
How do you write the following expression in standard form \[\dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}}\]?
Answer
453k+ views
Hint: Here we will firstly take the LCM of both the fraction and solve it to get a simplified fraction. Then we will simply multiply and divide the simplified fraction with the conjugate of the denominator of the fraction and solve it to get the standard form of the given equation.
Complete Step by Step Solution:
Given equation is \[\dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}}\].
Firstly we will take the LCM of the both the fraction in the given equation. Therefore, we get
\[ \dfrac{{\left( {1 + i} \right)\left( {4 - i} \right) - 3i}}{{i\left( {4 - i} \right)}}\]
Now we will simplify the above equation, we get
\[ \dfrac{{4 - i + 4i - {i^2} - 3i}}{{4i - {i^2}}}\]
\[ \dfrac{{4 - {i^2}}}{{4i - {i^2}}}\]
We know that the value of \[{i^2}\] is equal to \[ - 1\]. Therefore, we will put the value in the above equation, we get
\[ \dfrac{{4 - \left( { - 1} \right)}}{{4i - \left( { - 1} \right)}}\]
\[ \dfrac{5}{{1 + 4i}}\]
Now we will simply multiply and divide above the equation with the conjugate of the denominator of the fraction i.e. \[1 - 4i\]. Therefore, we get
\[ \dfrac{5}{{1 + 4i}} \times \dfrac{{1 - 4i}}{{1 - 4i}}\]
We know this algebraic property i.e. \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]. Therefore, by using this we get
\[ \dfrac{{5 - 20i}}{{{1^2} - {{\left( {4i} \right)}^2}}}\]
\[ \dfrac{{5 - 20i}}{{1 - 16{i^2}}}\]
Now we again put the value of \[{i^2}\] in the above equation, we get
\[ \dfrac{{5 - 20i}}{{1 + 16}}\]
\[ \dfrac{{5 - 20i}}{{17}}\]
Now we will write it in the split form. Therefore, we get
\[ \dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}} = \dfrac{5}{{17}} - \dfrac{{20i}}{{17}}\]
Hence, the expression \[\dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}}\] in the standard form is written as \[\dfrac{5}{{17}} - \dfrac{{20i}}{{17}}\].
Note:
We should know the standard form of writing an imaginary number. Imaginary numbers are generally written in the form of \[a + ib\] where \[a,b\] are the real numbers. In the imaginary number, the first term is the real part of the number and the second term is the imaginary part of the number.
Real value is a number which has some real or discrete or possible value. But imaginary value is the number with a real number multiplied with an imaginary part \[i\].
Complete Step by Step Solution:
Given equation is \[\dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}}\].
Firstly we will take the LCM of the both the fraction in the given equation. Therefore, we get
\[ \dfrac{{\left( {1 + i} \right)\left( {4 - i} \right) - 3i}}{{i\left( {4 - i} \right)}}\]
Now we will simplify the above equation, we get
\[ \dfrac{{4 - i + 4i - {i^2} - 3i}}{{4i - {i^2}}}\]
\[ \dfrac{{4 - {i^2}}}{{4i - {i^2}}}\]
We know that the value of \[{i^2}\] is equal to \[ - 1\]. Therefore, we will put the value in the above equation, we get
\[ \dfrac{{4 - \left( { - 1} \right)}}{{4i - \left( { - 1} \right)}}\]
\[ \dfrac{5}{{1 + 4i}}\]
Now we will simply multiply and divide above the equation with the conjugate of the denominator of the fraction i.e. \[1 - 4i\]. Therefore, we get
\[ \dfrac{5}{{1 + 4i}} \times \dfrac{{1 - 4i}}{{1 - 4i}}\]
We know this algebraic property i.e. \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]. Therefore, by using this we get
\[ \dfrac{{5 - 20i}}{{{1^2} - {{\left( {4i} \right)}^2}}}\]
\[ \dfrac{{5 - 20i}}{{1 - 16{i^2}}}\]
Now we again put the value of \[{i^2}\] in the above equation, we get
\[ \dfrac{{5 - 20i}}{{1 + 16}}\]
\[ \dfrac{{5 - 20i}}{{17}}\]
Now we will write it in the split form. Therefore, we get
\[ \dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}} = \dfrac{5}{{17}} - \dfrac{{20i}}{{17}}\]
Hence, the expression \[\dfrac{{1 + i}}{i} - \dfrac{3}{{4 - i}}\] in the standard form is written as \[\dfrac{5}{{17}} - \dfrac{{20i}}{{17}}\].
Note:
We should know the standard form of writing an imaginary number. Imaginary numbers are generally written in the form of \[a + ib\] where \[a,b\] are the real numbers. In the imaginary number, the first term is the real part of the number and the second term is the imaginary part of the number.
Real value is a number which has some real or discrete or possible value. But imaginary value is the number with a real number multiplied with an imaginary part \[i\].
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

A mixture of o nitrophenol and p nitrophenol can be class 11 chemistry CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE
