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

User MrKsn
by
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2 Answers

2 votes

Answer:

C

Explanation:

trust

User J Atkin
by
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0 votes

Answer:

(1 + 3i)(1 – 3i) gives real number product

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

User Gernot Ullrich
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