Maximum Marks:
10
Due Date: November 25, 2022
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Question
# 1:
Find real π₯ and π¦ if (π₯ – ππ¦) (3 + 5π) is the conjugate
of – 6 – 24π.
SOLUTION:
According to condition
(π₯ − ππ¦)(3 + 5π) = ̅−̅̅6̅̅−̅̅̅2̅̅4̅̅π
3π₯ + 5π₯π − 3π¦π − 5π¦π2 = −6 + 24π
3π₯ + 5π₯π −
3π¦π − 5π¦(−1) = −6 + 24π ∴ π2 = −1
3π₯ + 5π₯π − 3π¦π + 5π¦ = −6 + 24π (3π₯
+ 5π¦) +
π(5π₯
− 3π¦) = −6 + 24π By comparing
3π₯ + 5π¦ = −6 … … … … (1)
5π₯ − 3π¦ = 24 … … … . . (2)
Multiply equation (1) by 3 and
equation (2) by 5 then
add
(9π₯ + 15π¦) + (25π₯ − 15π¦) = −18 +
110
9π₯ + 15π¦ + 25π₯ − 15π¦ = 92
34π₯ = 92
π₯ = 92 = 3
34
π₯ = 3
Putting x
= 3 in equation (1)
3(3) + 5π¦ = −6
9 + 5π¦ = −6
5π¦ = −6 −
9
5π¦ = −15
π¦ = −15 = −3
5
π¦ = −3
Question # 2:
Simplify (√3 + 3π)31 by using De-Moiver’s theorem.
SOLUTION:
Here π§ = (√3 + 3) πππ π
= 31
We know that
π§π = ππ(πππ ππ + ππ ππππ)
So, first we find
r and π
π = √(√3)2 + (3)2 = √12 = 2√3
πΌ = π‘ππ−1 ( 3 ) = π‘ππ−1(√3) = π
π = πΌ = π
3
√3 3
(πππππ’π π π§
πππ ππ 1π π‘ ππ’ππππππ‘)
So, the above equation become
(√3 + 3)31 = (2√3 )31(cos (31) π + ππ ππ(31) π)
3 3
(√3 + 3)31 = (2√3 )31(cos 31π + ππ ππ 31π)
3 3
(√3 + 3)31 = (2√3 )31 (cos(10Ο + π) + ππ ππ(10π + π))
3 3
(√3 + 3)31 = (2√3 )31 (πππ π + ππ ππ π)
3 3
(√3 + 3)31 = (2√3 )31 (1 + π √3)
2 2
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