Question

Find a and b if (a+ib) (1+i) = 2+i

Answer 1

Answer:

a(1+¡)+b¡(1+¡)=2+¡

a+a¡+b¡+b¡^2=2+¡

a+(a+b)¡+b(-1)=2+¡

a+(a+b)¡-b=2+¡

(a+b)¡+(a-b)=2+¡

a+b=1,a-b=2

2a=3

a=3/2

puta=3/2in a+b=1

3/2-1=b

b=-1/2

Answer 2

Answer:  The required values of a and b are $a=\dfrac{3}{2},~~b=-\dfrac{1}{2}.$

Step-by-step explanation:  We are given to find the values of a and b from the following equality :

$(a+ib)(1+i)=2+i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the fact that $i^2=-1.$

From equation (i), we have

$(a+ib)(1+i)=2+i\\\\\Rightarrow a(1+i)+ib(1+i)=2+i\\\\\Rightarrow a+ia+ib+i^2b=2+i\\\\\Rightarrow a+ia+ib-b=2+i\\\\\Rightarrow (a-b)+i(a+b)=2+i.$

Equating the real and imaginary parts on both sides of the above equation, we get

$a-b=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\a+b=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Adding equations (ii) and (iii), we get

$2a=3\\\\\Rightarrow a=\dfrac{3}{2}.$

Substituting the value of a in equation (ii), we get

$\dfrac{3}{2}-b=2\\\\\Rightarrow b=\dfrac{3}{2}-2\\\\\Rightarrow b=-\dfrac{1}{2}.$

Thus, the required values of a and b are $a=\dfrac{3}{2},~~b=-\dfrac{1}{2}.$