Verify if the result of multiplication (2+5i)(4i-3) is a real number?
We need to verify if the product of (2+5i) and (4i-3) is a real number.
Multiply the two terms.
(2 + 5i)(4i - 3)
=> 2*4i + 20*i^2 - 6 - 15i
we know i^2 = -1
=> 8i - 20 - 6 - 15i
=> -7i - 26
This has an imaginary component. The product -7i - 26 of the two complex numbers given is not real.
To verify the nature of the result, we'll have to remove the brackets.
For this reason, we'll use the property of distributivity of multiplication over the addition.
(2+5i)(4i-3) = 2*(4i-3) + 5i(4i-3)
We'll remove the brackets from the right side:
(2+5i)(4i-3) = 8i - 6 + 20i^2 - 15i
We'll keep in mind that i^2 = -1 and we'll substitute in the expression above.
(2+5i)(4i-3) = 8i - 6- 20 - 15i
We'll combine like terms:
(2+5i)(4i-3) = -26 - 7i
We notice that the result of multiplication of the given complex numbers is also a complex number: (2+5i)(4i-3) = -26 - 7i.