Reduce the given expression $\left( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right)\left( \dfrac{3-4i}{5+i} \right)$ to standard form

Hint: If the fraction is given in form $\dfrac{a+ib}{c+id}$ then multiply with c – id to both numerator and denominator just like in the question $\dfrac{33+31i}{28-10i}$ and multiply with 28 + 10i and then apply fact that ${{i}^{2}}=-1$ and then separate constant terms and terms with I and get desired result.

Complete step-by-step answer:
 In the question an expression is given $\left( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right)\left( \dfrac{3-4i}{5+i} \right)$ and change or express it to a standard a +ib form.
Before doing so, we will learn what complex numbers are?.
A complex number is a number that can be written in form of a + bi, where a, b are real numbers and i is a solution of the equation ${{x}^{2}}=-1$ .This is because no real value satisfies for equation ${{x}^{2}}+1=0$ or ${{x}^{2}}=-1$ , hence i is called imaginary number. For the complex number a + ib, a is considered as real part and b as imaginary part. Despite the historical nomenclature “imaginary” complex numbers are regarded in the mathematical sciences as just as “real” as real numbers and are fundamental in any aspects of scientific description of the natural world
So, $\left( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right)\left( \dfrac{3-4i}{5+i} \right)$ can be written as by taking LCM,
\[\left( \dfrac{\left( 1+i \right)-2\left( 1-4i \right)}{\left( 1-4i \right)\left( 1+i \right)} \right)\left( \dfrac{3-4i}{5+i} \right)\]
On simplifying we get,
\[\left( \dfrac{1+i-2+8i}{1-4i+i-4{{i}^{2}}} \right)\left( \dfrac{3-4i}{5+i} \right)\]
Using fact ${{i}^{2}}=-1$ and further simplifying we get,
$\left( \dfrac{-1+9i}{5-3i} \right)\left( \dfrac{3-4i}{5+i} \right)$
Now by multiplying we get,
$\dfrac{-3+4i+27i-36{{i}^{2}}}{25-15i+5i-3{{i}^{2}}}$
Now again using fact ${{i}^{2}}=-1$ and further simplifying we get,
$\dfrac{33+31i}{28-10i}$
If a fraction is in form of $\dfrac{a+ib}{c+id}$ then we will rationalize it by multiplying the numerator and denominator with (c – id) so we get,
\[\dfrac{\left( a+ib \right)\left( c-id \right)}{\left( c+id \right)\left( c-id \right)}\]
So on simplification we get,
$\dfrac{ac+ibc-iad-{{i}^{2}}bd}{{{c}^{2}}-icd+icd-{{i}^{2}}{{d}^{2}}}$
Now substituting ${{i}^{2}}=-1$ we get,
$\dfrac{ac+bd+ibc-iad}{{{c}^{2}}+{{d}^{2}}}$
Now on rearranging we can write it as
$\dfrac{\left( ac+bd \right)+i\left( bc-ad \right)}{{{c}^{2}}+{{d}^{2}}}$
Hence it can be written as
$\left( \dfrac{ac+bd}{{{c}^{2}}+{{d}^{2}}} \right)+i\left( \dfrac{bc-ad}{{{c}^{2}}+{{d}^{2}}} \right)$
which is a complex number in form of $x+iy$ where $x=\dfrac{ac+bd}{{{c}^{2}}+{{d}^{2}}}$ and $y=\dfrac{bc-ad}{{{c}^{2}}+{{d}^{2}}}$
Now we will do same process in $\dfrac{33+31i}{28-10i}$
So we will multiply by (28 + 10i) to both numerator and denominator so we get,
$\dfrac{\left( 33+31i \right)\left( 28+10i \right)}{\left( 28-10i \right)\left( 28+10i \right)}$
On further simplifying we get,
$\dfrac{924+868+330i+310{{i}^{2}}}{784-280i+280i-100{{i}^{2}}}$
Now once again using fact ${{i}^{2}}=-1$ so we get,
$\dfrac{614+1198i}{884}$
Hence it can be written as,
$\dfrac{614}{884}+\dfrac{1198}{884}i$
which is in form of a + ib
So, $a=\dfrac{614}{884}$ and $b=\dfrac{1198}{884}$
Hence the answer is $\dfrac{614}{884}+\dfrac{1198}{884}i$

Note: There is a notion that in denominator of fraction, complex number should not be there so if it is there, we will rationalize it by multiplying it with its conjugate like for an example if a + ib is there in numerator then we should multiply with a – ib.