A complex number, (a + bi), multiplied by (2 + 3i) and added to -i gives the product of (-11 + 5i) and (1 – i).?

(a+bi)(2+3i) - i = (-11+5i)(1-i)

⇒ (a+bi)(2+3i) = [-11+ 5 + 5i + 11i] + i

⇒ a+bi

= (-6+17i) / (2+3i)

= [(-6+17i)(2-3i)] / [(2+3i)(2-3i)] .... .... multiply top and bottom by complex conjugate of 2+3i

= [-12 + 51 + 34i + 18i] / (2²+3²)

= (39+52i) / 13

= 3+4i

So a = 3, b = 4 by comparing real and imaginary parts respectively.

(a + bi)(2 + 3i) - i = (-11 + 5i)(1 - i)

2a + 3ai + 2bi + 3bi^2 - i = -11 + 11i + 5i - 5i^2

2a + 3ai + 2bi - 3b - i = -11 + 16i + 5

2a - 3b + i(3a + 2b) = -6 + 17i

so now you have two equations

2a - 3b = -6

3a + 2b = 17

4a - 6b = -12

9a + 6b = 51

13a = 39

a = 3

b = 4

(3 + 4i)(2 + 3i) - i = (-11 + 5i)(1 - i)

6 + 9i + 8i + 12i^2 - i = -11 + 11i + 5i - 5i^2

6 + 17i - 12 - i = -11 + 16i + 5

-6 + 16i = -6 + 16i

I think you mean:

−i + (a+bi)(2+3i) = (−11+5i)(1−i)

−i + 2a + 3ai + 2bi + 3bi² = −11 + 11i + 5i − 5i²

−i + 2a + 3ai + 2bi − 3b = −11 + 16i + 5

(2a−3b) + (3a+2b−1)i = −6 + 16i

2a − 3b = −6

3a + 2b − 1 = 16

a = 3

b = 4

whoa

A