26.3 - Scalar Multiplication of Vectors

In the previous section, we saw addition of vectors. In this  section, we will see multiplication.

Multiplication of a vector by a scalar

This can be explained in 3 steps:
1. Suppose that, we are given a vector $\small{\vec{a}}$.
• We can multiply $\small{\vec{a}}$ by a scalar $\small{\lambda}$
• The product obtained is denoted as $\small{\lambda\vec{a}}$

2. $\small{\lambda\vec{a}}$ is also a vector. It has the following four properties:
(i) $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ are collinear
(ii) $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ have the same direction if $\small{\lambda}$ is +ve  
(iii) $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ have opposite directions if $\small{\lambda}$ is −ve
(iv) Magnitude of $\small{\lambda\vec{a}}$ is $\small{\left|\lambda \right|}$ times the magnitude of $\small{\vec{a}}$. That is.,
$\small{\left|\lambda\vec{a} \right|~=~\left|\lambda \right|\,\left|\vec{a} \right|}$

3. Fig.26.17 below gives a geometrical explanation of the multiplication process.

Fig.26.17

• In fig(i), we have the original $\small{\vec{a}}$. The single perpendicular white line indicates that, the vector has a magnitude of two units.
• In fig(ii), the vector has a magnitude of one unit. That is., half the magnitude of the original $\small{\vec{a}}$. Direction of this vector is same as that of $\small{\vec{a}}$
• In fig(iii), the vector has a magnitude of four units. That is., two times the magnitude of the original $\small{\vec{a}}$. Direction of this vector is same as that of $\small{\vec{a}}$
• In fig(iv), the vector has a magnitude of one unit. That is., half the magnitude of the original $\small{\vec{a}}$. Direction of this vector is opposite of $\small{\vec{a}}$. This is because, $\small{\vec{a}}$ is multiplied by a −ve scalar.
• In fig(v), the vector has a magnitude of three unit. That is., (3/2) times the magnitude of the original $\small{\vec{a}}$. Direction of this vector is opposite of $\small{\vec{a}}$. This is because, $\small{\vec{a}}$ is multiplied by a −ve scalar.
• The vectors in all five figs are parallel to each other

---

Now we can discuss about additive inverse. It can be explained in 6 steps:
1. Consider the two vectors $\small{\vec{a}~\text{and}~\lambda \vec{a}}$
2. Suppose that, $\small{\lambda = -1}$. Then $\small{\lambda \vec{a} = -\vec{a}}$
3. Let us compare $\small{\vec{a}~\text{and}~- \vec{a}}$:
    ♦ $\small{-\vec{a}}$ has the same magnitude as $\small{\vec{a}}$
    ♦ $\small{-\vec{a}}$ has the direction opposite to that of $\small{\vec{a}}$
4. We can add the two vectors. We will get:
$\small{\vec{a} + \left(-\vec{a} \right)~=~\left(-\vec{a} \right) + \vec{a}~=~\vec{0}}$
5. So $\small{-\vec{a}}$ is called the additive inverse of $\small{\vec{a}}$
6. $\small{-\vec{a}}$ is also called the negative of $\small{\vec{a}}$

---

Next we will discus about unit vectors. It can be explained in 6 steps:

1. Consider the two vectors $\small{\vec{a}~\text{and}~\lambda \vec{a}}$

2. Suppose that, $\small{\lambda = \frac{1}{\left|\vec{a} \right|}}$. Then $\small{\lambda \vec{a} = \frac{\vec{a}}{\left|\vec{a} \right|}}$

3. We know that, $\small{\vec{a}}$ can be written as the product of two items:
(i) $\small{\left|\vec{a} \right|}$
(ii) A vector of magnitude one unit and direction same as that of $\small{\vec{a}}$

4. So step (2) can be modified as:

If $\small{\lambda = \frac{1}{\left|\vec{a} \right|}}$, then $\small{\lambda \vec{a} = \frac{\vec{a}}{\left|\vec{a} \right|}~=~\frac{\left|\vec{a} \right|[\text{A vector of magnitude one unit and direction same as that of}\,\vec{a}]}{\left|\vec{a} \right|}}$

$\small{\Rightarrow\frac{\vec{a}}{\left|\vec{a} \right|}~=~\text{A vector of magnitude one unit and direction same as that of}\,\,\vec{a}}$

5. $\small{\text{A vector of magnitude one unit and direction same as that of}\,\,\vec{a}}$ is called:

$\bf{\text{A unit vector in the direction of}\,\,\vec{a}}$

• Such a vector is denoted as: $\bf{\hat{a}}$

6. So we can modify step (4) as:
$\small{\frac{\vec{a}}{\left|\vec{a} \right|}~=~\hat{a}}$

---

In the next section, we will see components of a vector.

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com

26.2 - Addition of Vectors

In the previous section, we saw types of vectors. In this  section, we will see addition of vectors.

Some basic details about addition can be written in 4 steps:
1. Suppose that, a person moves from A to B. This displacement can be represented by $\small{\vec{AB}}$ in fig.26.12 below:

Fig.26.12

2. If after reaching B, the person moves to C, we will get a new vector $\small{\vec{BC}}$
3. So the net effect is that, the person has moved from A to C. This net displacement can be represented by a third vector $\small{\vec{AC}}$.
• The three vectors $\small{\vec{AB}, \vec{BC}~\text{and}~\vec{AC}}$ form the sides of a triangle ABC.
4. We say that, $\small{\vec{AC}}$ is the resultant of $\small{\vec{AB}~\text{and}~\vec{BC}}$
• Mathematically, it is written as:$\small{\vec{AC} = \vec{AB} + \vec{BC}}$
• This is known as the triangle law of vector addition.

---

Now we will see the general method of vector addition. It can be written in 3 steps:

1. Fig.26.13 (i) below shows two vectors $\small{\vec{a}~\text{and}~\vec{b}}$

Fig.26.13

• We want to find $\small{\vec{a} + \vec{b}}$
2. For that, we shift $\small{\vec{b}}$ in such a way that, the initial point of $\small{\vec{b}}$ coincides with the terminal point of $\small{\vec{a}}$. It is important not to change the magnitude or direction of $\small{\vec{b}}$ while making the shift. Fig(ii) shows the position after the shift. We see that, $\small{\vec{a}~\text{and}~\vec{b}}$ now forms the sides AB and BC of the triangle ABC
3. The resultant vector $\small{\vec{a} + \vec{b}}$ is represented by the third side AC of triangle ABC.
We can write: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$

---

From the above discussion, we get an interesting result. It can be written in 4 steps:

1. Suppose that the person traveled three segments:
• First from A to B, then from B to C and finally from C to A
• We want to know the resultant displacement.
• That is., we want to find $\small{\vec{AB} + \vec{BC} + \vec{CA}}$  
2. We have obtained: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$
Substituting this in the above result, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Resultant}}    & {~=~}    &{\vec{AB} + \vec{BC} + \vec{CA}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\vec{AC}+ \vec{CA}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-\vec{CA}+ \vec{CA}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$
◼ Remarks
3 (magenta color): Here we apply the fact that, $\small{-\vec{CA}}$ is same as $\small{\vec{AC}}$

3. So the resultant is a null vector. That means, the resultant does not have magnitude. That means, the net displacement is zero.

4. We can write:
When sides of a triangle are taken in order, the resultant will be a null vector. This is because, the initial and terminal points will be the same.

---

Now we will see subtraction. It can be written in 4 steps:

1. We want $\small{\vec{a} - \vec{b}}$. That is., we want the resultant of $\small{\vec{a}~\text{and}~-\vec{b}}$
2. For that, we draw $\small{-\vec{b}}$ at a convenient place and shift it. We shift it in such a way that, the initial point of $\small{-\vec{b}}$ coincides with the terminal point of $\small{\vec{a}}$. It is important not to change the magnitude or direction of $\small{-\vec{b}}$ while making the shift. Fig.26.13(iii) above, shows the position after the shift. We see that, $\small{\vec{a}~\text{and}~-\vec{b}}$ forms the sides AB and BC' of the triangle ABC'
3. Applying the triangle law of vector addition, we get:
$\small{\vec{a} + \left(-\vec{b} \right) = \vec{AB} + \vec{BC'} = \vec{AC'}}$
• $\small{\vec{AC'}}$ is said to represent the difference of $\small{\vec{a}}$ and $\small{\vec{b}}$
4. So, for subtraction, we make use of the negative of the vector to be subtracted.

---

Now we will see the parallelogram law of vector addition. It can be explained in 7 steps:

1. In fig.26.14(i) below, we have two vectors $\small{\vec{a}~\text{and}~\vec{b}}$

Fig.26.14

• We want to find $\small{\vec{a} + \vec{b}}$

2. For that, we shift $\small{\vec{b}}$ in such a way that, the initial point of $\small{\vec{b}}$ coincides with the initial point of $\small{\vec{a}}$. This is shown in fig(ii). [Recall that in the triangle method, initial point of $\small{\vec{b}}$ must coincides with the terminal point of $\small{\vec{a}}$. This is shown in fig(iii)] It is important not to change the magnitude or direction of $\small{\vec{b}}$ while making the shift.

3. Next step is to draw two parallel lines:
    ♦ Through the terminal point of $\small{\vec{b}}$, draw a dashed line parallel to $\small{\vec{a}}$
    ♦ Through the terminal point of $\small{\vec{a}}$, draw a dashed line parallel to $\small{\vec{b}}$

4. Let the two dashed lines intersect at C. Then the resultant of $\small{\vec{a}~\text{and}~\vec{b}}$ is given by $\small{\vec{AC}}$
• Note that AC is the diagonal of the parallelogram ABCD
• Note also that, the diagonal AC, which is drawn through the initial points of $\small{\vec{a}~\text{and}~\vec{b}}$ has to be considered. We must not consider the other diagonal BD

5. Now we can write the parallelogram law of vector addition:
If two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ can be represented in magnitude and direction by two adjacent sides of a parallelogram, then their sum $\small{\vec{a} + \vec{b}}$ is represented in magnitude and direction by the diagonal of the parallelogram through their common point.

6. Note that $\small{\vec{a} + \vec{b}}$ obtained in both fig(ii) and fig(iii) is the same. So we can write:
• Triangle law and parallelogram law are equivalent to each other.

7. The equivalence can be proved analytically also. It can be done in 5 steps:
(i) In fig.26.14(ii), $\small{\vec{BC} = \vec{AD}}$ because, they have the same magnitude and direction.
(ii) In triangle ABC, we can apply triangle law. We get: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$
(iii) So using parallelogram law, we have: $\small{\vec{AB} + \vec{AD} = \vec{AC}}$
(iv) Using triangle law, we have: $\small{\vec{AB} + \vec{BC} = \vec{AC}}$, which is same as: $\small{\vec{AB} + \vec{AD} = \vec{AC}}$  
(v) The results in (iii) and (iv) are the same. Hence the two laws are equivalent to each other.

---

Properties of vector addition

Property I (commutative property):
For any two vectors $\small{\vec{a}~\text{and}~\vec{b}}$, $\small{\vec{a}+\vec{b} = \vec{b} + \vec{a}}$
Proof can be written in 5 steps:
1. In fig.26.15 below, $\small{\vec{AB}=\vec{a}~\text{and}~\vec{AD}=\vec{b}}$

Fig.26.15

$\small{\vec{a}~\text{and}~\vec{b}}$ form the adjacent sides of the parallelogram ABCD. Also, the initial points of both vectors are at A.

2. In the parallelogram ABCD,
• $\small{\vec{BC}=\vec{AD}=\vec{b}}$ because, they have the same magnitude and direction.
• $\small{\vec{DC}=\vec{AB}=\vec{a}}$ because, they have the same magnitude and direction.

3. In the triangle ABC, we can apply the triangle law. We get:
$\small{\vec{AB}+\vec{BC}=\vec{AC}}$
$\small{\Rightarrow \vec{a}+\vec{b}=\vec{AC}}$

4. In the triangle ADC also, we can apply the triangle law. We get:
$\small{\vec{AD}+\vec{DC}=\vec{AC}}$
$\small{\Rightarrow \vec{b}+\vec{a}=\vec{AC}}$

5. From (3) and (4), we get: $\small{\vec{a}+\vec{b} = \vec{b} + \vec{a}}$

Property II (Associative property):
For any three vectors $\small{\vec{a}, \vec{b}~\text{and}~\vec{c}}$,
$\small{\left(\vec{a}+\vec{b} \right) + \vec{c}~=~\vec{a} + \left(\vec{b}+\vec{c} \right)~=~\left(\vec{a}+\vec{c} \right) + \vec{b}}$
Proof can be written in 4 steps:
1. In fig.26.16(i) below, $\small{\vec{PQ}=\vec{a}, \vec{QR}=\vec{b}~\text{and}~\vec{RS}=\vec{c}}$

Fig.26.16

• Consider triangle PQR. Applying triangle law, we get:
$\small{\vec{a}+\vec{b} = \vec{PQ} + \vec{QR} = \vec{PR}}$
• Consider triangle PRS. Applying triangle law, we get:
$\small{\left(\vec{a}+\vec{b} \right) + \vec{c}= \vec{PR} + \vec{RS} = \vec{PS}}$

2. In fig.26.16(ii) above, $\small{\vec{PQ}=\vec{a}, \vec{QR}=\vec{b}~\text{and}~\vec{RS}=\vec{c}}$
• Consider triangle QRS. Applying triangle law, we get:
$\small{\vec{b}+\vec{c} = \vec{QR} + \vec{RS} = \vec{QS}}$
• Consider triangle PQS. Applying triangle law, we get:
$\small{\vec{a} + \left(\vec{b}+\vec{c} \right)= \vec{PQ} + \vec{QS} = \vec{PS}}$

3. In fig.26.16(iii) above, $\small{\vec{PQ}=\vec{a}, \vec{RS}=\vec{b}~\text{and}~\vec{QR}=\vec{c}}$
• Consider triangle PQR. Applying triangle law, we get:
$\small{\vec{a}+\vec{c} = \vec{PQ} + \vec{QR} = \vec{PR}}$
• Consider triangle PRS. Applying triangle law, we get:
$\small{\left(\vec{a}+\vec{c} \right) + \vec{b} = \vec{PR} + \vec{RS} = \vec{PS}}$

4. In all of the above three steps, we get the resultant as $\small{\vec{PS}}$. So we can write:
$\small{\left(\vec{a}+\vec{b} \right) + \vec{c}~=~\vec{a} + \left(\vec{b}+\vec{c} \right)~=~\left(\vec{a}+\vec{c} \right) + \vec{b}}$

---

◼ Based on property I, we can write:
$\small{\vec{a}+\vec{0}~=~\vec{0} + \vec{a}~=~\vec{a}}$
So $\small{\vec{0}}$ is called the additive identity for vector addition

◼ Based on property II, we can add three vectors $\small{\vec{a}, \vec{b}, \vec{c}}$ in any order we like. There is no need to group any two of them.

---

In the next section, we will see scalar multiplication of vectors by scalars.

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com

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.

Previous

Contents

Next

Copyright©2022 Higher secondary mathematics.blogspot.com