21.4 - Solved Examples on Algebra of Continuous Functions

In the previous section, we saw algebra of continuous functions. In this section, we will see some solved examples.

Solved Example 21.16
Prove that every rational function is continuous.
Solution:
1. Every rational function will be of the form: $f(x) = \frac{p(x)}{q(x)}$
    ♦ p(x) and q(x) are polynomial functions.
    ♦ Also q(x) should not be zero.
2. From the algebra of limits, we know that:
If both numerator and denominator are continuous functions, then that fraction will be a continuous function.
3. In our present case, both numerator and denominator are polynomial functions. We have seen that all polynomial functions are continuous.
• So we can write: all rational functions are continuous.
4. We must consider the domain of a given rational function. That domain will not contain any real number which makes q(x) equal to zero.
• So for all values in the domain, the given rational function will be indeed continuous.

Solved Example 21.17
Discuss the continuity of the sine function.
Solution:
1. Fig.21.11 below shows the graph of the sine function f(x) = sin x.

Fig.21.11

2. An arbitrary point c is marked on the graph.
We want to find $\lim_{x\rightarrow c^{-}} f(x)$  and $\lim_{x\rightarrow c^{+}} f(x)$ 

3. First we will find $\lim_{x\rightarrow c^{-}} f(x)$. It can be done in 3 steps:
(i) Consider a point to the left of c. We can write it as x = c−h
(ii) When x approaches c, h will approach zero.
(iii) So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c^{-}} f(x)}    & {~=~}    &{\lim_{h\rightarrow 0} \sin (c-h)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h - \cos c \, \sin h]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h] ~-~\lim_{h\rightarrow 0} [\cos c \, \sin h]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \lim_{h\rightarrow 0} [\cos h] ~-~\cos c \, \lim_{h\rightarrow 0} [ \sin h]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \cos 0 ~-~\cos c \,  \sin 0}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, (1) ~-~\cos c \, (0)}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\sin c}    \\
\end{array}$

◼ Remarks:
• 2 (magenta color): Here we use the identity:
sin(a−b) = sin a cos b − cos a sin b
• 5 (magenta color):
    ♦ From the graph of the cosine function, it is clear that, when the input value approaches zero from left or right, the limiting cosine value is 1.
    ♦ From the graph of the sine function, it is clear that, when the input value approaches zero from left or right, the limiting sine value is 0.

4. Next we will find $\lim_{x\rightarrow c^{+}} f(x)$. It can be done in 3 steps:
(i) Consider a point to the right of c. We can write it as x = c+h. This is shown in the graph below:

Fig.21.12

(ii) When x approaches c, h will approach zero.
(iii) So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c^{+}} f(x)}    & {~=~}    &{\lim_{h\rightarrow 0} \sin (c+h)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h + \cos c \, \sin h]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h] ~+~\lim_{h\rightarrow 0} [\cos c \, \sin h]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \lim_{h\rightarrow 0} [\cos h] ~+~\cos c \, \lim_{h\rightarrow 0} [ \sin h]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \cos 0 ~+~\cos c \,  \sin 0}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, (1) ~+~\cos c \, (0)}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\sin c}    \\
\end{array}$

◼ Remarks:
• 2 (magenta color): Here we use the identity:
sin(a+b) = sin a cos b + cos a sin b

5. We see that, left side limit is equal to the right side limit. That means, limit exists.

6. We know that, the value of the function at x = c is:
f(c) = sin c

7. We can write: $\lim_{x\rightarrow c} f(x) ~=~f(c)$
So f(x) = sin x is a continuous function.

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In a similar way, we can prove that, the cosine function is also a continuous function. Steps are shown briefly below:

1. Left side limit can be calculated as:

◼ Remarks:
• 2 (magenta color): Here we use the identity:
cos(a−b) = cos a cos b + sin a sin b

2. Right side limit can be calculated as:

◼ Remarks:
• 2 (magenta color): Here we use the identity:
cos(a+b) = cos a cos b − sin a sin b

3. We know that, the value of the function at x = c is:
f(c) = cos c

4. We can write: $\lim_{x\rightarrow c} f(x) ~=~f(c)$
So f(x) = cos x is a continuous function.

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Solved Example 21.18
Discuss the continuity of the function f(x) = tan x.
Solution:
1. f(x) = tan x can be written as: $\rm{f(x) = \frac{\sin x}{\cos x}}$
2. From the algebra of limits, we know that:
If both numerator and denominator are continuous functions, then that fraction will be a continuous function.
3. In our present case, we just saw in the above solved example that, both sine and cosine functions are continuous functions.
• So we can write: f(x) = tan x is a continuous function.
4. We must consider the domain of this function. That domain will not contain any real number which makes 'cos x' equal to zero.
• So for all values in the domain, f(x) = tan x will be indeed continuous.
5. More details about the domain can be obtained from the graph below. It is the graph of f(x) = tan x.

Fig.21.13

• We see that:
When the input x value approaches $\rm{-{\frac{3 \pi}{2}},~-{\frac{\pi}{2}},~\frac{\pi}{2},~\frac{3 \pi}{2},\frac{5 \pi}{2}}$ etc., the f(x) value approaches infinity. This is because, the denominator becomes smaller and smaller and approaches zero. When the input x values are exact $\rm{-{\frac{3 \pi}{2}},~-{\frac{\pi}{2}},~\frac{\pi}{2},~\frac{3 \pi}{2},\frac{5 \pi}{2}}$ etc., the denominator becomes zero. Division by zero will give a number which is not defined. So we must avoid these input values in the domain. In short, the domain should not contain those x values which are given by $\rm{(2n+1){\frac{\pi}{2}}}$, where n is any integer, +ve or -ve.

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Now we will see continuity of composite functions. It can be written in 4 steps:
1. Consider the composite function (f∘g)(x). It can also be written as f(g(x)).
• We see that, the output of g is being used as the input for f.
2. Consider an arbitrary point c. When the input for g is c, the output will be g(c).
• Then the input for f will be g(c)
3. When we consider the two functions f and g together, we see that, two inputs are being made:
    ♦ c is the input for g
    ♦ g(c) is the input for f
4. Now we can write about the continuity of (f∘g).
    ♦ Suppose that, g is continuous at c.
    ♦ Also suppose that, f is continuous at g(c)
• Then we can write: (f∘g) is continuous at c.
We will see the proof in higher classes.

Solved Example 21.19
Show that the function defined by f(x) = sin (x²) is a continuous function.
Solution:
1. The given function is: f(x) = sin (x²)
2. We can write it as the composite of two functions: f(x) = g(h(x))
    ♦ Where h(x) = x² and g(h(x)) = sin (h(x))
3. Consider h(x) = x²
• At any arbitrary point c, we know that, h will be continuous.
• The output of h at c is h(c) = c².
4. So the input for g at c is c².
• c² is a real number. We know that sin x is continuous for all real numbers.
• So g is continuous at h(c)
• Therefore, f(x) = g(h(x)) = sin (x²) is a continuous function.

Solved Example 21.20
Show that the function defined by f(x) = |1− x + |x|| is a continuous function.
Solution:
1. The given function is: f(x) = |1 − x + |x||
2. We can write it as the composite of two functions: f(x) = g(h(x))
    ♦ Where h(x) = 1 − x + |x| and g(h(x)) = |h(x)|
3. Consider h(x) = 1 − x + |x|
• This is the sum of two continuous functions:
(1−x) and |x|
    ♦ (1−x) is a polynomial function. So it is continuous.
    ♦ |x| is the modulus function. It is also continuous.
• Thus h, which is the sum of two continuous functions, will be continuous.
• In other words, at any arbitrary point c, h will be continuous.
• The output of h at c is h(c) = 1 − c + |c|.
4. So the input for g at c is (1 − c + |c|).
• (1 − c + |c|) is a real number. We know that g(x) = |x| is continuous for all real numbers.
• So g is continuous at h(c)
• Therefore, f(x) = g(h(x)) = |1 − x + |x|| is a continuous function.

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The link to a few more solved examples is given below:

Exercise 21.1 (Parts 1 and 2)

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In the next section, we will see differentiability.

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Chapter 5.2 - Division of two Complex Numbers

In the previous section, we saw complex numbers and some of their algebraic operations like addition, subtraction and multiplication. We also saw additive inverse and multiplicative inverse. In this section, we will see some more algebraic operations.

D. Division of two complex numbers.
This can be written in 2 steps:
1. If z₁ and z₂ are two complex numbers,
Then the quotient ${z_1}\div{z_2}$ is defined as: ${z_1}\div{z_2}=z_1 × \frac{1}{z_2}$
(Recall that, for 'division of real numbers' also, we use the reciprocal.
For example: $5 \div 2=5 × \frac{1}{2}$)
2. We have seen that, $\frac{1}{z_2}$ is the multiplicative inverse of z₂
• So we can write:
    ♦ If we want to divide z₁ by z₂,
    ♦ We must multiply z₁ by the multiplicative inverse of z₂
• Let us see an example:
If z₁ = 6+3i  and  z₂ = 2-i, find $\frac{z_1}{z_2}$
Solution:
• We have:
Multiplicative inverse of a+bi = $\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i$
• So the multiplicative inverse of 2-i = $\frac{2}{2^2+(-1)^2}~+~\left(\frac{-(-1)}{2^2+(-1)^2} \right)i$
= $\frac{2}{4+1}~+~\left(\frac{1}{4+1} \right)i$
= $\frac{2}{5}~+~\left(\frac{1}{5} \right)i$
• Thus we get:
$\begin{array}{ll}
\frac{6+3i}{2-i}&{}={}& (6+3i) × \left[\frac{2}{5}~+~\left(\frac{1}{5} \right)i \right]& {} \\
\phantom{\frac{6+3i}{2-i}}&{}={}& 6×\frac{2}{5}~-~3×\frac{1}{5}~+~\left[6×\frac{1}{5}~+~3×\frac{2}{5}\right] i &{} & {} \\
\phantom{\frac{6+3i}{2-i}}&{}={}& \frac{12}{5}~-~\frac{3}{5}~+~\left[\frac{6}{5}~+~\frac{6}{5}\right] i &{}& {} \\
\phantom{\frac{6+3i}{2-i}}&{}={}& \frac{9}{5}~+~\left[\frac{12}{5}\right] i &{}& {} \\
\end{array}$

E. Power of i
◼ First we will consider positive powers. It can be written in 7 steps:
1. We know that: $i=\sqrt{-1}$
2. So $i^2=\left(\sqrt{-1}\right)^2 = -1$
3. Then $i^3 = (i^2) i = (-1)i = -i$
4. $i^4 = \left(i^2 \right)^2 = (-1)^2 = 1$
5. Then $i^5 = (i^4)i = (1)i = i$
6. Then $i^6 = (i^5)i = (i)i = i^2 = -1$
7. We see that the result can be -1, -i, 1 and i

• Based on the above results, we can write:
If k is any integer, the general form can be written as:
   ♦ $i^{4k}=1$
   ♦ $i^{4k+1}=i$
   ♦ $i^{4k+2}=-1$
   ♦ $i^{4k+3}=-i$
• That means, we have to divide the power by 4
   ♦ If the remainder is 0, the result is 1
   ♦ If the remainder is 1, the result is i
   ♦ If the remainder is 2, the result is -1
   ♦ If the remainder is 3, the result is -i
• Let us see an example:
Find the value of $i^{15}$
Solution:
(i) Dividing 15 by 4, we get: $\frac{15}{4}=3\frac{3}{4}$
   ♦ So the remainder is 3
   ♦ Thus we get: $i^{15}=-i$
(ii) Check:
$i^{15}=\left(i^{14}\right)i=\left(i^{2}\right)^7i=\left(-1\right)^7 i=\left[(-1)^6 × -1 \right]i=\left[\left(-1^2\right)^3 × -1 \right]i$

${}=\left[\left(1\right)^3 × -1 \right]i=\left[1 × -1 \right]i=\left[-1 \right]i=-i$

◼ Next we will consider negative powers.
Negative powers can be converted into positive powers by taking the reciprocals. Once the powers become positive, we can use the above method.
• Let us see an example:
Find the value of $i^{-18}$
Solution:
(i) $i^{-18}= \frac{1}{i^{18}}$
(ii) Dividing 18 by 4, we get: $\frac{18}{4}=4\frac{2}{4}$
   ♦ So the remainder is 2
   ♦ Thus we get: $i^{-18}= \frac{1}{i^{18}}=\frac{1}{-1}=-1$
(iii) Check:
$i^{-18}= \left[ i^{-1}\right]^{18}=(-i)^{18}=(-1)^{18}(i)^{18}=(1)(i)^{18}$

${}=i^{18}=\left(i^{2}\right)^9=\left(-1\right)^9=-1$

F. The square roots of negative real numbers
• In the first section of this chapter, we derived the result:
If '$a$' is a positive real number, $\sqrt{-a}=\sqrt{a} × \sqrt{-1}=\sqrt{a}\,i$
• Using this result, we can write the square root of any real number.
• However, there is a possible contradiction when we take the square root of the product of two -ve real numbers. This can be explained in 2 steps:
1. We know that, $\sqrt{a} × \sqrt{b}=\sqrt{ab}$
• This result is applicable when:
   ♦ Both a and b are greater than zero.
   ♦ a is greater than 0 but b is less than zero.
         ✰ $\sqrt{a} × \sqrt{-b}=\sqrt{a} × \sqrt{b}\,i = \sqrt{ab}\,i$
   ♦ b is greater than 0 but a is less than zero.
         ✰ $\sqrt{-a} × \sqrt{b}=\sqrt{a}\,i × \sqrt{b} = \sqrt{ab}\,i$
2. What happens if both a and b are less than zero?
Answer: If both are less than zero, it is possible to write as:
$\sqrt{-a} × \sqrt{-b}=\sqrt{a}\,i × \sqrt{b}\,i = \sqrt{ab} × i^2 =\sqrt{ab} × -1 =-\sqrt{ab}$
• But the square root of the product of two -ve real numbers is never negative. So there is a contradiction.
• Based on this, mathematicians have decided that:
The formula $\sqrt{a} × \sqrt{b}=\sqrt{ab}$ is not applicable when both a and b are less than zero.

G. Identities related to complex numbers.
• Identities help us to perform algebraic operations quickly. Some important identities related to complex numbers are given below:
1. $(z_1+z_2)^2=z_1^2 +2z_1 z_2 + z_2^2 $
• Let us write the proof:
$\begin{array}{ll}
(z_1+z_2)^2&{}={}& (z_1+z_2) × (z_1+z_2)& {} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& (z_1+z_2)z_1~+~(z_1+z_2)z_2 &\color {green}{\text{Distributive law}} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& z_1^2+z_2z_1+z_1z_2+z_2^2 &\color {green}{\text{Distributive law}} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& z_1^2+z_1z_2+z_1z_2+z_2^2 &\color {green}{\text{Commutative law}} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& z_1^2+2z_1z_2+z_2^2 &{} &{} \\
\end{array}$
• Two points can be written about this identity:
(i) We know how to multiply two complex numbers z₁ and z₂
• If z₁ = (a+bi) and z₂ = (c+di), then:
z₁z₂ = (ac-bd)+(ad+bc)i
• So the second term in the above identity can be easily calculated.
(ii) z₁² and/or z₂² can be calculated in a similar way:
$\begin{array}{ll}
(a+bi)(c+di)&{}={}& (ac-bd)~+~(ad+bc)i  & {} \\
(a+bi)(a+bi)&{}={}& a×a~-~b×b~+~(a×b~+~b×a)i &\color {green}{\text{Put c = a and d = b}} & {} \\
\Rightarrow (a+bi)^2 &{}={}& a^2~-~b^2~+~(ab+ba)i&{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& a^2~-~b^2~+~(ab+ab)i&{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& \left(a^2-b^2\right)+(2ab)i&{} & {} \\
\end{array}$
2. $(z_1-z_2)^2=z_1^2 -2z_1 z_2 + z_2^2 $

3. $(z_1+z_2)^3=z_1^3 +3z_1^2 z_2+3z_1 z_2^2 + z_2^3 $

4. $(z_1-z_2)^3=z_1^3 -3z_1^2 z_2+3z_1 z_2^2 - z_2^3 $

5. $z_1^2-z_2^2=(z_1+z_2)(z_1-z_2)$

• Identities (2) to (5) above can be proved using methods similar to that we used for (1).

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Now we will see some solved examples.
Solved example 5.1
Express the following in the form a+bi
(i) $(-5i)\left(\frac{1}{8}i\right)$  (ii) $(-i)(2i)\left(\frac{1}{8}i\right)^3$
Solution (i):
$\begin{array}{ll}
(-5i)\left(\frac{1}{8}i\right)&{}={}& -1 × 5 × i × \frac{1}{8} × i& {} &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& -1 × 5 × \frac{1}{8} × i^2 &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& -1 × 5 × \frac{1}{8} × -1 &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& \frac{5}{8} &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& \frac{5}{8}+0i &{} \\
\end{array}$

Solution (ii):
$\begin{array}{ll}
(-i)(2i)\left(\frac{1}{8}i\right)^3&{}={}& -1 × i × 2 × i × (-1)^3 × \frac{1}{8^3} × i^3& {} &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& -1 × 2 × (-1)^3 × \frac{1}{8^3} × i^5 &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& -1 × 2 × -1 × \frac{1}{8^3} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}&\color {green}{5\div4=1 \frac{1}{4}\text{    Remainder is 1. So }i^5=i} &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& 2 × \frac{1}{(2^3)^3} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& 2 × \frac{1}{2^9} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& \frac{1}{2^8} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& \frac{1}{256}i &{} \\
\end{array}$

Solved example 5.2
Express (5-3i)³ in the form a+bi
Solution:
We can use the identity: $(z_1-z_2)^3=z_1^3 -3z_1^2 z_2+3z_1 z_2^2 - z_2^3 $
$\begin{array}{ll}
(5-3i)^3&{}={}&5^3~-~3 × 5^2 × (3i)~+~3 × 5 × (3i)^2~-~ (3i)^3 & {} &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~75 × (3i)~+~15 × 3^2 × i^2~-~ 3^3 × i^3 &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~225i~+~135 × i^2~-~ 27 × i^3 &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~225i~+~135 × -1~-~ 27 × (-i) &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~225i~+~-135~-~ -27i &{} \\
\phantom{(5-3i)^3}&{}={}&125~-135~-~225i~+27i &{} \\
\phantom{(5-3i)^3}&{}={}&-10-198i &{} \\
\end{array}$

Solved example 5.3
Express $(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)$ in the form a+bi
Solution:
$\begin{array}{ll}
(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)&{}={}&(-\sqrt{3}+\sqrt{2}\,i)(2\sqrt{3}-i)& {} &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}&\color {green}{\because~\sqrt{-2}=\sqrt{2}\,i} &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[(-\sqrt{3} × 2\sqrt{3})-(\sqrt{2} × -1)]~+~[(-\sqrt{3} × -1)+(\sqrt{2} × 2\sqrt{3})]i &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}&\color {green}{\because~(a+bi)(c+di)=(ac-bd)~+~(ad+bc)i} &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[(-2(\sqrt{3})^2)--(\sqrt{2})]~+~[(\sqrt{3})+(2\sqrt{2}×\sqrt{3})]i &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[(-2 × 3)+\sqrt{2}]~+~[(\sqrt{3})i+(2\sqrt{2}×\sqrt{3})i] &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[-6+\sqrt{2}]~+~[\sqrt{3}\,i+(2\sqrt{2}×\sqrt{3}) × i] &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&(-6+\sqrt{2})+\sqrt{3}(1+2\sqrt{2})i &{} \\
\end{array}$

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• So we have completed a discussion on the algebra of complex numbers.
• In the next section, we will see Modulus and conjugate of a complex number.

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Chapter 5.1 - Complex Numbers And Their Algebra

In the previous section, we saw how square root of -ve numbers can be denoted using the symbol i. In this section, we will see complex numbers.

Complex numbers

Some basics about complex numbers can be written in 6 steps:
1. Consider the formula for finding the solutions of quadratic equation:
$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• In some cases, b²-4ac may work out to a number less than zero. In such cases, we used to wind up the calculations saying that 'there are no real solutions'.
2. But now we have a method to proceed with the calculations along an alternate path.
• Suppose that, we have the quadratic equation: $2x^2 + 3x + \frac{7}{4}$
    ♦ Here a = 2, b = 3 and c = 7/4
• Then we get:
$\begin{array}{ll}
b^2 - 4ac~=& 3^2 - 4 × 2 × \frac{7}{4} & {} & {} & {} \\
\phantom{b^2 - 4ac}~=&9 - 14 & {} &{} & {} \\
\phantom{b^2 - 4ac}~=& - 5 & {} &{} & {} \\
\end{array}$
3.Applying the formula, we get: $x=\frac{-3 \pm \sqrt{-5}}{4}$
• The right side can be split into two parts: $-\frac{3}{4}~\pm~\frac{\sqrt{-5}}{4}$
• We used to wind up the calculations due to the presence of '-5'.
• But now, based on what we have discussed in the previous section, we can write the right side as: $-\frac{3}{4}~\pm~\frac{\sqrt{5}i}{4}$
• So the solutions are: $-\frac{3}{4}~+~\frac{\sqrt{5}i}{4}$ and $-\frac{3}{4}~-~\frac{\sqrt{5}i}{4}$
4. A number of the form $a+bi$ is defined to be a complex number.
◼ ‘a’ and ‘b’ should be real numbers
• Some examples are:
    ♦ $3+5i$  
    ♦ $-2+ \sqrt{7}i$  
    ♦ $5+ \left(\frac{-3}{11} \right)i$  
5. Consider a complex number $z=a+bi$
    ♦ ‘a’ is called the real part. It is denoted by Re z
    ♦ ‘b’ is called the imaginary part. It is denoted by Im z 
• Let us see an example: If $z=3+7i$, then:
   ♦ Re z = 3
   ♦ Im z = 7
• For our present case of $-\frac{3}{4}~+~\frac{\sqrt{5}i}{4}$,
    ♦ Re z = $-\frac{3}{4}$
    ♦ Im z = $\frac{\sqrt{5}}{4}$
• For our present case of $-\frac{3}{4}~-~\frac{\sqrt{5}i}{4}$,
    ♦ Re z = $-\frac{3}{4}$
    ♦ Im z = $-\frac{\sqrt{5}}{4}$
6. Consider two complex numbers:
z₁ = a+bi and z₂ = c+di
• z₁ will be equal to z₂ if two conditions are satisfied:
    ♦ a = c
    ♦ b = d
• Let us see an example:
Given that the two complex numbers [(2x+3y) + 25i] and [20 + (3x+2y)i] are equal. If x and y are real numbers, find the actual values of x and y.
Solution:
• Given that the two complex numbers are equal. So we can equate the corresponding parts.
    ♦ Equating the real parts, we get: 2x+3y = 20
    ♦ Equating the imaginary parts, we get: 25 = 3x+2y
• So we have two simultaneous equations in two variables:
    ♦ 2x+3y = 20
    ♦ 3x+2y = 25
• They can be solved as follows:
$\begin{array}{ll}
2x+3y&{}={}& 20 & \color {green}{\text{- - - - (a)}} & {} \\
3x+2y&{}={}& 25 & \color {green}{\text{- - - - (b)}} & {} \\
6x+9y&{}={}& 60 & \color {green}{\text{- - - - (c)}} & \color {green}{\text{[Multiplying (a) by 3]}} \\
6x+4y&{}={}& 50 & \color {green}{\text{- - - - (d)}} & \color {green}{\text{[Multiplying (b) by 2]}} \\
5y&{}={}& 10 & {} & \color {green}{\text{[(c) - (d)]}} \\
\Rightarrow ~y&{}={}& 2 & {} & {} \\
2x+6&{}={}& 20 & {} & \color {green}{\text{[Substituting for y in (a)]}} \\
\Rightarrow ~x&{}={}& 7 & {} & {} \\
\end{array}$
• Thus we get: x = 7 and y = 2

Algebra of Complex Numbers

• Consider the quadratic equation that we saw in step (2) above. Let us write it again: $2x^2 + 3x + \frac{7}{4}$
    ♦ We saw that, it’s solutions are complex numbers:
        ✰ $-\frac{3}{4}~+~\frac{\sqrt{5}i}{4}$
        ✰ $-\frac{3}{4}~-~\frac{\sqrt{5}i}{4}$
• We will want to check whether those complex numbers indeed are the solutions.
• For that, we will want to input each of them into the original quadratic equation. We will be writing thus:
(i) Checking whether $-\frac{3}{4}~+~\frac{\sqrt{5}i}{4}$ is a solution:
$$2\left(-\frac{3}{4}~+~\frac{\sqrt{5}i}{4} \right)^2 + 3\left(-\frac{3}{4}~+~\frac{\sqrt{5}i}{4} \right) + \frac{7}{4}=0$$
(ii) Checking whether $-\frac{3}{4}~-~\frac{\sqrt{5}i}{4}$ is a solution:
$$2\left(-\frac{3}{4}~-~\frac{\sqrt{5}i}{4} \right)^2 + 3\left(-\frac{3}{4}~-~\frac{\sqrt{5}i}{4} \right) + \frac{7}{4}=0$$
• Consider (i):
In the first term, we squared the complex number. In the second term, we multiplied the complex number by a constant 3
• Consider (ii):
Here also, in the first term, we squared the complex number. In the second term, we multiplied the complex number by a constant 3
• In the same way, we may want to:
    ♦ Add two complex numbers.
    ♦ Subtract a complex number from another complex number.
    ♦ Multiply two complex numbers.
    ♦ Divide a complex number by another complex number.
• In short, we may want to perform various algebraic operations on complex numbers. So we must have a good knowledge on performing such operations.

A. Addition of two complex numbers
This can be written in 6 steps:
1. If z₁ = a+bi and z₂ = c+di, then z₁+z₂ is defined as:
z₁+z₂ = (a+c)+(b+d)i
• That is:
    ♦ Real parts are added together.
    ♦ Imaginary parts are added together.
• Let us see an example:
    ♦ (4+7i)+(-3+11i) = (4-3)+(7+11)i = 1+18i
• Note that, sum is also a complex number.
2. The addition of complex numbers satisfy the closure law.
• According to this law, the sum of two complex numbers is a complex number.
• That is:
For any two complex numbers z₁ and z₂, the sum (z₁+z₂) will be a complex number.
3. Addition of two complex numbers satisfy the commutative law.
• That is:
For any two complex numbers z₁ and z₂, (z₁+z₂) = (z₂+z₁).
4. Addition of two complex numbers satisfy the associative law.
• That is:
For any three complex numbers z₁, z₂ and z₃, (z₁+z₂)+z₃ =  z₁+(z₂+z₃).
5. Any complex number z added to (0+0i) will give the same z.
• That is:
z+(0+0i) = z
• This property is called existence of additive identity.
    ♦ The complex number (0+0i) is called the additive identity.
    ♦ It is also called zero complex number.
    ♦ It is denoted as 0
6. Any complex number z = a+bi added to [-a+(-b)i] will give a zero complex number.
• That is:
a+bi + [-a+(-b)i] = (a-a)+(b-b)i = 0+0i
• This property is called existence of additive inverse.
    ♦ The complex number [-a+(-b)i] is called the additive inverse.
    ♦ It is also called negative of z.
    ♦ If a complex number is denoted as z, it’s additive inverse is denoted as -z.
• The sum of any complex number and it’s additive inverse will always be a zero complex number.

B. Difference of two complex numbers.
This can be written as follows:
1. If z₁ = a+bi and z₂ = c+di, then the difference z₁-z₂ is defined as:
z₁-z₂ = z₁+(-z₂) = a+bi+[-c+(-d)i] = (a-c)+(b-d)i
• That is:
    ♦ We add the additive inverse of z₂ to z₁.
• Let us see two examples:
    ♦ (4+7i)-(-3+11i) = (4+7i)+[3+(-11)i] = (4+3)+(7-11)i = 7-4i
    ♦ (2-3i)-(6+9i) = (2-3i)+[-6+(-9)i] = (2-6)+(-3-9)i = -4-12i
• Note that, difference is also a complex number.

C. Multiplication of two complex numbers.
This can be written in 7 steps:
1. If z₁ = a+bi and z₂ = c+di, then the product z₁z₂ is defined as:
z₁z₂ = (ac-bd)+(ad+bc)i
• This result can be derived as follows:
$\begin{array}{ll}
(a+bi)(c+di)&{}={}& a × c~+~a × di~+~bi × c~+~bi × di &{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& ac~+~adi~+~bci~+~bd × i^2 &{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& ac~+~(ad+bc)i~+~bd × (\sqrt{-1})^2 &{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& ac~+~(ad+bc)i~+~bd × (-1) &{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& ac~+~(ad+bc)i~+~(-bd) &{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& (ac-bd)~+~(ad+bc)i &{} & {} \\
\end{array}$
• Let us see an example:
    ♦ (2+9i)(3+7i) = (2 × 3 - 9 × 7)+(2 × 7 + 9 × 3)i = (6-63)+(14+27)i = -57+41i
2. The multiplication of complex numbers satisfy the closure law.
• According to this law, the product of two complex numbers is a complex number.
• That is:
For any two complex numbers z₁ and z₂, the product z₁z₂ will be a complex number.
3. Multiplication of two complex numbers satisfy the commutative law.
• That is:
For any two complex numbers z₁ and z₂, (z₁z₂) = (z₂z₁).
4. Multiplication of two complex numbers satisfy the associative law.
• That is:
For any three complex numbers z₁, z₂ and z₃, (z₁z₂)z₃ =  z₁(z₂z₃).
5. Any complex number z multiplied by (1+0i) will give the same z.
• That is:
z × (1+0i) = z
• This property is called existence of multiplicative identity.
    ♦ The complex number (1+0i) is called the multiplicative identity.
    ♦ It is denoted as 1
• The proof can be written as follows:
$\begin{array}{ll}
(a+bi)(c+di)&{}={}& (ac-bd)~+~(ad+bc)i &\color {green}{\text{This is already proved above.}} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& a×1~-~b×0~+~(a×0~+~b×1)i &\color {green}{\text{Put c = 1 and d = 0.}} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& a~-~0~+~(0+b)i&{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& a+bi&{} & {} \\
\end{array}$
6. Any complex number z = a+bi multiplied by $\left[\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i\right]$ will give 1.
(Remember that 1, which actually is the complex number (1+0i), is the multiplicative identity)
• That is:
$(a+bi) × \left[\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i \right]~=~1$
• This property is called existence of multiplicative inverse.
    ♦ The complex number $\left[\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i \right]$ is called the multiplicative inverse.
    ♦ If a complex number is denoted as $z$, it’s multiplicative inverse is denoted as $\frac{1}{z}$.
• We can write:
If $z=a+bi$, then $\frac{1}{z}=\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i$
• The product of any complex number and it’s multiplicative inverse will always be 1. That is: $z × \frac{1}{z}=1$
• The proof can be written as follows:
$\begin{array}{ll}
(a+bi)(c+di)&{}={}& (ac-bd)~+~(ad+bc)i &\color {green}{\text{This is already proved above.}} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& a×\frac{a}{a^2+b^2}~-~b×\frac{-b}{a^2+b^2}~+~\left[a×\frac{-b}{a^2+b^2}~+~b×\frac{a}{a^2+b^2}\right] i &\color {green}{\text{Changing c and d.}} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& \frac{a^2}{a^2+b^2}~+~\frac{b^2}{a^2+b^2}~+~\left[\frac{-ab}{a^2+b^2}~+~\frac{ab}{a^2+b^2}\right] i &{}& {} \\
\phantom{(a+bi)(c+di)}&{}={}& \frac{a^2+b^2}{a^2+b^2}~+~\left[\frac{-ab+ab}{a^2+b^2}\right] i &{}& {} \\
\phantom{(a+bi)(c+di)}&{}={}& 1~+~0i&{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& 1&{} & {} \\
\end{array}$
◼ For the existence of multiplicative inverse, both a and b should be non-zero real numbers.
7. Multiplication of two complex numbers satisfy the distributive law.
• That is:
For any three complex numbers z₁, z₂ and z₃,
(a) z₁(z₂+z₃) =  z₁z₂ + z₁z₃.
(a) (z₁+z₂)z₃ =  z₁z₃ + z₂z₃.

---

In the next section, we will see more algebraic operations.

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