Assignment # 01 MTH403 (Fall 2022)

 

 

 

Maximum Marks: 10

Due Date: November 25, 2022

 

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Question # 1:                                                                                                     

Find real 𝑥 and 𝑦 if (𝑥 – 𝑖𝑦) (3 + 5𝑖) is the conjugate of – 6 – 24𝑖.

SOLUTION:                                                                                                     

According to condition

(𝑥 − 𝑖𝑦)(3 + 5𝑖) = ̅−̅̅6̅̅−̅̅̅2̅̅4̅̅𝑖

 

3𝑥 + 5𝑥𝑖 − 3𝑦𝑖 − 5𝑦𝑖² = −6 + 24𝑖

 

3𝑥 + 5𝑥𝑖 − 3𝑦𝑖 − 5𝑦(−1) = −6 + 24𝑖                                     ∴ 𝑖² = −1

 

3𝑥 + 5𝑥𝑖 − 3𝑦𝑖 + 5𝑦 = −6 + 24𝑖 (3𝑥 + 5𝑦) + 𝑖(5𝑥 − 3𝑦) = −6 + 24𝑖 By comparing

3𝑥 + 5𝑦 = −6 … … … … (1)

 

5𝑥 − 3𝑦 = 24 … … … . . (2)

 

Multiply equation (1) by 3 and equation (2) by 5 then add

 

(9𝑥 + 15𝑦) + (25𝑥 − 15𝑦) = −18 + 110

 

9𝑥 + 15𝑦 + 25𝑥 − 15𝑦 = 92

 

34𝑥 = 92

 

𝑥 = 92 = 3

34

 

𝑥 = 3

 

Putting x = 3 in equation (1)

 

3(3) + 5𝑦 = −6

 

9 + 5𝑦 = −6

 

5𝑦 = −6 − 9

 

5𝑦 = −15

 

𝑦 = −15 = −3

5

 

𝑦 = −3

 

 Question # 2:                                                                                                    

Simplify (√3 + 3𝑖)³¹ by using De-Moiver’s theorem.

SOLUTION:                                                                                                     

Here 𝑧 = (√3 + 3)     𝑎𝑛𝑑 𝑛 = 31

We know that

𝑧^𝑛 = 𝑟^𝑛(𝑐𝑜𝑠𝑛𝜃 + 𝑖𝑠𝑖𝑛𝑛𝜃)

So, first we find r and 𝜃

 

𝑟 = √(√3)² + (3)² = √12 = 2√3

 

𝛼 = 𝑡𝑎𝑛⁻¹ ( 3 ) = 𝑡𝑎𝑛⁻¹(√3) = 𝜋

 

𝜃 = 𝛼 = 𝜋

3

√3                                                3

 

(𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑧 𝑙𝑖𝑒 𝑖𝑛 1𝑠𝑡 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡)

So, the above equation become

(√3 + 3)³¹ = (2√3 )31(cos (31) 𝜋 + 𝑖𝑠𝑖𝑛(31) 𝜋)

3                                3

(√3 + 3)³¹ = (2√3 )31(cos 31𝜋 + 𝑖𝑠𝑖𝑛 31𝜋)

3                         3

 

(√3 + 3)³¹ = (2√3 )31 (cos(10π + 𝜋) + 𝑖𝑠𝑖𝑛(10𝜋 + 𝜋))

3                                          3

(√3 + 3)³¹ = (2√3 )31 (𝑐𝑜𝑠 𝜋 + 𝑖𝑠𝑖𝑛 𝜋)

3                    3

 

(√3 + 3)³¹ = (2√3 )31 (1 + 𝑖 √3)

2            2

 

 

 

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