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

(a+bi)(2+3i)-i=(-11+5i)(1-i)
2a+3ai+2bi+3bi²-i=-11+11i+5i-5i²
2a+(3a+2b-1)i-3b=-11+16i+5
(2a-3b)+(3a+2b-1)i=-6+16i
Therefore; we have the next system of equations:
2a-3b=-6
3a+2b-1=16

Finally, the system of equations is:
2a-3b=-6
3a+2b=17
We can solve this system of equations by reduction method
 2(2a-3b=-6) 
 3(3a+2b=17)

   13a=39  ⇒a=39/13=3

 3(2a-3b=-6)
-2(3a+2b=17)

       -13b=-52   ⇒ b=52/13=4

a=3;    b=4