Question

Which values of x and y would make the following expression represent a real number?

(6 + 3i)(x + yi)
x = 6, y = 0
x= -3, y = 0
x = 6, y=-3
x = 0, y = -3

Answer 1

The answer is C on edge

Explanation:

just took the test

Answer 2

x = 6, y = - 3

Explanation:

Given the product of a complex number and its conjugate, that is

(x + yi)(x - yi) → where x and y are real

= x² + y² ← a real number

Given

(6 + 3i), then we require the conjugate 6 - 3i , that is

(6 + 3i)(6 - 3i)

= 6² + 3² = 36 + 9 = 45 ← a real number

Thus

x = 6 and y = - 3 are the required values