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

a = and b =

User Roledenez
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2 Answers

5 votes
(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

User Berriel
by
7.4k points
3 votes

Answer:

Hence, we have:

a= 3 and b= 4

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* 3-3b=-6\\\\i.e.\\\\6-3b=-6\\\\i.e.\\\\-3b=-6-6\\\\i.e.\\\\-3b=-12\\\\i.e.\\\\b=4

User Prabakaran Raja
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6.8k points