Sure, let's determine which pair of complex numbers has a real-number product by going through each pair step-by-step.
1. [tex]\((1 + 2i)(8i)\)[/tex]
- Expanding the product: \
[tex]\((1 + 2i)(8i) = 1 \cdot 8i + 2i \cdot 8i\)[/tex]
- Simplifying the terms: \
[tex]\(1 \cdot 8i = 8i\)[/tex] \
[tex]\(2i \cdot 8i = 16i^2 = 16(-1) = -16\)[/tex] \
Combining these: [tex]\(8i - 16\)[/tex]
- The result can be written as: [tex]\(-16 + 8i\)[/tex], which is not a real number
2. [tex]\((1 + 2i)(2 - 5i)\)[/tex]
- Expanding the product: \
[tex]\((1 + 2i)(2 - 5i) = 1 \cdot 2 + 1 \cdot (-5i) + 2i \cdot 2 + 2i \cdot (-5i)\)[/tex]
- Simplifying the terms: \
[tex]\(1 \cdot 2 = 2\)[/tex] \
[tex]\(1 \cdot (-5i) = -5i\)[/tex] \
[tex]\(2i \cdot 2 = 4i\)[/tex] \
[tex]\(2i \cdot (-5i) = -10i^2 = -10(-1) = 10\)[/tex] \
Combining these: [tex]\(2 - 5i + 4i + 10\)[/tex]
- The result can be written as: [tex]\(12 - i\)[/tex], which is not a real number
3. [tex]\((1 + 2i)(1 - 2i)\)[/tex]
- Expanding the product: \
[tex]\((1 + 2i)(1 - 2i) = 1 \cdot 1 + 1 \cdot (-2i) + 2i \cdot 1 + 2i \cdot (-2i)\)[/tex]
- Simplifying the terms: \
[tex]\(1 \cdot 1 = 1\)[/tex] \
[tex]\(1 \cdot (-2i) = -2i\)[/tex] \
[tex]\(2i \cdot 1 = 2i\)[/tex] \
[tex]\(2i \cdot (-2i) = -4i^2 = -4(-1) = 4\)[/tex] \
Combining these: [tex]\(1 - 2i + 2i + 4\)[/tex]
- The result can be written as: [tex]\(5\)[/tex], which is a real number
4. [tex]\((1 + 2i)(4i)\)[/tex]
- Expanding the product: \
[tex]\((1 + 2i)(4i) = 1 \cdot 4i + 2i \cdot 4i\)[/tex]
- Simplifying the terms: \
[tex]\(1 \cdot 4i = 4i\)[/tex] \
[tex]\(2i \cdot 4i = 8i^2 = 8(-1) = -8\)[/tex] \
Combining these: [tex]\(4i - 8\)[/tex]
- The result can be written as: [tex]\(-8 + 4i\)[/tex], which is not a real number
From the above calculations, we find that the pair [tex]\((1 + 2i) \cdot (1 - 2i)\)[/tex] gives us a real number product, specifically [tex]\(5\)[/tex]. Thus, it is the pair of complex numbers that has a real-number product, and it is the third pair in the list.
Answer: The third pair of complex numbers, [tex]\((1 + 2i)(1 - 2i)\)[/tex], has a real-number product.
