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

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sapon | Student, Undergraduate | (Level 2) Honors

Posted April 24, 2011 at 4:11 PM via web

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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted April 24, 2011 at 4:15 PM (Answer #1)

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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.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 24, 2011 at 4:50 PM (Answer #2)

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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.

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