MTH403 Assignment 1 Fall 2022 Solution

Question #1:                                                                                                      

Find real𝑥and𝑦if(𝑥–𝑖𝑦)(3+5𝑖)istheconjugate of–6–24𝑖.

SOLUTION:                                                                                                     

Accordingtocondition

(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 + 24𝑖Bycomparing

3𝑥+ 5𝑦=−6…………(1)

 

5𝑥−3𝑦=24………..(2)

 

Multiplyequation(1)by3 andequation (2)by5 thenadd

 

(9𝑥+ 15𝑦)+ (25𝑥− 15𝑦)=−18+ 110

 

9𝑥+15𝑦+25𝑥−15𝑦=92

 

34𝑥=92

 

𝑥=92=3

34

 

𝑥=3

 

Puttingx =3 inequation(1)

 

3(3) + 5𝑦=−6

 

9 + 5𝑦=−6

 

5𝑦=−6− 9

 

5𝑦=−15

 

𝑦=−15=−3

5

 

𝑦=−3

 

Question#2:                                                                                                       

Simplify(√3 +3𝑖)31byusingDe-Moiver’stheorem.

SOLUTION:                                                                                                     

Here 𝑧=(√3+3)         𝑎𝑛𝑑𝑛=31

Weknowthat

𝑧𝑛=𝑟𝑛(𝑐𝑜𝑠𝑛𝜃+𝑖𝑠𝑖𝑛𝑛𝜃)

So,firstwefind rand𝜃

 

𝑟=√(√3)2+(3)2=√12=2√3

 

𝛼=𝑡𝑎𝑛−1(3)=𝑡𝑎𝑛−1(√3)=𝜋

 

𝜃=𝛼=𝜋

3

√3                                                3

 

(𝑏𝑒𝑐𝑎𝑢𝑠𝑒𝑧𝑙𝑖𝑒𝑖𝑛1𝑠𝑡𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡)

So,theaboveequationbecome

(√3+3)31=(2√3)31(cos(31)𝜋+𝑖𝑠𝑖𝑛(31)𝜋)

3                                3

(√3+3)31=(2√3)31(cos31𝜋+𝑖𝑠𝑖𝑛31𝜋)

3                         3

 

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

3                                          3

(√3+3)31=(2√3)31(𝑐𝑜𝑠𝜋+𝑖𝑠𝑖𝑛𝜋)

3                    3

 

(√3+3)31=(2√3)31(1+𝑖√3)

2            2














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