 110°)
𝐼 = 40sin(120𝜋𝑡 − 110°) 𝐴
d) 𝑉 =
6000
√2
𝑉∠ − 180°
𝑉 = 4242.64sin(2𝜋 × 60𝑡 − 180°)
𝐼 = 4242.64cos(120𝜋𝑡 − 180°) 𝑉