which pair of complex factors results in a real - number product?\n15(-15i)\n3i(1 - 3i)\n(8 + 20i)(-8…

Explanation:

Step1: Recall the formula for multiplying complex - numbers

The product of two complex numbers ((a + bi)(c+di)=ac + adi + bci+bdi^{2}), and since (i^{2}=- 1), it becomes ((a + bi)(c + di)=(ac - bd)+(ad + bc)i). A complex - conjugate pair ((a + bi)(a - bi)) has a product of (a^{2}+b^{2}) (because ((a + bi)(a - bi)=a^{2}-abi + abi - b^{2}i^{2}=a^{2}+b^{2}), which is a real number).

Step2: Analyze each option

Option 1: (15(-15i)=-225i), which is a pure imaginary number.

Option 2: (3i(1 - 3i)=3i-9i^{2}=9 + 3i), which is a complex number with non - zero imaginary part.

Option 3: ((8 + 20i)(-8 - 20i)=-64-160i-160i - 400i^{2}=-64 + 400-320i=336-320i), which is a complex number with non - zero imaginary part.

Option 4: ((4 + 7i)(4 - 7i)=4^{2}-(7i)^{2}=16-49i^{2}=16 + 49=65), which is a real number.

Answer:

((4 + 7i)(4 - 7i))