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

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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:

...

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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` .

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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.

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