Question

What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di)
choices: A)c = 2, d = 3 B)c = –2, d = 3 C)c = 3, d = –2 D)c = –3, d = –2

Answer 1

The correct answer is:

D)c = –3, d = –2

Explanation:

To simplify this, we first distribute the i through the first set of parentheses:

i(2+3i) = i(2) + i(3i) = 2i + 3i²

Since i² = -1, this means we have

2i + 3(-1) = 2i - 3

Now we multiply this by the second set of parentheses:

(2i-3)(c+di) = 2i(c) + 2i(di) - 3(c) - 3(di)

= 2ci + 2di² - 3c - 3di

= 2ci + 2d(-1) - 3c - 3di

= 2ci - 2d - 3c - 3di

We need 2ci and -3di to cancel. This means we want the product 2c and the product 3d to be the same; this happens if c = -3 (2*-3 = -6) and d = -2 (3*-2 = -6).

Answer 2

The values of c and d are required such that the expression results in a real number.

The required values are D)c = –3, d = –2

The expression is

$i(2+3i)(c+di)\\ =i(2c+2di+3ci+3di^2)\\ =2ci+2di^2+3ci^2+3di^3\\ =2ci-2d-3c-3di\\ =(2c-3d)i-2d-3c$

Now the coefficient of $i$ has to be zero in order for the expression to result in a real number.

Let us check each option

$2c-3d=2\times 2-3\times 3=-5$

$2\times -2-3\times 3=-13$

$2\times 3-3\times -2=12$

$2\times -3-3\times -2=0$

Hence option D is correct.

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