Verify if the result of multiplication (2+5i)(4i-3) is a real number?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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'llĀ 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.

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