Question

What values of c and d would make
the following expression represent a real number?
i(2 + 3i)(c + di)

c = 2, d = 3

c = –2, d = 3

c = 3, d = –2

c = –3, d = –2

Answer 1

Compute i(2+3i)(c+di) = (i(2+3i))(c+di) = (-3+2i)(c+di) = (-3c-2d) + (2c-3d)i.

For the product to be real, the imaginary part must be 0: 2c - 3d = 0 → c = 3d/2.

Checking the choices, only c = −3, d = −2 satisfies 2c − 3d = 0 (2·(−3) − 3·(−2) = −6 + 6 = 0). The result is then real (13).