Answer
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Hint: In this question, we are given a complex expression and we have been asked to simplify the given expression. We will start by rationalizing the given expression. Once we have rationalized the expression, we will simplify it and put the desired values. Then, lastly, write the final simplified expression in the standard form.
Complete step-by-step solution:
We have been given a complex expression $\dfrac{{3 - i}}{{4 - i}}$.
Let us start by rationalizing the expression.
In order to rationalize, we divide and multiply the entire given expression by the denominator (but after changing the signs.) Look below how it is done.
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}}$
Now, we will solve it. If you look at the denominator, you will notice $\left( {a + b} \right)\left( {a - b} \right)$. We can simply solve it using the identity$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.
Simplifying the above expression,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 + 3i - 4i - {i^2}}}{{{4^2} - {i^2}}}$
On simplifying RHS we get,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 - i - {i^2}}}{{16 - {i^2}}}$
Now, we know that $i = \sqrt { - 1} $
If we square this expression, we get, ${i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1$
Hence, putting ${i^2} = - 1$ in the expression,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 - i - \left( { - 1} \right)}}{{16 - \left( { - 1} \right)}}$
Simplifying the expression,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 - i + 1}}{{17}}$
Let us add the numerator term and we get
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{13 - i}}{{17}}$
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} = \dfrac{{13}}{{17}} - \dfrac{i}{{17}}$
Hence, our final answer is $\dfrac{{3 - i}}{{4 - i}} = \dfrac{{13}}{{17}} - \dfrac{i}{{17}}$.
Also, notice that our answer is already in the standard form $x + iy$.
The given complex expression is equal to $\dfrac{{13}}{{17}} - \dfrac{i}{{17}}$.
Note: 1) Complex numbers are expressions in the form $x + iy$, where $x$ is the real part and $y$ is the imaginary part. These numbers cannot be marked on the number line. (Note that the imaginary part is $y$, and not $iy$.)
2) While rationalizing, we multiplied the entire expression with the denominator, but after changing the signs. This is called the conjugate of the complex number.
What is the conjugate of a complex number?
For a complex number, ${\rm Z} = x + iy$, its conjugate is $x - iy$, denoted by - $\bar {\rm Z}$. Conjugate, basically means changing the sign of the imaginary part.
Complete step-by-step solution:
We have been given a complex expression $\dfrac{{3 - i}}{{4 - i}}$.
Let us start by rationalizing the expression.
In order to rationalize, we divide and multiply the entire given expression by the denominator (but after changing the signs.) Look below how it is done.
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}}$
Now, we will solve it. If you look at the denominator, you will notice $\left( {a + b} \right)\left( {a - b} \right)$. We can simply solve it using the identity$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.
Simplifying the above expression,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 + 3i - 4i - {i^2}}}{{{4^2} - {i^2}}}$
On simplifying RHS we get,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 - i - {i^2}}}{{16 - {i^2}}}$
Now, we know that $i = \sqrt { - 1} $
If we square this expression, we get, ${i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1$
Hence, putting ${i^2} = - 1$ in the expression,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 - i - \left( { - 1} \right)}}{{16 - \left( { - 1} \right)}}$
Simplifying the expression,
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{12 - i + 1}}{{17}}$
Let us add the numerator term and we get
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} \times \dfrac{{4 + i}}{{4 + i}} = \dfrac{{13 - i}}{{17}}$
$ \Rightarrow \dfrac{{3 - i}}{{4 - i}} = \dfrac{{13}}{{17}} - \dfrac{i}{{17}}$
Hence, our final answer is $\dfrac{{3 - i}}{{4 - i}} = \dfrac{{13}}{{17}} - \dfrac{i}{{17}}$.
Also, notice that our answer is already in the standard form $x + iy$.
The given complex expression is equal to $\dfrac{{13}}{{17}} - \dfrac{i}{{17}}$.
Note: 1) Complex numbers are expressions in the form $x + iy$, where $x$ is the real part and $y$ is the imaginary part. These numbers cannot be marked on the number line. (Note that the imaginary part is $y$, and not $iy$.)
2) While rationalizing, we multiplied the entire expression with the denominator, but after changing the signs. This is called the conjugate of the complex number.
What is the conjugate of a complex number?
For a complex number, ${\rm Z} = x + iy$, its conjugate is $x - iy$, denoted by - $\bar {\rm Z}$. Conjugate, basically means changing the sign of the imaginary part.