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.

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

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