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Precisodeajudaeajudo
@Precisodeajudaeajudo
December 2019
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2) Sejam os numeros complexos
z₁ = 9 + 5i, z₂ = 15 - 2i, z₃ = 6i e z₄ = -8 calcule
A) z₁ + z₂ - z₃
B) z₁ - z₂ + z₃ - z₄
C) z₃ - z₁ - z₂ + 5i
D) ( z₄ + z₁) - (z₃ - z₂)
3) calcule:
A) ( 2 + 4i ).(1 + 3i)
B) (6 + 3i).(3 + 4i)
C) ( 5 - i).(1 + 3i)
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rosanev
z₁ = 9 + 5i, z₂ = 15 - 2i, z₃ = 6i e z₄ = -8 calcule
A) z₁ + z₂ - z₃
(9 + 5i) + (15 - 2i) - 6i =
(5i - 2i - 6i) + (9 + 15) = -3i + 24
B) z₁ - z₂ + z₃ - z₄
(9 + 5i) - (15 - 2i) + 6i -(-8) =
9 + 5i - 15 + 2i + 6i + 8 =
(5i + 2i + 6i) + (9 - 15 + 8) = 13i + 2
C) z₃ - z₁ - z₂ + 5i
6i - (9 + 5i) - (15 - 2i) + 5i =
(6i - 5i + 2i + 5i) + (-9 - 15) =
8i + (-24) = 8i - 24
D) ( z₄ + z₁) - (z₃ - z₂)
(-8 + 9 + 5i) - (6i - (15 - 2i)) =
1 + 5i - (6i -15 + 2i) =
1 + 5i - 8i + 15 = -3i + 16
3)
A) ( 2 + 4i ).(1 + 3i) =
2 + 2.3i + 4i + 4i.3i =
2 + 6i + 4i + 12i² = i² = -1
2 + 10i + 12.(-1) =
2 - 12 + 10i = -10 + 10i
B) (6 + 3i).(3 + 4i)
6.3 + 6.4i + 3i.3 + 3i.4i =
18 + 24i + 9i + 12i² =
18 + 33i + 12(-1) =
18 - 12 + 33i = 6 + 33i
C) ( 5 - i).(1 + 3i)
5.1 + 5.3i - 1.i - 3i.i = i.i = i² = -1
5 + 15i - i -3(-1) =
5 + 14i + 3 = 8 + 14i
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A) z₁ + z₂ - z₃
(9 + 5i) + (15 - 2i) - 6i =
(5i - 2i - 6i) + (9 + 15) = -3i + 24
B) z₁ - z₂ + z₃ - z₄
(9 + 5i) - (15 - 2i) + 6i -(-8) =
9 + 5i - 15 + 2i + 6i + 8 =
(5i + 2i + 6i) + (9 - 15 + 8) = 13i + 2
C) z₃ - z₁ - z₂ + 5i
6i - (9 + 5i) - (15 - 2i) + 5i =
(6i - 5i + 2i + 5i) + (-9 - 15) =
8i + (-24) = 8i - 24
D) ( z₄ + z₁) - (z₃ - z₂)
(-8 + 9 + 5i) - (6i - (15 - 2i)) =
1 + 5i - (6i -15 + 2i) =
1 + 5i - 8i + 15 = -3i + 16
3)
A) ( 2 + 4i ).(1 + 3i) =
2 + 2.3i + 4i + 4i.3i =
2 + 6i + 4i + 12i² = i² = -1
2 + 10i + 12.(-1) =
2 - 12 + 10i = -10 + 10i
B) (6 + 3i).(3 + 4i)
6.3 + 6.4i + 3i.3 + 3i.4i =
18 + 24i + 9i + 12i² =
18 + 33i + 12(-1) =
18 - 12 + 33i = 6 + 33i
C) ( 5 - i).(1 + 3i)
5.1 + 5.3i - 1.i - 3i.i = i.i = i² = -1
5 + 15i - i -3(-1) =
5 + 14i + 3 = 8 + 14i