Question

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

Answer 1

Hint: 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.