Knowing (2 - i)*(a - bi) = 2 + 9i, where i is the imaginary unit and a and b are real numbers, what is a?
3 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
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
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
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)
We'll add (1) + (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.