Question

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

Answer 1

(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

Answer: a=3;    b=4

Answer 2

Answer:

Hence, we have:

               a= 3 and b= 4

Step-by-step explanation:

It is given that:

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

i.e.

$(a+ib)(2+3i)-i=(-11+5i)(1-i)\\\\i.e.\\\\a(2+3i)+ib(2+3i)-i=-11(1-i)+5i(1-i)\\\\i.e.\\\\2a+3ai+2bi+3bi^2-i=-11+11i+5i-5i^2$

Since, we know that :

$i^2=-1$

Hence, we have:

$2a+3ai+2bi-3b-i=-11+11i+5i+5\\\\i.e.\\\\(2a-3b)+i(3a+2b-1)=-11+5+11i+5i\\\\i.e.\\\\(2a-3b)+i(3a+2b-1)=-6+16i$

i.e. we have:

$2a-3b=-6---------(1)$

and

$3a+2b-1=16\\\\i.e.\\\\3a+2b=16+1\\\\i.e.\\\\3a+2b=17------------(2)$

On multiplying equation (1) by 2 and equation (2) by 3 we get:

$13a=39\\\\i.e.\\\\a=3$

on putting the value of a into equation (1) we get:

$2\times 3-3b=-6\\\\i.e.\\\\6-3b=-6\\\\i.e.\\\\-3b=-6-6\\\\i.e.\\\\-3b=-12\\\\i.e.\\\\b=4$