Question

Solve the following: $\left( a+ib \right)\left( 1+i \right)=\left( 2+i \right)$

Answer 1

Hint: The given question is solved using the concept of complex numbers. Firstly, we multiply and simplify the left-hand side of the equation. Then, we equate the terms on the left-hand side to the corresponding terms on the right-hand side to get the desired result.

Complete step by step answer:
We are given an expression in the question and find the value of a, b. We will be using the concept of complex numbers to find the values of a, b.
Complex numbers, in mathematics, are represented in the form $p+iq$
Here,
p, q are the real numbers
$i$ is the unit imaginary number
According to the question,
We need to find the values of a, b in the equation $\left( a+ib \right)\left( 1+i \right)=\left( 2+i \right)$
Firstly, we need to simplify the left-hand side of the equation.
$\Rightarrow \left( a+ib \right)\left( 1+i \right)$
Multiplying through the brackets, we get,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=\left( a\times 1 \right)+\left( a\times i \right)+\left( ib\times 1 \right)+\left( ib\times i \right)$
Simplifying the terms in the brackets, we get,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=a+ia+ib+{{i}^{2}}b$
From the concept of complex numbers, we know that the unit imaginary number $i$ is equal to the square root of minus one.
The value of $i$ is given by,
$\Rightarrow i=\sqrt{-1}$
Squaring the above equation on both sides, we get,
$\Rightarrow {{i}^{2}}=-1$
Substituting the same in the above equation, we get,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=a+ia+ib+\left( -1 \right)b$
Simplifying the above equation, we get,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=a+ia+ib+\left( -b \right)$
From the rules of arithmetic, we know that $\left( + \right)\times \left( - \right)=\left( + \right)$
Following the same, we get,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=a+ia+ib-b$
Writing the above equation in the form of $p+iq$ , we get,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=\left( a-b \right)+i\left( a+b \right)$
As per the question,
$\Rightarrow \left( a+ib \right)\left( 1+i \right)=\left( 2+i \right)$
Substituting the value of the $\left( a+ib \right)\left( 1+i \right)$ , we get,
$\Rightarrow \left( a-b \right)+i\left( a+b \right)=\left( 2+i \right)$
Equating the corresponding terms, we get,
$\Rightarrow a-b=2$
$\Rightarrow a+b=1$
Adding the above two equations, we get,
$\Rightarrow \left( a-b \right)+\left( a+b \right)=2+1$
The above equation can be written as follows,
$\Rightarrow a+a+b-b=3$
Simplifying the above equation, we get,
$\Rightarrow 2a=3$
Moving the number 2 to the other side of the equation, we get,
$\therefore a=\dfrac{3}{2}$
From the above, we know that
$\Rightarrow a-b=2$
Substituting the value of a in the equation, we get,
$\Rightarrow \dfrac{3}{2}-b=2$
Moving the term $2$ to the other side of the equation, we get,
$\Rightarrow \dfrac{3}{2}-2=b$
Taking the LCM, we get,
$\Rightarrow \dfrac{3-4}{2}=b$
Simplifying the above equation, we get,
$\therefore b=\dfrac{-1}{2}$

Note: The result of the given question can be cross-checked by substituting the values of a, b in the equation. If the left-hand side of the equation is equal to the right-hand side the result is correct.
LHS:
 $\Rightarrow \left( a+ib \right)\left( 1+i \right)$
Substituting the values of a, b in the above expression, we get,
$\Rightarrow \left( \dfrac{3}{2}+i\left( \dfrac{-1}{2} \right) \right)\left( 1+i \right)$
Simplifying the above equation, we get,
$\Rightarrow \left( \dfrac{3}{2}-i\dfrac{1}{2} \right)\left( 1+i \right)$
Multiplying through the brackets, we get,
$\Rightarrow \left( \dfrac{3}{2}\times 1 \right)+\left( \dfrac{3}{2}\times i \right)-\left( \dfrac{i}{2}\times 1 \right)-\left( \dfrac{i}{2}\times i \right)$
Simplifying the terms, we get,
$\Rightarrow \dfrac{3}{2}+i\dfrac{3}{2}-i\dfrac{1}{2}-{{i}^{2}}\dfrac{1}{2}$
We know that ${{i}^{2}}=-1$
Substituting the same, we get,
$\Rightarrow \dfrac{3}{2}+i\dfrac{3}{2}-i\dfrac{1}{2}+\dfrac{1}{2}$
Let us evaluate it further.
$\Rightarrow \left( \dfrac{3}{2}+\dfrac{1}{2} \right)+i\left( \dfrac{3}{2}-\dfrac{1}{2} \right)$
$\Rightarrow \dfrac{4}{2}+i\dfrac{2}{2}$
$\Rightarrow 2+i$
RHS:
$\Rightarrow 2+i$
LHS = RHS. The result attained is correct.