Express the following in the form a + i b : ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i )

Given, ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 3 × 2 + 3 × 3 i - 2 i 2 - 2 i 3 i 1 2 - 1 × i + 2 i × 2 - 2 i × i = 6 + 9 i - 4 i - 6 i 2 2 - i + 4 i - 2 i 2 = 6 + 5 i - 6 - 1 2 + 3 i - 2 - 1 ∵ i 2 = - 1 = 6 + 6 + 5 i 2 + 2 + 3 i = 12 + 5 i 4 + 3 i Multiply and divide the numerator and denominator by 4 - 3 i ⇒ ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 12 + 5 i 4 + 3 i × 4 - 3 i 4 - 3 i = 12 + 5 i 4 - 3 i 4 2 - 3 i 2 = 48 - 36 i + 20 i - 15 i 2 16 - 9 i 2 = 48 - 16 i - 15 - 1 16 - 9 - 1 = 48 + 15 - 16 i 16 + 9 = 63 - 16 i 25 Hence, ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 63 25 - i 16 25

Express the following in the form a + i b : ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i )

Given, ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 3 × 2 + 3 × 3 i - 2 i 2 - 2 i 3 i 1 2 - 1 × i + 2 i × 2 - 2 i × i = 6 + 9 i - 4 i - 6 i 2 2 - i + 4 i - 2 i 2 = 6 + 5 i - 6 - 1 2 + 3 i - 2 - 1 ∵ i 2 = - 1 = 6 + 6 + 5 i 2 + 2 + 3 i = 12 + 5 i 4 + 3 i Multiply and divide the numerator and denominator by 4 - 3 i ⇒ ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 12 + 5 i 4 + 3 i × 4 - 3 i 4 - 3 i = 12 + 5 i 4 - 3 i 4 2 - 3 i 2 = 48 - 36 i + 20 i - 15 i 2 16 - 9 i 2 = 48 - 16 i - 15 - 1 16 - 9 - 1 = 48 + 15 - 16 i 16 + 9 = 63 - 16 i 25 Hence, ( 3 - 2 i ) ( 2 + 3 i ) ( 1 + 2 i ) ( 2 - i ) = 63 25 - i 16 25

Express the following in the form a + i b : ( 3 - 4 i ) ( 4 - 2 i ) ( 1 + i )

Given, ( 3 - 4 i ) ( 4 - 2 i ) ( 1 + i ) = 3 - 4 i 4 + 4 i - 2 i - 2 i 2 = 3 - 4 i 4 + 2 i - 2 - 1 ∵ i 2 = - 1 = 3 - 4 i 6 + 2 i Multiply and divide the numerator and denominator by 6 - 2 i = 3 - 4 i 6 + 2 i × 6 - 2 i 6 - 2 i = 3 - 4 i 6 - 2 i 6 + 2 i 6 - 2 i = 18 - 6 i - 24 i + 8 i 2 6 2 - 2 i 2 = 18 - 8 - 30 i 36 - 4 i 2 = 10 - 30 i 40 ∵ i 2 = - 1 Hence, ( 3 - 4 i ) ( 4 - 2 i ) ( 1 + i ) = 1 4 - i 3 4

Express the following in the form a + i b : ( 3 + 2 i ) ( 2 - 3 i ) + ( 3 - 2 i ) ( 2 + 3 i )

Given, ( 3 + 2 i ) ( 2 - 3 i ) + ( 3 - 2 i ) ( 2 + 3 i ) Rationalizing the denominators, we have = 3 + 2 i 2 - 3 i × 2 + 3 i 2 + 3 i + 3 - 2 i 2 + 3 i × 2 - 3 i 2 - 3 i = 3 + 2 i 2 + 3 i 2 - 3 i 2 + 3 i + 3 - 2 i 2 - 3 i 2 + 3 i 2 - 3 i = 6 + 9 i + 4 i + 6 i 2 4 - 9 i 2 + 6 - 9 i - 4 i + 6 i 2 4 - 9 i 2 ∵ a + b a - b = a 2 - b 2 = 13 i 13 + - 13 i 13 ∵ i 2 = - 1 = i - i = 0 Hence, ( 3 + 2 i ) ( 2 - 3 i ) + ( 3 - 2 i ) ( 2 + 3 i ) = 0 + 0 i

Show that ( 7 + i 3 ) ( 7 - i 3 ) + ( 7 - i 3 ) ( 7 + i 3 ) is real.

To Prove: ( 7 + i 3 ) ( 7 - i 3 ) + ( 7 - i 3 ) ( 7 + i 3 ) is real. Consider ( 7 + i 3 ) ( 7 - i 3 ) + ( 7 - i 3 ) ( 7 + i 3 ) Taking the L.C.M, we get = ( 7 + i 3 ) ( 7 + i 3 ) + ( 7 - i 3 ) ( 7 - i 3 ) ( 7 - i 3 ) ( 7 + i 3 ) = ( 7 + i 3 ) 2 + ( 7 - i 3 ) 2 ( 7 ) 2 - ( i 3 ) 2 → i ∵ ( a + b ) ( a - b ) = a 2 - b 2 Now, we know that ( a + b ) 2 + ( a - b ) 2 = 2 a 2 + b 2 So, by applying the formula in equation i , we get = 2 ( 7 ) 2 + ( i 3 ) 2 7 - 3 i 2 = 2 7 + 3 i 2 7 - 3 ( - 1 ) ∵ i 2 = - 1 = 2 [ 7 - 3 ] 10 = 8 10 = 8 10 + 0 i = 4 5 + 0 i Hence, the given equation is purely real as there is no imaginary part.

Find real values of θ for which 3 + 2 i sin θ 1 - 2 i sin θ is purely real.

To Prove: 3 + 2 i sin θ 1 - 2 i sin θ is purely real Consider 3 + 2 i sin θ 1 - 2 i sin θ Rationalizing the denominator, we have = 3 + 2 i sin θ 1 - 2 i sin θ × 1 + 2 i sin θ 1 + 2 i sin θ = 3 + 2 i sin θ 1 + 2 i sin θ 1 - 2 i sin θ 1 + 2 i sin θ = 3 + 6 i sin θ + 2 i sin θ + 4 i 2 sin 2 θ 1 2 - 2 i sin θ 2 = 3 + 8 i sin θ - 4 sin 2 θ 1 - 4 i 2 sin 2 θ ∵ i 2 = - 1 = 3 - 4 sin 2 θ + 8 i sin θ 1 - 4 - 1 sin 2 θ = 3 - 4 sin 2 θ + 8 i sin θ 1 + 4 sin 2 θ ⇒ 3 + 2 i sin θ 1 - 2 i sin θ = 3 - 4 sin 2 θ 1 + 4 sin 2 θ + i 8 sin θ 1 + 4 sin 2 θ Since 3 + 2 i sin θ 1 - 2 i sin θ is purely real Hence, imaginary part is equal to zero. ⇒ 8 sin θ 1 + 4 sin 2 θ = 0 Since - 1 ≤ sin θ ≤ 1 ∴ 1 + 4 sin 2 θ ≠ 0 ⇒ 8 sin θ = 0 ⇒ sin θ = 0 We know that sin θ = 0 at θ = . . . , - 2 π , - π , 0 , π , 2 π , 3 π , . . . ⇒ θ = n π , n ∈ Z Hence, the required values are θ = n π , n ∈ Z

Find the real values of x and y for which: ( x - 1 ) ( 3 + i ) + ( y - 1 ) ( 3 - i ) = i

Given, ( x - 1 ) ( 3 + i ) + ( y - 1 ) ( 3 - i ) = i ⇒ x - 1 3 - i + y - 1 3 + i 3 + i 3 - i = i ⇒ 3 x - i x - 3 + i + 3 y + i y - 3 - i 3 + i 3 - i = i ⇒ 3 x + 3 y - 6 + i - x + y 3 2 - i 2 = i ∵ a + b a - b = a 2 - b 2 ⇒ 3 x + 3 y - 6 + i - x + y 9 - - 1 = i ⇒ 3 x + 3 y - 6 + i - x + y 10 = i ⇒ 3 x + 3 y - 6 10 + i - x + y 10 = 0 + i We know that two complex numbers are equal if their real and imaginary parts are respectively equal to each other. ⇒ 3 x + 3 y - 6 10 = 0 and - x + y 10 = 1 ⇒ 3 x + 3 y = 6 and - x + y = 10 ⇒ x + y = 2 → i and - x + y = 10 → i i Adding equations i and i i , we have 2 y = 12 ⇒ y = 6 Substituting the value of y in equation i , we get x + 6 = 2 ⇒ x = 2 - 6 = - 4 Hence, x = - 4 and y = 6