Question

Which pair of complex numbers has a real-number product? (1 + 2i)(8i) (1 + 2i)(2 – 5i) (1 + 2i)(1 – 2i) (1 + 2i)(4i)

Answer 1

Answer:

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

Step-by-step explanation:

Answer 2

Answer:

(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.