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Perform the operation and get your answer in reduced form of a + bi. (2+i)/(-3-4i) 

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted September 18, 2013 at 3:33 AM via web

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Perform the operation and get your answer in reduced form of a + bi.

(2+i)/(-3-4i) 

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Zaca | Student, Undergraduate | (Level 1) Salutatorian

Posted September 18, 2013 at 3:57 AM (Answer #1)

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When there are "i's" in the denominator, the easiest way to get rid of them is to multiply by the conjugate. 

In this case, our expression is:

`(2+i)/(-3-4i)`

The congugate is a term to multiply the denominator with to get a perfect square. (Or, to put in simple terms, switch the sign in the middle of the two numbers)

In this case: `(-3+4i)`

Mutiply by the conjugate:

` ` `((2+i)/(-3-4i))*((-3+4i)/(-3+4i))`

Combine: 

`(-6 + 8i -3i -4)/(9-12i+12i+16)`

Simplify:

`(-10 + 5i)/25`

Divide to turn into a + bi form:

`-2/5 + (1/5)i`

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flbyrne | (Level 3) Assistant Educator

Posted September 18, 2013 at 4:32 AM (Answer #2)

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To divide complex numbers, multiply both numerator and denominator by the conjugate of the denominator.

The conjugate of -3-4i is -3+4i, then

`(2+i)/(-3-4i) * (-3+4i)/(-3+4i)`

Multiply complex numbers as two binomials:

`((2)(-3)+(2)(4i)+(i)(-3)+(i)(4i))/((-3)(-3)+(-3)(4i)+(-4i)(-3)+(-4i)(4i))`

`=(-6+8i-3i-4)/(9-12i+12i+16)`

`=(-10+5i)/25`

`=-0.4+0.2i`

Thus the answer in reduced form is: -0.4+0.2i

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