Problem 1
Slide 83
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Slide 83 • Problem 1
(a+bi)² = 2i
a² + 2abi - b² = 2i
a = b
2abi = 2i
a = 1, b = 1
Answer: m = ±(1+i)
Problem 2
Slide 83
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Slide 83 • Problem 2
a² + 2abi - b² = -5 + 12i
a² - b² = -5
2abi = 12i
ab = 6
a = 6/b
(6/b)² - b² = -5
36/b² - b² = -5
36 - b⁴ = -5b²
b⁴ - 5b² - 36 = 0
Let b² = u
u² - 5u - 36 = 0
(u - 2.5)² = 42.25
u - 2.5 = ±6.5
u = 2.5 ± 6.5
u = 9 (negative is extraneous)
b² = 9, b = ±3
a = ±2
Answer: n = ±(2+3i)
Problem 3
Slide 83
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Slide 83 • Problem 3
Let u = b²
a² + 2abi - b² = 24 - 10i
2abi = -10i
ab = -5
a² - b² = 24
a = -5/b
(-5/b)² - b² = 24
25/b² - b² = 24
25 - b⁴ = 24b²
25 = b⁴ + 24b²
(u + 12)² = 169
u + 12 = ±13
u = -12 ± 13
u = 1 (u = -25 is extraneous)
b² = 1, b = ±1
a = ±5
Answer: z = 5 - i or -5 + i
Problem 1
Slide 84
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Slide 84 • Problem 1
A group of 4 always equals 2 - 2i
22 groups of 4
22(2 - 2i) = 44 - 44i
44 - 44i + 89i = 44 + 45i
Problem 1
Slide 109
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Slide 109 • Problem 1
(11 - 13i)(13 + 11i) / 290
= 143 + 143i + 121i - 169i² / 290
= (143 + 264i + 169) / 290
= (286 + 216i) / 290
c = 286/290 = 143/145
d = -48/290 = -24/145
Problem 1
Slide 110
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Slide 110 • Problem 1
(a + bi)³ = a³ + 3a²bi + 3ab²i² + b³i³
c = a³ - 3ab²
107 = 3a²b - b³
107 = b(3a² - b²)
Since 107 is prime: b = 1 or b = 107
If b = 1: 3a² - 1 = 107
a² = 36, a = 6
c = 216 - 3(6)(1) = 216 - 18 = 198
Answer: a = 6, b = 1, c = 198