math problemWhat is the complex number a-bi if (2-i)(a-bi) = 2+9i

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to determine a-bi if (2-i)(a-bi) = 2+9i

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

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

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

Equating the real and imaginary coefficients we get

2a - b = 2 and 2b + a = -9

=> 2(-2b - 9) - b = 2

=> -4b - 18 - b = 2

=> -5b = 20

=> b = -4

a = 8 - 9 = -1

The required number a + bi = -1 - 4i

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll start by removing the brackets from the left side

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

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

But i^2 = -1

We'll re-write the identity replacing i^2 by -1.

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

We'll combine real parts and imaginary parts:

(2a - b) + i(-2b - a) = 2 + 9i

The real and imaginary parts from both sides have to be equal:

2a - b = 2 (1)

-a - 2b = 9 (2)

We'll multiply (2) by 2:

-2a - 4b = 18

(1)+(2):

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

We'll combine and eliminate like terms:

-5b = 20 => b = -4

2a + 4 = 2 => 2a = 2 - 4 => 2a = -2 => a = -1

The complex number is: z = -1 + 4i

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