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 =
(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)
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13a=39 ⇒a=39/13=3
3(2a-3b=-6)
-2(3a+2b=17)
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-13b=-52 ⇒ b=52/13=4
Answer: a=3; b=4
Hence, we have:
a= 3 and b= 4
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:
\(2\times 3-3b=-6\\\\i.e.\\\\6-3b=-6\\\\i.e.\\\\-3b=-6-6\\\\i.e.\\\\-3b=-12\\\\i.e.\\\\b=4\)