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Question: Reduce \[\left( {\dfrac{1}{{1 - 4i}} - \dfrac{2}{{1 + i}}} \right)\left( {\dfrac{{3 - 4i}}{{5 + i}}}...

Reduce (114i21+i)(34i5+i)\left( {\dfrac{1}{{1 - 4i}} - \dfrac{2}{{1 + i}}} \right)\left( {\dfrac{{3 - 4i}}{{5 + i}}} \right) to the standard form.

Explanation

Solution

Here we are asked to reduce the given expression to standard form. The standard form of a complex number is nothing but a+iba + ib here aa is called the real part and bibi is called the imaginary part
(ii is called the imaginary number). We are supposed to reduce the given expression to the form a+iba + ib . This can be done by simplifying the expression by using conjugate multiplication since the terms are complex numbers.

Complete step-by-step answer:
In this question, we aim to reduce the given expression to the standard form.
The standard form is nothing but a+iba + ib where aa be a real part and bb be the complex part.
Let XX be (114i21+i)\left( {\dfrac{1}{{1 - 4i}} - \dfrac{2}{{1 + i}}} \right) and YY be (34i5+i)\left( {\dfrac{{3 - 4i}}{{5 + i}}} \right)
First, we will reduce XX consider the term
X=(114i21+i)X = \left( {\dfrac{1}{{1 - 4i}} - \dfrac{2}{{1 + i}}} \right)
Use the cross-multiplication method so that the second part denominator is multiplied to the first part on the numerator and the first denominator is multiplied on the second part on the numerator. Then multiply on both part denominator

=((1+i)(2(14i))(1+i)(14i)) =(1+i2+8i14i+i+4) X=(1+9i53i) = \left( {\dfrac{{\left( {1 + i} \right) - \left( {2(1 - 4i)} \right)}}{{\left( {1 + i} \right)\left( {1 - 4i} \right)}}} \right) \\\ = \left( {\dfrac{{1 + i - 2 + 8i}}{{1 - 4i + i + 4}}} \right) \\\ X = \left( {\dfrac{{ - 1 + 9i}}{{5 - 3i}}} \right) \\\

Finally, get the simplest form of XX
Now let us simplify the YY part consider the term
Y=(34i5+i)Y = \left( {\dfrac{{3 - 4i}}{{5 + i}}} \right)
There is no further simplification is need for this term so let us keep as it is.
Now multiply the XXand YYpart,
XY=(1+9i53i)(34i5+i)XY = \left( {\dfrac{{ - 1 + 9i}}{{5 - 3i}}} \right)\left( {\dfrac{{3 - 4i}}{{5 + i}}} \right)
On multiply the numerator to numerator and denominator to the denominator we get
XY=3+4i+27i+3625+5i15i+3XY = \dfrac{{ - 3 + 4i + 27i + 36}}{{25 + 5i - 15i + 3}}
XY=33+31i2810iXY = \dfrac{{33 + 31i}}{{28 - 10i}}
On further simplification we get
XY=33+31i2(145i)XY = \dfrac{{33 + 31i}}{{2\left( {14 - 5i} \right)}}
=12(33+31i145i×14+5i14+5i)= \dfrac{1}{2}\left( {\dfrac{{33 + 31i}}{{14 - 5i}} \times \dfrac{{14 + 5i}}{{14 + 5i}}} \right)
=12(462+434i+165i155196+25)= \dfrac{1}{2}\left( {\dfrac{{462 + 434i + 165i - 155}}{{196 + 25}}} \right)
Putting i2=1{i^2} = - 1 and adding the real parts and complex parts respectively we get
=12(307+599i221)= \dfrac{1}{2}\left( {\dfrac{{307 + 599i}}{{221}}} \right)
On simplifying this we get
=307+599i442= \dfrac{{307 + 599i}}{{442}}
=307442+599i442= \dfrac{{307}}{{442}} + \dfrac{{599i}}{{442}}
Now we get two perfect parts, real and imaginary.
307442+599i442\dfrac{{307}}{{442}} + \dfrac{{599i}}{{442}}
Thus, this is the reduced standard form.

Note: The complex number is nothing but the combination of the real and imaginary number and it will be in the form a+iba + ib this form is also known as standard form. To simplify a fraction of a complex number we cannot use the usual method we have to use conjugate multiplication that is nothing but multiplying and dividing the fraction by the conjugate term of its denominator. If a+iba + ib complex number then its conjugate is aiba - ib .