Which pair of complex numbers has a real-number product?
de
de llegada de ti
(1 + 3)(61)
(1 + 3)(2 – 31)
(1 + 31)(1 – 31)
(1 + 31)(3)

Answers

The answer is (1 + 3i)(1 – 3i).

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)

     

Step-by-step explanation:

the complex number and its conjugate

calculating each of the products

noting that i² = (√- 1 )² = - 1

(1 + 2i)(8i) ( distribute by 8i )

= 8i + 16i² = - 16 + 8i ← complex number

(1 + 2i)(2 - 5i) ( expand using FOIL )

= 2 - 5i + 4i - 10i²

= 2 - i + 10 = 12 - i ← complex number

(1 + 2i)(1 - 2i) ( expand using FOIL )

= 1 - 2i + 2i - 4i²

= 1 + 4 = 5 ← real number

(1 + 2i)(4i) ( distribute by 4i )

= 4i +8i² = - 8 + 4i ← complex number

(1 + 2i)(1 - 2i) is the only product which results in a real number

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

Step-by-step explanation:

(1+2i)(1-2i)

Step-by-step explanation:

Following are the pairs of the complex number:

(1+2i)(8i),

(1 + 2i)(2 – 5i)

(1+2i)(1-2i) and (1+2i)(4i)

We have to check which pair out of these is a real number product, which means which pair do not contain terms consisting of "i".

A. (1+2i)(8i)= 8i+16i^{2}

                        =8i-16

B. (1+2i)(2-5i)=2-i-10i^{2}

                           =12-i

C. (1+2i)(1-2i)=1^{2}-4i^{2}

                          =5

D. (1+2i)(4i)=4i+8i^{2}

                        =4i-8

Since, A,B,D contains the term "i" which means they are not real valued, therefore option C that is (1+2i)(1-2i) has a real number product.

(1+2i)(1-2i)

Step-by-step explanation:

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

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



Do you know the answer?

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