Chapter 6.1 - Linear Inequalities in One Variable

In the previous section, we saw the disadvantages of using trial and error method for solving inequalities. In this section, we will see a systematic method.

For developing a systematic method, we must first learn some basic properties of inequalities. We can use those basic properties as rules for solving inequalities. This can be explained in 7 steps:

1. Let us recall the two rules that we used while solving linear equations:
First rule:
• Any number can be added to any side of the equation. The same number should be added to the other side also.
• Any number can be subtracted from any side of the equation. The same number should be subtracted from the other side also.
Second rule:
• Any side of the equation can be multiplied by a non-zero number. The other side also should be multiplied by the same number.  
• Any side of the equation can be divided by a non-zero number. The other side also should be divided by the same number.
2. In the same way, three rules can be developed for inequalities.
• If we add the same number to both sides of the inequality, there will be no change for the sign.
    ♦ For example, consider the inequality 2 < 8.
    ♦ Let us add 7 on both sides. We get 9 < 15.
    ♦ The ‘<’ sign has not changed.   
• If we subtract the same number from both sides of the inequality, there will be no change for the sign.
    ♦ For example, consider the inequality 12 < 21.
    ♦ Let us subtract 4 from both sides. We get 8 < 17.
    ♦ The ‘<’ sign has not changed.
3. So we can write the first rule for solving inequalities.
Rule 1 for solving inequalities:
• Any number can be added to any side of the inequality. The same number should be added to the other side also. Then the sign will not change.
• Any number can be subtracted from any side of the inequality. The same number should be subtracted from the other side also. Then the sign will not change.
4. If we multiply both sides of an inequality by the same +ve number, there will be no change for the sign.
    ♦ For example, consider the inequality 2 < 8.
    ♦ Let us multiply both sides by 3. We get 6 < 24.
    ♦ The ‘<’ sign has not changed.
• If we divide both sides of an inequality by the same +ve number, there will be no change for the sign.
    ♦ For example, consider the inequality 9 < 12.
    ♦ Let us divide both sides by 4. We get 2.25 < 3.
    ♦ The ‘<’ sign has not changed.
5. So we can write the second rule for solving inequalities.
Rule 2 for solving inequalities:
• Any side of the inequality can be multiplied by a +ve number. The other side also should be multiplied by the same number. The sign will not change.  
• Any side of the inequality can be divided by a +ve number. The other side also should be divided by the same number. The sign will not change.
6. If we multiply both sides of an inequality by the same -ve number, the sign will be reversed.
    ♦ For example, consider the inequality 2 < 8.
    ♦ Let us multiply both sides by -3. We get -6 > -24.
    ♦ The ‘<’ sign has become ‘>’.
• If we divide both sides of an inequality by the same -ve number, the sign will be reversed.
    ♦ For example, consider the inequality 9 < 12.
    ♦ Let us divide both sides by -4. We get -2.25 > -3.
    ♦ The ‘<’ sign has become ‘>'.
7. So we can write the third rule for solving inequalities.
Rule 3 for solving inequalities:
• Any side of the inequality can be multiplied by a -ve number. The other side also should be multiplied by the same number. The sign will be reversed.  
• Any side of the inequality can be divided by a -ve number. The other side also should be divided by the same number. The sign will be reversed.
• Reversal of sign means:
    ♦ < will become >      
    ♦ > will become <
    ♦ ≤ will become ≥     
    ♦ ≥ will become ≤

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Now we will see some solved examples

Solved example 6.1
Solve 30x < 200 when (i) x is a natural number  (ii) x is an integer
Solution:
• Given inequality is: 30x < 200
• Applying Rule 2, we get:
$\frac{30x}{30}~<~\frac{200}{30}$
$\Rightarrow x~<~\frac{20}{3}$
$\Rightarrow~x~<~6.667$
• So whichever value we select for x, must be less than 6.667
Part (i):
• The set of natural number is {1, 2, 3, 4, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {1, 2, 3, 4, 5, 6}
Part (ii):
• The set of integers is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

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Why do we specifically say natural numbers, integers etc.,?
The answer can be written in 2 steps:
1. In some problems, x may represent number of pens, number of books, number of cars etc., In such cases, x must be a natural number (or whole number).
2. In some problems, x may represent the floor level of a building.
    ♦ The floor immediately below the ground level is indicated by -1
    ♦ The floor below that is indicated by -2 and so on . . .
• In such cases, x must be an integer.

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Solved example 6.2
Solve 5x-3 < 3x+1 when (i) x is an integer  (ii) x is a real number
Solution:
• Given inequality is: 5x-3 < 3x+1
• This can be simplified as follows:
$\begin{array}{ll}
{}&5x-3 &{}<{}& {3x+1} &{} \\
\Rightarrow &5x-3+3&{}<{}& 3x+1+3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &5x&{}<{}& 3x+4 &{} \\
\Rightarrow &5x-3x&{}<{}& 3x+4-3x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 4 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{4}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 2 &{} \\
\end{array}$

Part (i):
• The set of integers is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {. . . , -4, -3, -2, -1, 0, 1}
Part (ii):
• We know that the set of real numbers contain all types of numbers.
    ♦ They include natural numbers, whole numbers, integers
    ♦ They also include rational numbers and irrational numbers.
    ♦ So there will be numbers like -√3, -√2, √3, √2, -π, π etc.,
• Since they do not occur at regular intervals, we pick out the "relevant portion of the number line" and write it as an interval.
• In our present case, the interval will be: (-∞,2)
• That means, the relevant portion on the number line in our case, begins from -∞ at the extreme left end. It extends upto +2.
• But the exact +2 should not be included. Numbers like 1.999, 1.9999 etc., are allowed. This is indicated by the ')' on the right side of 2.
• This interval can be shown graphically as in fig.6.1 below:

Fig.6.1

• The yellow line represents the number line. The red line is the graph.
• The arrow at the left end of the red line indicates that, the left boundary of the interval is at -∞.
• The circle at the right end of the red line indicates that, the right boundary of the interval is at 2.
   ♦ Since 2 is not included in the interval, it is an ordinary circle.
   ♦ If 2 is also included, we give a filled circle.
• All points on the red line is a solution of the inequality. We can write:
When x is a real number, the solution set is: (-∞,2)

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Why do we specifically say integers, real numbers etc.,?
The answer can be written in 2 steps:
1. In solved example 6.1 above, we saw the situations where integers are specified. So we need not discuss about it again.
2. In some problems, x may represent lengths or distances. Such quantities are not always available as integers.
• We may encounter lengths like 3.5 cm, 121.667 meter etc.,
• Some times distances towards the left are considered -ve and those towards the right are considered +ve
   ♦ Then we may encounter lengths like -3.5 cm, -121.667 meter etc.,
• Also it is possible to encounter lengths like -2√3 meter, 5√2 meter, -3π cm etc.,
• In some problems, x may represent temperature.
• In some problems, x may represent volume.
• In all such cases, x must be a real number.

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We have seen the significance of specifying whether x is natural number, integer or real number. For the rest of our discussion in this chapter, the variables x, y etc., will be considered as representing real numbers.

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Solved example 6.3
Solve 4x+3 < 6x+7
Solution:
• Given inequality is: 4x+3 < 6x+7
• This can be simplified as follows:
$\begin{array}{ll}
{}&4x+3 &{}<{}& {6x+7} &{} \\
\Rightarrow &4x+3-3&{}<{}& 6x+7-3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &4x&{}<{}& 6x+4 &{} \\
\Rightarrow &4x-6x&{}<{}& 6x+4-6x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-2x&{}<{}& 4 &{} \\
\Rightarrow &\frac{-2x}{-2}&{}<{}& \frac{4}{-2} &\color {green}{\text{(Rule 3)}} \\
\Rightarrow &x&{}>{}& -2 &{} \\
\end{array}$
• So all real numbers greater than -2 are the solutions of this inequality.
• The solution set is: (-2,∞)
• The graph is shown in fig.6.2 below:

Fig.6.2

Solved example 6.4
Solve $\frac{5-2x}{3}\le\frac{x}{6} -5$
Solution:
• Given inequality is: $\frac{5-2x}{3}\le\frac{x}{6} -5$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{5-2x}{3}&{}\le{}& \frac{x}{6} -5 &{} \\
\Rightarrow &6 \left(\frac{5-2x}{3}\right)&{}\le{}& 6\left(\frac{x}{6} -5\right) &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &2 \left(5-2x \right)&{}\le{}& x-30 &{} \\
\Rightarrow &10-4x&{}\le{}& x-30 &{} \\
\Rightarrow &10-4x+4x&{}\le{}& x-30+4x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &10&{}\le{}& 5x-30 &{} \\
\Rightarrow &10+30&{}\le{}& 5x-30+30 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &40&{}\le{}& 5x &{} \\
\Rightarrow &\frac{40}{5}&{}\le{}& \frac{5x}{5} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &8&{}\le{}& x &{} \\
\Rightarrow &x&{}\ge{}& 8 &{} \\
\end{array}$
• So all real numbers greater than or equal to 8 are the solutions of this inequality.
• The solution set is: [8,∞)
• The graph is shown in fig.6.3 below:

Fig.6.3

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• In the next section, we will see a few more solved examples.

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Chapter 6 - Linear Inequalities

In the previous section, we completed a discussion on complex numbers and quadratic equations. In this chapter, we will see Linear Inequalities.

In our earlier classes, we have seen statements which can be converted into mathematical equations. Let us see some examples:
Example 1:
• If the cost of one book is Rs 20/-, then the cost of n books will be Rs 20n
• We can write it in the form of an equation:
Total cost = 20n
• A student purchasing books will have to make the payment based on this equation.
Example 2:
• If the cost of one book is Rs 20 and that of a pen is Rs 8, then the cost of x books and y pens will be Rs (20x + 8y)
• We can write it in the form of an equation:
Total cost = 20x + 8y
• A student purchasing books and pens will have to make the payment based on this equation.

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• In the above two examples, we effectively converted the statements into mathematical equations ('Equation' indicates 'equality'. Note the ‘=’ sign in both examples).
• But in our day to day life, we will come across some types of statements which are impossible to convert into equations. Let us see some examples:

Example 3:
• If the cost of one book is Rs 20/-, then the cost of n books will be Rs 20n
• If a student has Rs 90/- with him, he will have to make the purchase in such a way that, the total cost is less than 90
• We can write it in a mathematical form:
20n < 90
Example 4:
• If the cost of one book is Rs 20 and that of a pen is Rs 8, then the cost of x books and y pens will be Rs (20x + 8y)
• If a student has Rs 120/- with him, he will have to make the purchase in such a way that, the total cost is less than 120
• We can write it in a mathematical form:
20x + 8y ≤ 120

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• In this situation, a doubt will arise in the mind of the reader:
In the third example, the sign is '<'. But in the fourth example, the sign is '≤'. Why is it so?
• The answer can be written in 2 steps:
1. In the third example, we count the number of books using whole numbers 0, 1, 2, 3, 4 etc.,
   ♦ No whole number when multiplied with 20, will give 90.
   ♦ In other words, 90 is not a multiple of 20.
   ♦ So the total cost in this case will never become equal to 90.
   ♦ Consequently, we cannot use '≤'. We can use only '<'
2. In the fourth example also, we count the number of books and pens using whole numbers.
   ♦ If the number of books (x) is 4 and number of pens (y) is 5, we get:
   ♦ Total cost = (20 × 4 + 8 × 5) = (80 + 40) =120
   ♦ So a total cost equal to 120 is possible.
   ♦ Consequently, we cannot use '<'. We can use only '≤'

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• Consider the four signs given below:
   ♦ < (less than)
   ♦ > (greater than)
   ♦ ≤ (less than or equal)
   ♦ ≥ (greater than or equal)
• If a statement contains any one of the above four signs, then that statement is called an inequality.
• The statement that we saw in example 3 is: 20n < 90
   ♦ This statement is an inequality.  
• The statement that we saw in example 4 is: 20x + 8y ≤ 120
   ♦ This statement is also an inequality.  
   ♦ In fact, this statement is a combination of two statements:
         ✰ (i) 20x +8y < 120
         ✰ (ii) 20x +8y = 120
   ♦ Statement (i) is an inequality.
   ♦ Statement (ii) is an equation.

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Definition 1
This definition can be written in 11 steps:
1. If a statement contains any one of the four signs <, >, ≤ or ≥, then that statement is called an inequality.
2. Inequalities can be broadly classified into two categories:
◼ Numerical inequalities
◼ Literal inequalities
• Two real numbers related by <, >, ≤ or ≥ is called a numerical inequality.
For example:
   ♦ 5 < 8
   ♦ 11 > 9
• Two algebraic expressions related by <, >, ≤ or ≥ is called a literal inequality.
For example:
   ♦ x < y
   ♦ (x+y) > z
• A real number and an algebraic expression related by <, >, ≤ or ≥ is also called a literal inequality.
For example:
   ♦ x < 8
   ♦ (x+y) ≥ 2
3. All inequalities that we meet in mathematics, science, engineering, statistics, economics, psychology etc., will fall in one of the above two categories.
4. If an inequality has two of the four signs , it is called a double inequality.
For example:
   ♦ 2 < 7 < 11
         ✰ This is read as: 7 is greater than 2 and less than 11
   ♦ 4 ≤ x < 15
         ✰ This is read as: x is greater than or equal to 4 and less than 15
   ♦ 3 < y ≤ 8
         ✰ This is read as: y is greater than 3 and less than or equal to 8
5. If the inequality involves only < or >, it is called a strict inequality.
For example:
   ♦ ax+b < 0
   ♦ ax+b > 0
6. If the inequality involves ≤ or ≥, it is called a slack inequality.
For example:
   ♦ ax+b ≤ 0
   ♦ ax+b ≥ 0
7. If the algebraic expression has only one variable, it is called an inequality in one variable.
For example:
   ♦ ax+b ≤ 0
         ✰ a and b are constants. x is the variable.
   ♦ ay+b > c
         ✰ a, b and c are constants. y is the variable.
8. If the algebraic expression has two variables, it is called an inequality in two variables.
For example:
   ♦ ax+by < c
         ✰ a, b and c are constants. x and y are the variables.
   ♦ ax+by ≥ c
         ✰ a, b and c are constants. x and y are the variables.
9. If in an "inequality in one variable", the highest exponent of the variable is '1', it is called a linear inequality in one variable.
For example:
   ♦ ax+b ≤ 0
         ✰ x is the variable. It's highest exponent is 1
   ♦ ay+b > c
         ✰ y is the variable. It's highest exponent is 1
(Recall that if the highest exponent is '1', the graph will be a line. That is why, it is called 'linear')
10. If in an "inequality in two variables", the highest exponent of both variables is '1', it is called a linear inequality in two variables.
For example:
   ♦ ax+by ≤ 0
         ✰ x and y are the variables.Highest exponent is 1 for both of them.
   ♦ ay+by > c
         ✰ x and y are the variables.Highest exponent is 1 for both of them.
11. If in an "inequality in one variable", the highest exponent of the variable is '2', it is called a quadratic inequality in one variable. It's graph will not be a line.
For example:
   ♦ ax²+bx + c ≤ 0
         ✰ x is the variable. It's highest exponent is 2
   ♦ ax²+bx > c
         ✰ x is the variable. It's highest exponent is 2

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In this chapter, we will learn about inequalities that fall in the following two categories:
(i) Linear inequality in one variable.
(ii) Linear inequality in two variables.

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Algebraic solutions of linear inequalities in one variable

• Consider the inequality (20n < 90) that we saw in example 3 above. An analysis about this inequality, can be written in 5 steps:

1. We want to find the value of n (the number of books).
• If the correct value of n can be calculated, the cost of purchase will not go above Rs 90/-
• n cannot be a -ve number because, we cannot buy -ve number of books.
• Also n cannot be decimals or fractions like 2.5, 3/2 etc., because, a book cannot be made into fractions.
2. So it is obvious that, n must be a whole number. Let us try various whole numbers:
(i) Put n = 0,
    ♦ the left hand side (LHS) become (20 × 0) = 0
    ♦ the right hand side (RHS) is always 90
    ♦ 0 < 90
`• So when n = 0, the statement is true. 
(ii) Put n = 1,
    ♦ LHS become (20 × 1) = 20
    ♦ RHS is always 90
    ♦ 20 < 90
`• So when n = 1, the statement is true. 
(iii) Put n = 2,
    ♦ LHS become (20 × 2) = 40
    ♦ RHS is always 90
    ♦ 40 < 90
`• So when n = 2, the statement is true. 
(iv) Put n = 3,
    ♦ LHS become (20 × 3) = 60
    ♦ RHS is always 90
    ♦ 60 < 90
`• So when n = 3, the statement is true. 
(v) Put n = 4,
    ♦ LHS become (20 × 4) = 80
    ♦ RHS is always 90
    ♦ 80 < 90
`• So when n = 4, the statement is true.
(vi) Put n = 5,
    ♦ LHS become (20 × 5) = 100
    ♦ RHS is always 90
    ♦ 100 ≮ 90
`• So when n = 5, the statement is false.
3. We can write:
The values of n which make the inequality true, are: 0, 1, 2, 3, 4
4. Now we can define solutions:
◼ The values which make the inequality true are called solutions of inequality.
5. Also we can define solution set:
◼ The set containing all the solutions of an inequality is called the solution set of that inequality.
◼ Such a set must contain only the solutions. Numbers which are not solutions must not be included in the solution set.
• So in our present case, the solution set is {0, 1, 2, 3, 4}

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Let us see the analysis of another example. It can be written in 7 steps:
1. A car parking area can accommodate a maximum of 15 cars. There are 9 cars already parked. What is the additional number of cars that can be allowed?
2. We can write the statement as an inequality: 9+x ≤ 15
Where x is the possible number of additional cars.
3. We want to find the value of x (the number of cars).
• If the correct value of x can be calculated, the parking will be OK.
• x cannot be a -ve number because, number of cars cannot be -ve.
• Also x cannot be decimals or fractions like 2.5, 3/2 etc., because, a car cannot be made into fractions.
4. So it is obvious that, x must be a whole number. Let us try various whole numbers:
(i) Put x = 0,
    ♦ the left hand side (LHS) become (9+0) = 9
    ♦ the right hand side (RHS) is always 15
    ♦ 9 ≤ 15
`• So when x = 0, the statement is true. 
(ii) Put x = 1,
    ♦ LHS become (9+1) = 10
    ♦ RHS is always 15
    ♦ 10 ≤ 15
`• So when x = 1, the statement is true. 
(iii) Put x = 2,
    ♦ LHS become (9+2) = 11
    ♦ RHS is always 15
    ♦ 11 ≤ 15
`• So when x = 2, the statement is true. 
(iv) Put x = 3,
    ♦ LHS become (9+3) = 12
    ♦ RHS is always 15
    ♦ 12 ≤ 15
`• So when x = 3, the statement is true. 
(v) Put x = 4,
    ♦ LHS become (9+4) = 13
    ♦ RHS is always 15
    ♦ 13 ≤ 15
`• So when x = 4, the statement is true. 
(vi) Put x = 5,
    ♦ LHS become (9+5) = 14
    ♦ RHS is always 15
    ♦ 14 ≤ 15
`• So when x = 5, the statement is true. 
(vii) Put x = 6,
    ♦ LHS become (9+6) = 15
    ♦ RHS is always 15
    ♦ 15 ≤ 15
`• So when x = 6, the statement is true. 
(viii) Put x = 7,
    ♦ LHS become (9+7) = 16
    ♦ RHS is always 15
    ♦ 16 ≰ 15
`• So when x = 7, the statement is false.
5. We can write:
The values of x which make the inequality true, are: 0, 1, 2, 3, 4, 5, 6
6. Now we can define solutions:
◼ The values which make the inequality true are called solutions of inequality.
7. Also we can define solution set:
◼ The set containing all the solutions of an inequality is called the solution set of that inequality.
◼ Such a set must contain only the solutions. Numbers which are not solutions must not be included in the solution set.
• So in our present case, the solution set is {0, 1, 2, 3, 4, 5, 6}

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• We have seen two examples. In both those examples, we found out the solution set using trial and error method.
• But the trial and error method is time consuming. Also it may not work in all cases.
• So we must develop a more systematic method. We will see such a method in the next section.

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Chapter 5.7 - Miscellaneous Examples

In the previous section, we completed a discussion on quadratic equations. We saw some miscellaneous examples also. In this section, we will see a few more miscellaneous examples.

Solved example 5.13
If $x+yi~=~\frac{a+bi}{a-bi}$, prove that $x^2+y^2=1$
Solution:
1. We will change the right side into the general form:
$\begin{array}{ll}
\frac{a+bi}{a-bi}&{}={}&\frac{a+bi}{a-bi} × \frac{a+bi}{a+bi}& {} &{} \\
\phantom{\frac{a+bi}{a-bi}}&{}={}&\frac{a^2+2abi-b^2}{a^2+b^2}&{} \\
\phantom{\frac{a+bi}{a-bi}}&{}={}&\frac{a^2-b^2}{a^2+b^2}~+~\frac{2abi}{a^2+b^2}&{} \\
\phantom{\frac{a+bi}{a-bi}}&{}={}&\frac{a^2-b^2}{a^2+b^2}~+~\left(\frac{2ab}{a^2+b^2}\right)i&{} \\
\end{array}$
2. So the given expression becomes: $x+yi~=~\frac{a^2-b^2}{a^2+b^2}~+~\left(\frac{2ab}{a^2+b^2}\right)i$
3. Equating the corresponding real and imaginary parts, we get:
$x~=~\frac{a^2-b^2}{a^2+b^2}$
$y~=~\frac{2ab}{a^2+b^2}$
4. Now we can find $x^2+y^2$:
$\begin{array}{ll}
x^2+y^2&{}={}&\left(\frac{a^2-b^2}{a^2+b^2} \right)^2~+~\left(\frac{2ab}{a^2+b^2} \right)^2& {} &{} \\
\phantom{x^2+y^2}&{}={}&\frac{(a^2-b^2)^2~+~(2ab)^2}{(a^2+b^2)^2}&{} \\
\phantom{x^2+y^2}&{}={}&\frac{a^4-2a^2 b^2+b^4~+~4a^2 b^2}{(a^2+b^2)^2}&{} \\
\phantom{x^2+y^2}&{}={}&\frac{a^4+2a^2 b^2+b^4}{(a^2+b^2)^2}&{} \\
\phantom{x^2+y^2}&{}={}&\frac{(a^2+b^2)^2}{(a^2+b^2)^2}&{} \\
\phantom{x^2+y^2}&{}={}&1&{} \\
\end{array}$

Solved example 5.14
Find real 𝜽 such that $\frac{3+2i\sin \theta}{1-2i\sin \theta}$ is purely real
Solution:
1. We will change the given expression into the general form:
$\begin{array}{ll}
\frac{3+2i\sin \theta}{1-2i\sin \theta}&{}={}&\frac{3+2i\sin \theta}{1-2i\sin \theta}~ × ~\frac{1+2i\sin \theta}{1+2i\sin \theta}& {} &{} \\
\phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3+6i \sin \theta +2i \sin \theta - 4 \sin^2 \theta}{1+4 \sin^2 \theta}&{} \\
\phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3+8i \sin \theta - 4 \sin^2 \theta}{1+4 \sin^2 \theta}&{} \\
\phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3- 4 \sin^2 \theta}{1+4 \sin^2 \theta}~+~\frac{8i \sin \theta}{1+4 \sin^2 \theta}&{} \\
\phantom{\frac{3+2i\sin \theta}{1-2i\sin \theta}}&{}={}&\frac{3- 4 \sin^2 \theta}{1+4 \sin^2 \theta}~+~\left(\frac{8 \sin \theta}{1+4 \sin^2 \theta}\right)i&{} \\
\end{array}$
2. So the given complex number is: $\frac{3- 4 \sin^2 \theta}{1+4 \sin^2 \theta}~+~\left(\frac{8 \sin \theta}{1+4 \sin^2 \theta}\right)i$
3. If this complex number is to be purely real, the imaginary part must be zero.
• That is., $\frac{8 \sin \theta}{1+4 \sin^2 \theta}$ must be zero.
• Denominator cannot be zero because, division by zero will give a number which does not exist.
• So we can write:
If the given complex number is to be purely real, $8 \sin \theta$ must be zero.
   ♦ That is., $8 \sin \theta~=~0$
   ♦ This is true only when sin 𝜽 = 0
4. sin 𝜽 = 0 is a trigonometrical equation.
• We know that, this equation is true whenever 𝜽 is a multiple of 𝞹
5. So we can write:
The given complex number will be purely real when 𝜽 = n𝞹
   ♦ Where n is member of the set of integers Z.

Solved example 5.15
Convert the complex number $\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ in the polar form.
Solution:
1. We will change the given expression into the general form:
$\begin{array}{ll}
\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}&{}={}&\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}~ × ~\frac{\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}}{\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}}& {} &{} \\
\phantom{\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}}&{}={}&\frac{(i-1)(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3})}{\cos^2 \frac{\pi}{3}+ \sin^2 \frac{\pi}{3}}&{} \\
\phantom{\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}}&{}={}&\frac{(i-1)( \frac{1}{2}- \frac{\sqrt{3}}{2}i)}{1}&{} \\
\phantom{\frac{i-1}{\cos \frac{\pi}{3}+ \frac{\pi}{3}}}&{}={}&(i-1)( \frac{1}{2}- \frac{\sqrt{3}}{2}i)&{} \\
\phantom{\frac{i-1}{\cos \frac{\pi}{3}+ \frac{\pi}{3}}}&{}={}&{\frac{1}{2}}i+\frac{\sqrt{3}}{2}-\frac{1}{2}+{\frac{\sqrt{3}}{2}}i&{} \\
\phantom{\frac{i-1}{\cos \frac{\pi}{3}+ \frac{\pi}{3}}}&{}={}&\frac{\sqrt{3}-1}{2}~+~\left(\frac{\sqrt{3}+1}{2} \right)i&{} \\
\end{array}$
2. Now the complex number is in the form x+yi
We have to convert it into the form r[cos 𝜽+ i sin 𝜽]
3. We know that the modulus r is given by: $r=\sqrt{x^2+y^2}$
So in our present case,
$\begin{array}{ll}
r&{}={}&\sqrt{x^2+y^2}& {} &{} \\
\phantom{r}&{}={}&\sqrt{\left(\frac{\sqrt{3}-1}{2} \right)^2+\left(\frac{\sqrt{3}+1}{2} \right)^2}&{} \\
\phantom{r}&{}={}&\sqrt{\frac{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2}{4}}&{} \\
\phantom{r}&{}={}&\sqrt{\frac{3-2\sqrt{3}+1~+~3+2\sqrt{3}+1}{4}}&{} \\
\phantom{r}&{}={}&\sqrt{\frac{8}{4}}&{} \\
\phantom{r}&{}={}&\sqrt{2}&{} \\
\end{array}$
4. Since x+yi and r[cos 𝜽+ i sin 𝜽] represent the same complex number, we can equate the corresponding real and imaginary parts. We get:
    ♦ x = r cos 𝜽
    ♦ y = r sin 𝜽
5. From this we get:
(i) $\frac{\sqrt{3}-1}{2}=\sqrt{2} \cos \theta$
(ii) $\frac{\sqrt{3}+1}{2}=\sqrt{2} \sin \theta$
• sin 𝜽 is +ve. cos 𝜽 is also +ve. So it is in the first quadrant.
6. Taking ratios, (ii) to (i), we get:
$\frac{\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta}=\frac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}}$
$\Rightarrow \tan \theta = \frac{\sqrt{3}+1}{\sqrt{3}-1}$
• This is a trigonometrical equation. We learned to solve them in chapter 3. (Details here)
7. We have, $\tan \frac{5\pi}{12}~=~ \tan 75^o~=~\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Proof:
$\begin{array}{ll}
\tan \frac{5 \pi}{12}&{}={}&\tan \left(\frac{5 \pi}{12}~\times ~\frac{2}{2}\right)~=~\tan \frac{10 \pi}{24}& {} &{} \\
\phantom{\tan \frac{5 \pi}{12}}&{}={}&\tan \left(\frac{\pi}{4}+\frac{\pi}{6} \right)&{} \\
\phantom{\tan \frac{5 \pi}{12}}&{}={}&\frac{\tan \frac{\pi}{4}~+~\tan \frac{\pi}{6}}{1-\tan \frac{\pi}{4}\tan \frac{\pi}{6}}&{} \\
\phantom{\tan \frac{5 \pi}{12}}&{}&\color {green}{\because \tan(p+q)=\frac{\tan p + \tan q}{1-\tan p \tan q}}&{} \\
\phantom{\tan \frac{5 \pi}{12}}&{}={}&\frac{1~+~\frac{1}{\sqrt3}}{1-1 \times \frac{1}{\sqrt3}}&{} \\
\phantom{\tan \frac{5 \pi}{12}}&{}={}&\frac{\sqrt3+1}{\sqrt3-1}&{} \\
\end{array}$
• From this we get: $\theta = \frac{5\pi}{12}$
• This 𝜽 is in the first quadrant. So there is no need to find the other value of 𝜽.
8. Thus we get the values of r and 𝜽:
   ♦ From (3), we get: $r= \sqrt{2}$
   ♦ From (7),we get: $\theta = \frac{5\pi}{12}$
• So the polar form is: $\sqrt{2}\left[\cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right]$
• The red dot in fig.5.9 below represents the complex number:

Fig.5.9

Solved example 5.16
Find the complex number z if $|z+1|=z+2(1+i)$.
Solution:
1. Let z = a+bi. Then the given equation becomes:
$\begin{array}{ll}
|a+bi+1|&{}={}&a+bi+2(1+i)& {} &{} \\
\Rightarrow |(a+1)+bi|&{}={}&a+bi+2+2i& {} &{} \\
\Rightarrow \sqrt{(a+1)^2+b^2}&{}={}&a+2+(b+2)i& {} &{} \\
\end{array}$
2. Equating the corresponding real and imaginary parts, we get:
(i) $\sqrt{(a+1)^2+b^2}~=~a+2$
(ii) $0 ~=~b+2$
3. From (ii), we get: b = -2
• Substituting for b in (i), we get:
$\begin{array}{ll}
\sqrt{(a+1)^2+b^2}&{}={}&a+2& {} &{} \\
\Rightarrow \sqrt{(a+1)^2+(-2)^2}&{}={}&a+2& {} &{} \\
\Rightarrow \sqrt{a^2+2a+1+4}&{}={}&a+2& {} &{} \\
\Rightarrow \sqrt{a^2+2a+5}&{}={}&a+2& {} &{} \\
\Rightarrow a^2+2a+5&{}={}&(a+2)^2& {} &{} \\
\phantom{\frac{1}{2-3i}}&{}&\color {green}{\text{(Squaring both sides)}} &{} \\
\Rightarrow a^2+2a+5&{}={}&a^2+4a+4& {} &{} \\
\Rightarrow 1&{}={}&2a& {} &{} \\
\Rightarrow a&{}={}&\frac{1}{2}& {} &{} \\
\end{array}$
4. Thus the complex number z = a+bi is $\frac{1}{2}-2i$

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The link below gives a PDF file with more miscellaneous examples

Miscellaneous exercise on chapter 5

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In the next chapter, we will see linear inequalities.

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