Question

which pair of complex numbers has a real - number product?
(1 + 2i)(8i)
(1 + 2i)(2 - 5i)
(1 + 2i)(1 - 2i)
(1 + 2i)(4i)

Explanation:

Step1: Recall the formula for multiplying complex numbers and the property of conjugates.

The product of a complex number \(a + bi\) and its conjugate \(a - bi\) is \(a^2 + b^2\), which is a real number. Also, we can expand each product using the distributive property (FOIL method) for complex numbers.

Step2: Expand \((1 + 2i)(8i)\)

Using the distributive property: \(1\times8i+2i\times8i = 8i + 16i^2\). Since \(i^2=-1\), this becomes \(8i-16=-16 + 8i\), which is not a real number.

Step3: Expand \((1 + 2i)(2 - 5i)\)

Using FOIL: \(1\times2+1\times(-5i)+2i\times2+2i\times(-5i)=2-5i + 4i-10i^2\). Combine like terms: \(2 - i+10 = 12 - i\), which is not a real number.

Step4: Expand \((1 + 2i)(1 - 2i)\)

Using the formula \((a + bi)(a - bi)=a^2-(bi)^2=a^2 + b^2\) (since \(i^2 = - 1\), so \(-(bi)^2=-b^2i^2=b^2\)). Here \(a = 1\) and \(b = 2\), so \(1^2-(2i)^2=1 - 4i^2=1+4 = 5\), which is a real number.

Step5: Expand \((1 + 2i)(4i)\)

Using the distributive property: \(1\times4i+2i\times4i=4i + 8i^2\). Since \(i^2=-1\), this becomes \(4i-8=-8 + 4i\), which is not a real number.

Answer:

\((1 + 2i)(1 - 2i)\) (the third option)