Question

(2+i)
111) (3-i) (1+2i)​

Answer 1

Answer

`(i) ((2+3i)/(3+4i))((2-3i)/(3-4i)) = [(4-9i^2)/(9-16i^2)]`...(Using `(a+b)(a-b) = a^2-b^2`)

Now, using `i^2 = -1`,

`=[(4+9)/(9+16)] = 13/25`

`:. ((2+3i)/(3+4i))((2-3i)/(3-4i)) = 13/25`

So, it is purely real.

`(ii) ((3+2i)/(2-3i))((3-2i)/(2+3i)) = [(9-4i^2)/(4-9i^2)]`...(Using `(a+b)(a-b) = a^2-b^2`)

Now, using `i^2 = -1`,

`=[(9+4)/(4+9)] = 13/13 = 1`

`:.((3+2i)/(2-3i))((3-2i)/(2+3i)) = 1`

So, it is purely real.