# Algebraic Operations On Complex Numbers

In Mathematics, algebraic operations on complex numbers are given by four basic arithmetic operations which include addition, subtraction, multiplication, and division. A complex number is the combination of a real number and an imaginary number.

The algebraic operations on complex numbers are defined purely by the algebraic methods. Some basic algebraic laws like associative, commutative, and distributive law are used to explain the relationship between the number of operations. By the use of these laws, the algebraic expressions are solved in a simple way. Since algebra is a concept based on known and unknown values (variables), its own rules are created to solve the problems.  In this article, let us discuss the basic algebraic operations on complex numbers with examples.

## What are Complex Numbers?

In Maths, basically, a complex number is defined as the combination of a real number and an imaginary number. Real numbers are the numbers that we usually work on to do mathematical calculations. But the imaginary numbers are not generally used for calculations but only in the case of complex numbers.

## Equality of Complex Numbers

Assume that z1 and z2 are the two complex numbers.

Here z1 = a1+i b1 and z2 = a2+ib2

If both the complex numbers z1 and z2 are equal (i.e) z1 = z2, then we can say that the real part of the first complex number is equal to the real part of the second complex number, whereas the imaginary part of the first complex number is equal to the imaginary part of the second complex number.

(i.e) Re(z1) = Re(z2) and Im(z1) = Im(Z2)

Thus, the equality of complex number states that,

if a1+ib1 = a2+ib2, then a1 = a2 and b1 = b2.

## Operations on Complex Numbers

The basic algebraic operations on complex numbers discussed here are:

• Addition of Two Complex Numbers
• Subtraction(Difference) of Two Complex Numbers
• Multiplication of Two Complex Numbers
• Division of Two Complex Numbers.

## Addition of Two Complex Numbers

We know that a complex number is of the form z=a+ib where a and b are real numbers.

Consider two complex numbers z1 = a1 + ib1 and z2 = a2 + ib2

Then the addition of the complex numbers z1 and z2 is defined as,

 z1+z2 =( a1+a2 )+i( b1+b2 )

We can see that the real part of the resulting complex number is the sum of the real part of each complex number and the imaginary part of the resulting complex number is equal to the sum of the imaginary part of each complex number.

That is, Re(z1+z2 )= Re( z1 )+Re( z1 )

Im( z1+z2 )=Im( z1)+Im(z2)

For the complex numbers,

z1 = a1+ib1

z2 = a1+ib2

z3 = a3+ib3

………..

………..

zn = an+ibn

a1+a2+a3+….+an = (a1+a2+a3+….+an )+i(b1+b2+b3+….+bn)

 Let’s Work Out: Example: z1 = a+3i, z2 = 4+bi, z3 = 6+10i Find the value of a and b if z3 = z1+z2 Solution: By the definition of addition of two complex numbers, Re(z3 ) = Re(z1 )+Re(z2 ) 6 = a + 4 a = 6 – 4 = 2 Im(z3 ) = Im(z1 ) + Im(z2 ) 10 = 3+b b = 10-3 =7

### Conjugate of Complex number

Conjugate of a complex number z=a+ib is given by changing the sign of the imaginary part of z which is denoted as $$\bar z$$

$$\bar z = a-ib \\ z+ \bar z =2a \\ z- \bar z =2bi$$

### Properties of Addition of Complex Numbers

 Name of the Property Description Expression Closure property Addition of two complex numbers is a complex number z1 + z2 = z Commutative property Order of addition of two complex numbers, does not change the result z1 + z2 = z2 + z1 Associative property Regrouping three complex numbers, while adding them, does not change the result (z1+z2)+z3 = z1+(z2+z3) Additive inverse property If z = a+ib is a complex number, then its additive inverse will be -z = -a – ib z+(-z) = 0 Additive identity If a value added to complex number results in the same complex number, then it becomes the additive identity (a+ib) + (0 + i0) = a + ib

## Difference of Two Complex Numbers

Consider the complex numbers z1 = a1+ib1 and z2 = a2+ib2, then the difference of z1 and z2, z1-z2 is defined as,

 z1-z2 = (a1-a2)+i(b1-b2)

From the definition, it is understood that,

Re(z1-z2)=Re(z1)-Re(z2)

Im(z1-z2)=Im(z1)-ImRe(z2)

 Example: z1 =4+ai,z2=2+4i,z3 =2. Find the value of a if z3=z1-z2 Solution: By the definition of difference of two complex numbers, Im3=Im1-Im2 0 = a – 4 a = 4

Note: All real numbers are complex numbers with imaginary part as zero.

## Multiplication of Two Complex Numbers

We know the expansion of (a+b)(c+d)=ac+ad+bc+bd

Similarly, consider the complex numbers z1 = a1+ib1 and z2 = a2+ib2

Then, the product of z1 and z2 is defined as:

z1 z2=(a1+ib1)(a2+ib2)

z1 z2 = a1 a2+a1 b2 i+b1 a2 i+b1 b2 i2

Since,  i2 = -1, therefore,

 z1 z2 = (a1 a2 – b1 b2 ) + i(a1 b2 + a2 b1 )

Let us see here a solved example based on the multiplication of complex numbers.

 Example: z1=6-2i, z2=4+3i. Find z1 z2 Solution: z1 z2 = (6-2i) (4+3i) = 6 × 4 + 6 × 3i + (-2i) × 4 + (-2i)(3i) = 24 + 18i – 8i – 6i2 = 24 + 10i + 6 = 30 + 10i

### Multiplicative inverse of a complex number

Definition: For any non-zero complex number z=a+ib(a≠0 and b≠0) there exists another complex number z-1 or 1/z, which is known as the multiplicative inverse of z such that zz-1 = 1.

z = a+ib, then,

$$z^{-1} = \frac{a}{a^2 + b^2} + i \frac{(-b)}{a^2 + b^2}\\ Re(z^{-1}) = \frac{a}{a^2 + b^2}\\ Im(z^{-1}) = \frac{-b}{a^2 + b^2}$$

 Example: z = 3 + 4i Solution: $$z^{-1} ~of ~a + ib = \frac{a}{a^2 + b^2} +i \frac{(-b)}{a^2 + b^2}= \frac{(a-ib)}{a^2 + b^2}\\$$ The numerator of z-1 is conjugate of z, that is a – ib Denominator of z-1 is sum of squares of the Real part and imaginary part of z Here, z =  3 + 4i  $$z^{-1} = \frac{3-4i}{3^2 + 4^2} = \frac{3-4i}{25}\\ z^{-1}= \frac{3}{25} – \frac{4i}{25}$$

### Properties of Multiplication of Complex Numbers

 Name of the Property Description Expression Closure property Product of two complex number is a complex number only z1 x z2 = z Commutative property Change of order of complex numbers, does not change the result of their product z1.z2 = z2.z1 Associative property Regrouping of complex numbers, does not change the result of their product z1(z2.z3) = (z1.z2)z3 Distributive property Multiplication of a complex number with the sum of two complex numbers is given by: z1(z2+z3) = z1.z2 + z1.z3

## Division of Complex Numbers

Consider the complex number z1 = a1 + ib1 and z2 = a2 + ib2, then the quotient of z1/z2 is defined as,

$$\frac{z_1}{z_2}= z_1 \times \frac{1}{z_2}$$

Therefore, to find z1/z2, we have to multiply z1 with the multiplicative inverse of z2.

Now, let us discuss in detail about the division of complex numbers:

Let z1 = a1+ib1 and z2 = a2+ib2, then z1/z2 is given as:

z1/z2 = (a1+ib1)/(a2+ib2)

Hence, (a1+ib1)/(a2+ib2) = [(a1+ib1)(a2-ib2)]/[(a2+ib2)(a2-ib2)]

(a1+ib1)/(a2+ib2) = [(a1a2)-(a1b2i)+(a2b1i)+b1b2)]/[(a22+b22)]

(a1+ib1)/(a2+ib2) = [(a1a2)+(b1b2) +i(a2b1-a1b2)]/(a22+b22)

Hence, $$\frac{z_{1}}{z_{2}} = \frac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+ i\frac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}$$

 Example: If z1 = 2 + 3i and z2 = 1 + i, find z1/z2. Solution: $$\frac{z_1}{z_2} = z_1 \times \frac{1}{z_2}$$ $$\frac{2+3i}{1+i} = (2+3i) \times \frac{1}{1+i}$$ $$Since, \frac{1}{1+i} = \frac{1-i}{1^2 + 1^2} = \frac{1-i}{2}$$ $$\frac{2 + 3i}{1 + i} = 2+3i \times \frac{1-i}{2} = \frac{(2+3i)(1-i)}{2}$$ $$=\frac{2 – 2i + 3i – 3i^2}{2} = \frac{5+i}{2}$$

### Algebraic Operations on Complex Numbers Summary

Assume that z1 = a1+ib1 and z2 = a2+ib2 are the two complex numbers

Addition: z1+z2 =( a1+a2 )+i( b1+b2 )

Subtraction: z1-z2 = (a1-a2)+i(b1-b2)

Multiplication: z1 z2 = (a1 a2 – b1 b2 ) + i(a1 b2 + a2 b1 )

Division: $$\frac{z_{1}}{z_{2}} = \frac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+ i\frac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}$$

### Solved Examples

Question 1:

Solution:

Given the two complex numbers are:

2+4i and -1+3i

(2+4i)+(-1+3i)

⇒ (2-1)+(4i+3i)

⇒ 1 + 7i

Question 2:

Simplify: 7 + i + 4 + 4.

Solution:

7 + i + 4 + 4

⇒ (7+4+4) + i

⇒ 15 + i

Example 3:

Multiply the complex numbers: (5+3i). (3+4i)

Solution:

Given:(5+3i). (3+4i)

(5+3i). (3+4i) = 15+20i+9i-12

(5+3i). (3+4i) = (15-12) + i(20+9)

(5+3i). (3+4i) = 3+29i

Hence, the product of (5+3i) and (3+4i) is 3+29i.

Example 4:

Subtract (2+5i) from (7+15i).

Solution:

We know that (a+bi) – (c+di) = (a-c) + i(b-d).

Hence,

(7+15i) – (2+5i) = (7-2)+i(15-5)

(7+15i) – (2+5i) =5+10i

Hence, (7+15i) – (2+5i) is 5+10i

## Practice Questions

Simplify the following:

• −3 + 6i − (−5 − 3i) − 8i
• 4i(−2 − 8i)
• (−2 − i)(4 + i)
• (−2 − 2i)(−4 − 3i)(7 + 8i)
• −3i ⋅ 6i − 3(−7 + 6i)
• (6i)3

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## Frequently Asked Questions on Algebraic operations on complex numbers

### What are complex numbers?

Complex numbers are the combination of real numbers and imaginary numbers. It is represented by z = a + ib, where a is real part and ib is imaginary part.

### How do we add two complex numbers?

To add two complex numbers, add the real part and imaginary part separately.

### How do we subtract two complex numbers?

Subtract the real part from real part and imaginary part from imaginary part, to subtract two complex numbers.

### What are the four algebraic operations?

The four basic algebraic operations are addition, subtraction, multiplication and division.

### How do we multiply two complex numbers?

If a+bi and c+di are the two complex numbers, then the product is given as: