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Question -

Find the real values of x and y, if

(i) (x + iy) (2 – 3i) = 4 + i

(ii) (3x – 2i y) (2 + i)2 = 10(1 + i)

(iv) (1 + i) (x + iy) = 2 – 5i



Answer -

(i) (x + iy) (2 – 3i) = 4+ i

Given:

(x + iy) (2 – 3i) = 4+ i

Let us simplify theexpression we get,

x(2 – 3i) + iy(2 – 3i)= 4 + i

2x – 3xi + 2yi – 3yi=4 + i

2x + (-3x+2y)i – 3y(-1) = 4 + i [since, i= -1]

2x + (-3x+2y)i + 3y =4 + i [since, i= -1]

(2x+3y) + i(-3x+2y) =4 + i

Equating Real andImaginary parts on both sides, we get

2x+3y = 4… (i)

And -3x+2y = 1… (ii)

Multiply (i) by 3 and(ii) by 2 and add

On solving we get,

6x – 6x – 9y + 4y = 12+ 2

13y = 14

y = 14/13

Substitute the valueof y in (i) we get,

2x+3y = 4

2x + 3(14/13) = 4

2x = 4 – (42/13)

= (52-42)/13

2x = 10/13

x = 5/13

x = 5/13, y = 14/13

 The real valuesof x and y are 5/13, 14/13

(ii) (3x – 2i y) (2 + i)2 =10(1 + i)

Given:

(3x – 2iy) (2+i)=10(1+i)

(3x – 2yi) (22+i2+2(2)(i))= 10+10i

(3x – 2yi) (4 +(-1)+4i) = 10+10i [since, i= -1]

(3x – 2yi) (3+4i) =10+10i

Let us divide with3+4i on both sides we get,

(3x – 2yi) =(10+10i)/(3+4i)

= Now multiply anddivide with (3-4i)

= [10(3-4i) +10i(3-4i)] / (32 – (4i)2)

= [30-40i+30i-40i2]/ (9 – 16i2)

= [30-10i-40(-1)] /(9-16(-1))

= [70-10i]/25

Now, equating Real andImaginary parts on both sides we get

3x = 70/25 and -2y =-10/25

x = 70/75 and y = 1/5

x = 14/15 and y = 1/5

 The real valuesof x and y are 14/15, 1/5

(4+2i) x-3i-3 +(9-7i)y = 10i

(4x+9y-3) + i(2x-7y-3)= 10i

Now, equating Real andImaginary parts on both sides we get,

4x+9y-3 = 0 … (i)

And 2x-7y-3 = 10

2x-7y = 13 … (ii)

Multiply (i) by 7 and(ii) by 9 and add

On solving theseequations we get

28x + 18x + 63y – 63y= 117 + 21

46x = 117 + 21

46x = 138

x = 138/46

= 3

Substitute the valueof x in (i) we get,

4x+9y-3 = 0

9y = -9

y = -9/9

= -1

x = 3 and y = -1

 The real valuesof x and y are 3 and -1

(iv) (1 + i) (x + iy) = 2 –5i

Given:

(1 + i) (x + iy) = 2 –5i

Divide with (1+i) onboth the sides we get,

(x + iy) = (2 –5i)/(1+i)

Multiply and divide by(1-i)

= (2 – 5i)/(1+i) ×(1-i)/(1-i)

= [2(1-i) – 5i (1-i)]/ (12 – i2)

= [2 – 7i + 5(-1)] / 2[since, i= -1]

= (-3-7i)/2

Now, equating Real andImaginary parts on both sides we get

x = -3/2 and y = -7/2

 Thee real valuesof x and y are -3/2, -7/2

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