3 years ago

Which pair of complex numbers has a real-number product? (1 + 3i)(6i) (1 + 3i)(2 – 3i) (1 + 3i)(1 – 3i) (1 + 3i)(3i)

Mathematics

1 answer:

Kobotan [32]3 years ago

6 0

Answer:

<h2>(1 + 3i)(1 – 3i) gives real number product</h2>

Step-by-step explanation:

Given the expressions

$(1 + 3i)(6i) ,(1 + 3i)(2 -3i), (1 + 3i)(1 - 3i), (1 + 3i)(3i)$

From analysis one of the following pairs has real-number products

$(1 + 3i)(2 -3i), (1 + 3i)(1 - 3i)$

Performing operations on

$(1 + 3i)(2 -3i)= 2-3i+6-6i^2 \\= 2-3i+6-6(-1) \\=2-3i+6+6 \\=14-3i$

Performing operations on

$(1 + 3i)(1 - 3i) \\= 1-3i+3i-9(i)^2 \\= 1+0-9(-1) \\= 1+9=10$

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