# Knowing (2 - i)*(a - bi) = 2 + 9i, where i is the imaginary unit and a and b are real numbers, what is a?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to open the brackets such that:

`2a - 2bi - ai + bi^2 = 2 + 9i`

You need to substitute -1 for `i^2` (complex number theory) such that:

`2a - 2bi - ai - b = 2 + 9i`

You need to factor out i such that:

`2a - b + i(-2b - a) = 2 + 9i`

Equating the coefficients of like parts yields:

`2a - b = 2 =gt b = 2a - 2`

`-a - 2b = 9 `

You need to substitute `2a - 2`  for b in equation `-a - 2b = 9`  such that:

`-a - 2(2a - 2) = 9`

`-a - 4a + 4 = 9 =gt -5a = 9 - 4 =gt -5a = 5 =gt a = -1`

Hence, evaluating the value of a yields `a = -1` .

justaguide | College Teacher | (Level 2) Distinguished Educator

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It is given that (2 - i)*(a - bi) = 2 + 9i. a can be determined in the following manner:

(2 - i)*(a - bi) = 2 + 9i

=> 2*(a - bi) - i(a - bi) = 2 + 9i

=> 2a - 2bi - ai + bi^2 = 2 + 9i

=> 2a - i(2b + a) - b = 2 + 9i

Equating the real and imaginary coefficients gives:

2a - b = 2...(1)

-2b - a = 9 ...(2)

2*(1) - (2)

=> 4a - 2b + 2b + a = 4 - 9

=> 5a = -5

=> a = -1

The required value of a is -1.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll remove the brackets from the left side. For this reason, we'll apply FOIL method:

2*a - 2*bi - i*a - i*(-bi) =  2 + 9i

2a - 2bi - ia + b*i^2 =  2 + 9i

But i^2 = -1

2a - 2bi - ia - b = 2 + 9i

We'll combine real parts and imaginary parts:

(2a-b) + i*(-2b - a) = 2 + 9i

Comparing, we'll get:

2a - b = 2 (1)

-a - 2b = 9 (2)

We'll multiply by 2 the relation (2):

-2a - 4b = 18 (3)

2a - b - 2a - 4b = 2 + 18

We'll eliminate like terms:

-5b = 20

b = -4

2a - (-4) = 2

2a + 4 = 2

2a = -2

a = -1

The requested value of a is: a = -1.