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

Explanation:

Step1: Expand ((1 + 2i)(8i))

[ \begin{align*} (1 + 2i)(8i)&=1\times8i+2i\times8i\ &=8i + 16i^{2}\ &=8i-16 \end{align*} ]

Step2: Expand ((1 + 2i)(2 - 5i))

[ \begin{align*} (1 + 2i)(2 - 5i)&=1\times2-1\times5i+2i\times2-2i\times5i\ &=2-5i + 4i-10i^{2}\ &=2 - i+10\ &=12 - i \end{align*} ]

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

[ \begin{align*} (1 + 2i)(1 - 2i)&=1^{2}-(2i)^{2}\ &=1-4i^{2}\ &=1 + 4\ &=5 \end{align*} ]

Step4: Expand ((1 + 2i)(4i))

[ \begin{align*} (1 + 2i)(4i)&=1\times4i+2i\times4i\ &=4i+8i^{2}\ &=4i - 8 \end{align*} ]

Answer:

((1 + 2i)(1 - 2i))