Answer:
(1 + 3i)(1 – 3i) gives real number product
Step-by-step explanation:
Given the expressions
[tex](1 + 3i)(6i) ,(1 + 3i)(2 -3i), (1 + 3i)(1 - 3i), (1 + 3i)(3i)[/tex]
From analysis one of the following pairs has real-number products
[tex](1 + 3i)(2 -3i), (1 + 3i)(1 - 3i)[/tex]
Performing operations on
[tex](1 + 3i)(2 -3i)= 2-3i+6-6i^2 \\= 2-3i+6-6(-1) \\=2-3i+6+6 \\=14-3i[/tex]
Performing operations on
[tex](1 + 3i)(1 - 3i) \\= 1-3i+3i-9(i)^2 \\= 1+0-9(-1) \\= 1+9=10[/tex]
