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)

Answer 1

(i) (x + iy) (2 – 3i) = 4 + i

Given as

(x + iy) (2 – 3i) = 4 + i

Now let us simplify the expression we get,

x(2 – 3i) + iy(2 – 3i) = 4 + i

2x – 3xi + 2yi – 3yi² = 4 + i

2x + (-3x + 2y)i – 3y(-1) = 4 + i [since, i² = -1]

2x + (-3x + 2y)i + 3y = 4 + i [since, i² = -1]

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

By equating real and imaginary parts on both sides, we get

2x + 3y = 4… (i)

And -3x + 2y = 1… (ii)

Now multiply (i) by 3 and (ii) by 2 and add

By solving we get,

6x – 6x – 9y + 4y = 12 + 2

13y = 14

y = 14/13

Then substitute the value of 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

Thus the real values of x and y are 5/13, 14/13

(ii) (3x – 2i y) (2 + i)² = 10(1 + i)

Given as

(3x – 2i y) (2 + i)² = 10(1 + i)

(3x – 2yi) (2² + i² + 2(2)(i)) = 10 + 10i

(3x – 2yi) (4 + (-1) + 4i) = 10 + 10i [since, i² = -1]

(3x – 2yi) (3 + 4i) = 10 + 10i

Now let us divide with 3 + 4i on both sides we get,

(3x – 2yi) = (10 + 10i)/(3 + 4i)

Then multiply and divide 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

Then, equating real and imaginary 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

Thus the real values of x and y are 14/15, 1/5