Question

How do you simplify $\dfrac{9-i}{2-i}$?

Answer 1

This type of problem is based on the concept of rationalising complex numbers. We have to first multiply the numerator and the denominator by 2+i. Here, we have to substitute ${{i}^{2}}=-1$. Then, simplify the given expression in such a manner that we get a constant in the denominator. Then, use the distributive property $\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd$ to solve further. Thus, we get a simplified expression which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to simplify $\dfrac{9-i}{2-i}$.
We have been given the expression $\dfrac{9-i}{2-i}$. ---------(1)
First, we have to rationalise the term so that we get a constant term in the denominator.
Let us multiply the numerator and denominator with 2+i.
We get $\dfrac{9-i}{2-i}=\dfrac{9-i}{2-i}\times \dfrac{2+i}{2+i}$.
On further simplification, we get
$\dfrac{9-i}{2-i}=\dfrac{\left( 9-i \right)\left( 2+i \right)}{\left( 2-i \right)\left( 2+i \right)}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Using this identity, we get
$\dfrac{9-i}{2-i}=\dfrac{\left( 9-i \right)\left( 2+i \right)}{{{2}^{2}}-{{i}^{2}}}$
We know that ${{i}^{2}}=-1$ and ${{2}^{2}}=4$.
Therefore, we get $\dfrac{9-i}{2-i}=\dfrac{\left( 9-i \right)\left( 2+i \right)}{5}$.
Now, let us use distributive property, that is, $\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd$ and simplify the numerator.
$\Rightarrow \dfrac{9-i}{2-i}=\dfrac{9\times 2+9i-2i-{{i}^{2}}}{5}$
On simplifying further, we get
$\Rightarrow \dfrac{9-i}{2-i}=\dfrac{18+9i-2i-{{i}^{2}}}{5}$
We know that ${{i}^{2}}=-1$, substituting the value in the above expression, we get
$\Rightarrow \dfrac{9-i}{2-i}=\dfrac{18+9i-2i-\left( -1 \right)}{5}$
Now let us group the i terms.
$\dfrac{9-i}{2-i}=\dfrac{18+\left( 9-2 \right)i-\left( -1 \right)}{5}$
On further simplification, we get
$\Rightarrow \dfrac{9-i}{2-i}=\dfrac{18+7i+1}{5}$
$\therefore \dfrac{9-i}{2-i}=\dfrac{7i+19}{5}$
Therefore, the simplified form of the expression $\dfrac{9-i}{2-i}$ is $\dfrac{7i+19}{5}$.

Note: We should not substitute the value of ${{i}^{2}}$ as 1 which will lead to an incorrect answer. Group the constants and i terms separately and then solve. Also, we can solve this type of problem only by rationalising the denominator and convert the denominator into a constant. The numerator can have i terms. We should not make calculation mistakes based on sign conventions.