Which pair of complex numbers has a real-number product?

Asked by maham237 @ 25/12/2020 in Mathematics viewed by 345 People

(1 + 3i)(6i)
(1 + 3i)(2 – 3i)
(1 + 3i)(1 – 3i)
(1 + 3i)(3i)

Answered by maham237 @ 25/12/2020

we will proceed to verify each case to determine the solution

remember that

$i^{2} =-1$

case A) $(1 + 3i)(6i)$

applying distributive property

$(1 + 3i)(6i)=1*6i+3i*6i \\=6i+18i^{2}\\=6i+18*(-1)\\=6i-18$

$-18+6i$ ------> is not a real number

therefore

the case A) is not a real number product

case B) $(1 + 3i)(2-3i)$

applying distributive property

$(1 + 3i)(2-3i)=1*2+1*(-3i)+3i*(2)+3i*(-3i)\\=2-3i+6i-9i^{2}\\=2+3i-9*(-1) \\=11+3i$

$11+3i$ ------> is not a real number

therefore

the case B) is not a real number product

case C) $(1 + 3i)(1-3i)$

applying difference of square

$(1 + 3i)(1-3i)=(1)^{2}-(3i)^{2}\\=1-9i^{2}\\=1-9*(-1) \\=10$

$10$ ------> is a real number

therefore

the case C) is a real number product

case D) $(1 + 3i)(3i)$

applying distributive property

$(1 + 3i)(3i)=1*3i+3i*3i \\=3i+9i^{2}\\=3i+9*(-1) \\=3i-9$

$-9+3i$ ------> is not a real number

therefore

the case D) is not a real number product

the answer is

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

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