Question

Which pair of complex factors results in a real-number product?

Answer 1

we will proceed to solve each case to determine the solution of the problem

we know that

$i^{2} =-1$

case A. $15*(-15i)$

$15*(-15i)=-225i$

The result of case A is not a real number product

case B. $3i*(1-3i)$

$3i*(1-3i)=3i*(1)-(3i)*(3i)\\=3i-9i^{2}\\=3i-9*(-1)\\=3i+9$

The result of case B is not a real number product

case C. $(8+20i)*(-8-20i)$

$(8+20i)*(-8-20i)=8*(-8)+8*(-20i)+20i*(-8)-20i*(20i)\\=-64-160i-160i-400i^{2}\\=-64-320i-400*(-1)\\=-64-320i+400\\=336-320i$

The result of case C is not a real number product

case D. $(4+7i)*(4-7i)$

$(4+7i)*(4-7i)=4^{2} -(7i)^{2}\\=16-49i^{2}\\=16-49*(-1)\\=65$

The result of case D is a real number product

therefore

the answer is the option

$(4+7i)*(4-7i)$

Answer 2

To answer the problem above, we must first define what "i" means. i means imaginary number which means square root of - 1, to get rid of i, you simply need to square, in this problem, we need to look for an equation where "i" is squared. The answer is D. (4+7i)(4-7i) to justify:
(4+7i)(4-7i)
(16-28i+28i-49i^2)
(16-49i^2)
16-49 = -33