Question

If $\dfrac{2+3i}{\left( 7-i \right)\left( 4+2i \right)}=A+iB$, then $A+B$
A. $\dfrac{13}{100}$
B. $\dfrac{13}{1000}$
C. $\dfrac{16}{100}$
D. $\dfrac{16}{1000}$

Answer 1

In the given question, we are given a question based on complex numbers, that is, having $i$ (iota). We have to solve the given expression and choose the correct option from the given possible options. So, firstly we will have to simplify the expression as much as possible. Also, applying the value ${{i}^{2}}=-1$ in the expression wherever applicable. So, finally we will have the simplified form of the given expression and from that simplified form of the expression we will take the values of $A$ and $B$. Then, we will add up the values of $A$ and $B$, and have the value of $A+B$.

Complete step by step answer:
According to the given question, we are given a question with $i$ (iota). We have to solve the given expression and find the equivalent value of the expression.
The expression we have is,
$\dfrac{2+3i}{\left( 7-i \right)\left( 4+2i \right)}$
We will begin by multiplying and dividing the given expression by the conjugate of the denominator. So, we have,
$\Rightarrow \dfrac{2+3i}{\left( 7-i \right)\left( 4+2i \right)}\times \dfrac{\left( 7+i \right)\left( 4-2i \right)}{\left( 7+i \right)\left( 4-2i \right)}$
Multiplying the above terms, we will get,
$\Rightarrow \dfrac{2+3i\left( 28-14i+4i-2{{i}^{2}} \right)}{\left( {{7}^{2}}-{{i}^{2}} \right)\left( {{4}^{2}}-{{\left( 2i \right)}^{2}} \right)}$
We know that the value of ${{i}^{2}}=-1$, so we will apply this in the above expression and we will get,
$\Rightarrow \dfrac{\left( 2+3i \right)\left( 28+2-10i \right)}{\left( 49+1 \right)\left( 16-4{{i}^{2}} \right)}$
Solving the denominator in the above expression, we will have,
$\Rightarrow \dfrac{\left( 2+3i \right)\left( 30-10i \right)}{\left( 50 \right)\left( 16+4 \right)}$
Multiplying the brackets, we will get,
$\Rightarrow \dfrac{\left( 60-20i+90i-30{{i}^{2}} \right)}{\left( 50 \right)\left( 20 \right)}$
Again, substituting the value of ${{i}^{2}}=-1$ and also simplifying the terms further, we get,
$\Rightarrow \dfrac{\left( 60+30+70i \right)}{\left( 1000 \right)}$
Adding up the constants and the terms with iota separately, we will get,
$\Rightarrow \dfrac{\left( 90+70i \right)}{\left( 1000 \right)}$
Now, we will write the above expression in terms of $A+iB$, so we get,
$\Rightarrow \dfrac{90}{1000}+i\dfrac{70}{1000}$
So, here $A=\dfrac{90}{1000}$ and $B=\dfrac{70}{1000}$.
The value of the expression,
$A+B$
$\Rightarrow \dfrac{90}{1000}+\dfrac{70}{1000}$
Adding up the terms here, we get,
$\Rightarrow \dfrac{160}{1000}$
Reducing the fraction further, we get,
$\Rightarrow \dfrac{16}{100}$

So, the correct answer is “Option C”.

Note: The expression should not mix the constant terms and the terms with iota. It is observed that even after substituting the value of ${{i}^{2}}=-1$, the iota symbol is not removed from the expression, hence leading to a residual iota factors in the expression, hence giving us a wrong answer.