• Written By Gnanambigai G S
  • Last Modified 23-01-2025

Operations with Complex Numbers: Properties and Examples

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Operations with complex numbers: Do you know how to find the square root of a negative number? We know that no real number when squared gives a negative number. This is achieved by the use of complex numbers. Any number of the form a+ib is called a complex number, where a and b are real numbers. For example, 214i is a complex number, where 2 is the real number, 14i is the imaginary number, and i is the imaginary unit.

Like real numbers, the four arithmetic operations can also be performed on complex numbers. These calculations that use complex numbers play a significant role in the fields of quantum mechanics, electrical circuit analysis, and fluid dynamics. The arithmetic operations are:

  • Addition
  • Subtraction
  • Multiplication
  • Division

In this article, let us learn to perform these operations with complex numbers and their various properties.

Adding and Subtracting Complex Numbers

What do you think happens when two complex numbers are added or subtracted?

They behave similarly to real numbers. The difference here is that since complex numbers have two parts, real and imaginary, they are operated on separately.

Let z1=a+ib and z2=c+id be two complex numbers. Then their sum and difference is written as,

Adding and Subtracting Complex Numbers

Steps to Add and Subtract Complex Numbers in Polar Form

Follow these steps to add or subtract two complex numbers in polar form:

Step 1: Convert the complex numbers from the given polar form to rectangular form.

Step 2: Add or subtract the complex numbers in rectangular form, as described above.

Step 3: Convert the resulting complex number to polar form again. This is the required answer.

Properties of Addition of Complex Numbers

Name of the PropertyDefinitionProof/ExampleResult
Closure PropertyA set is closed for some arithmetic operation(4+i2)+(6+i3)
=(4+6)+(i2+i3)
=10+i5
TRUE
Commutative PropertyThe order of numbers does not alter the result(6+i3)+(4+i2)
=(6+4)+(i3+i2)
=10+i5
TRUE
Associative PropertyRearranging the parenthesis in an expression will not alter the resultz1+(z2+z3)=(α1+iβ1)+{(α2+iβ2)+(α3+iβ3)}
=(α1+iβ1)+{(α2+α3)+i(β2+β3)}
={α1+(α2+α3)}+{i(β1+β2+β3)}
=(α1+α2+α3)+i(β1+β2+β3)
={α1+(α2+α3)}+{i(β1+β2+β3)}
=((α1+α2)+α3)+i((β1+β2)+β3)
=((α1+α2)+i(β1+β2))+(α3+iβ3)
=(z1+z2)+z3
TRUE
Additive Inverse A number, when added to the given number, gives zeroz1+AIn.=0
AIn.=0z1
AIn.=z1
Additive inverse of z1 is z1
Additive IdentityA number, when added to the number, does not alter itz1+AId.=z1
AId.=z1z1
AId.=0
Additive Identity of z1 is 0

Properties of Subtraction of Complex Numbers

Name of the PropertyDefinitionProof/ExampleResult
Closure PropertyA set is closed for some arithmetic operation(4+i2)(6+i3)
=(46)+(i2i3)
=2i
TRUE
Commutative PropertyThe order of numbers does not alter the result(6+i3)(4+i2)
=(64)+(i3i2)
=2+i
FALSE
Associative PropertyRearranging the parenthesis in an expression will not alter the resultz1(z2z3)=(α1+iβ1){(α2+iβ2)(α3+iβ3)}
=(α1+iβ1){(α2α3)+i(β2β3)}
={α1+(α2+α3)}+{i(β1β2+β3)}
=(α1α2+α3)+i(β1β2+β3)
=((α1α2)+α3)+i((β1β2)+β3)
=((α1α2)+i(β1β2))+(α3+iβ3)
=(z1z2)+z3
FALSE

Multiplying Complex Numbers

The process of multiplying two complex numbers is more complicated than the multiplication of real numbers. As there are two parts to a complex number, we find the product of four different pairs of numbers as shown below when multiplying two of them.

Multiplying Complex Numbers

For two complex numbers z1=a+ib and z2=c+id, their product is given by,

z1z2=(a+ib)(c+id)
=ac+iad+ibc+i2bd
z1z2=(acbd)+i(ad+bc)

Note that i2=1

Multiplying Complex Numbers in Polar Form

For any two complex numbers, z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2), the formula for multiplication in polar form is given by,

Multiplying Complex Numbers in Polar Form

Properties of Multiplication of Complex Numbers

Name of the PropertyProof/ExampleResult
Closure Property(3+i2)×(4+i5)
=34+3i5+i24+i5i2
=12+i15+i8+i22
=122+i23
=10+i23
TRUE
Commutative Property(4+i5)×(3+i2)
=12+i15+i8+i22
=122+i23
=10+i23
TRUE
Associative Propertyz1×(z2×z3)=(α1+iβ1)×{(α2+iβ2)+(α3+iβ3)}
=(α1+iβ1)×{(α2α3β2β3)+i(β3α2+β2α3)}
=(α1(α2α3β2β3)β1(β3α2+β2α3))+i{α1(β3α2+β2α3)+(α2α3β2β3)β1}
=(α1α2α3α1β2β3β1β3α2β1β2α3)+i{β3α1α2+α1β2α3+β1α2α3β1β2β3}
={(α1α2β1β2)α3(β1α2+α1β2)β3}+i{(α1α2β1β2)α3+(β1α2+α1β2)β3}
={(α1α2β1β2)+i(β1α2+α1β2)}×(α3+iβ3)
={(α1+iβ1)×(α2+iβ2)}×(α3+iβ3)
=(z1×z2)×z3
TRUE
Distributive Propertyz1×(z2+z3)=(α1+iβ1)×{(α2+iβ2)+(α3+iβ3)}
=(α1+iβ1)×{(α2+α3)+i(β2+β3)}
={(α1+iβ1)(α2+α3)+i(α1+iβ1)(β2+β3)}
={((α1α2+iα2β1)+(α1α3+iα3β1))+i((β2α1+β3α1)+i(β2β1+β3β1))} On further simplification, z1×(z2+z3) becomes,
z1×(z2+z3)={(α1α2+iα2β1+α1α3+iα3β1)+i(β2α1+β3α1)(β2β1+β3β1)}
={(α1α2+α1α3β2β1β3β1)+i(β2α1+α2β1+β3α1+α3β1)}
=(α1α2β1β2+α1α3β1β3)+i(α3β1+α2β1+α1β2+α1β3)
=(α1α2β1β2)+i(α2β1+α1β2)+(α1α3β1β3)+i(α3β1+α1β3)
=(α1+iβ1)(α2+iβ2)+(α1+iβ1)(α3+iβ3)
=(z1×z2)+(z1×z3)
Applicable
Multiplicative Inverse (a1+iβ1)1=1a1+iβ1
=1a1+iβ1×a1iβ1a1iβ1
=a1iβ1(a1)2(iβ1)2
=a1iβ1a12+β12
Multiplicative inverse of a1+iβ1 is a1iβ1a12+β12
Multiplicative Identityz1×A.Id.=z1
A.Id.=z1z1
A.Id.=1
Multiplicative Identity of z1 is 1

Dividing Complex Numbers

Mathematically, although the division of complex numbers is similar to real numbers, it is a lengthier process. This is because it is difficult to perform division using an imaginary number.

Method 1: Rationalising the Denominator

Step 1: Write the division of complex numbers as a fraction.

Step 2: Rationalise the denominator to remove the imaginary part of the divisor. This is done by multiplying the conjugate of the denominator with the numerator and the denominator.

Step 3: Simplify the expression using the algebraic identity (a+b)(ab)=a2b2.

Step 4: Substitute i2=1.

Step 5: Simplify the numerator using the distributive property.

Step 6: Separate the real and imaginary parts in the resulting complex number.

Method 2: Using the Division Formula for Complex Numbers

The alternate method is to use the division of complex numbers formula directly.

Formula for the division of complex numbers z1=a+ib and z2=c+id is given by,

Using the Division Formula for Complex Numbers

Derivation of the Division Formula for Complex Numbers

z1z2=a+ibc+id

=(a+ib)(cid)(c+id)(cid)

=(ac+bd)+i(bcad)c2+d2

z1z2=ac+bdc2+d2+i(bcadc2+d2)

Dividing Complex Numbers in Polar Form

For any two complex numbers, z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2), the formula for division in polar form is given by,

Dividing Complex Numbers in Polar Form

Solved Examples of Operations with Complex Numbers

Q.1. Write z=1+i32+i5 in the form of a+ib
Ans
:
z1z2=ac+bdc2+d2+i(bcadc2+d2)
Here,
z1=1+i3
z2=2+i5
z1z2=[(1)×2]+[3×5]22+52+i[3×2][(1)×2]22+52
=[2]+154+25+i6[2]4+25
1+i32+i5=1329+i829

Q.2. Simplify: (3+i2)(4+i5)
Ans:

(32i)(4+5i)=3×(4+i5)+(i2)(4+i5)
=12+i15+(i8)i210
=12+7i+10×1
=12+7i+10
=22+7i

Q.3. Simplify: (1i)31i3
Ans:

(1i)31i3=(1i)3(1i)(1+i+i2)
=(1i2)i
=1+i2i2i
=i2i
(1i)31i3=2

Q.4. Expand and simplify: (4i3)(2+i3)(2i3)
Ans:

(4i3)(2+i3)(2i3)=(4i3)(4+i6i6i29)
=(4i3)(49(1))
=(4i3)(4+9)
=(4i3)×13
(4i3)(2+i3)(2i3)=52i39

Q5. Subtract (10i8) from (13i7).
Ans:

(13i7)(10i8)=13i710+i8
=(1310)i7+i8
(13i7)(10i8)=3+i

Q6. If (1+i)22i=x+iy, then find the value of x+iy
Ans:

x+iy=(1+i)22i
=1+2i+i22i
=2i2i
=2i(2+i)(2i)(2+i)
=4i+2i24i2
=4i24+1
=25+4i5
x=25,y=45
x+y=25+45
x+y=25

Summary of Operations with Complex Numbers

Complex numbers can be represented in rectangular and polar forms. Similar to real numbers, arithmetic operations such as addition, subtraction, multiplication and division can also be performed on complex numbers.

Frequently Asked Questions (FAQs)

Q.1. What are the mathematical operations of complex numbers?
Ans:
 The four operations that can be performed with complex numbers are addition, subtraction, multiplication, and division. However, subtraction is defined in terms of addition, and division in terms of multiplication.

Q.2. What is the addition of complex numbers?
Ans:
While adding two or more complex numbers, the real parts and the imaginary numbers are added separately.

Q.3. Can you multiply real and imaginary numbers?
Ans:
 The short answer is yes. Mathematically, the multiplication happens only with the real number, but the imaginary unit (i) is carried forward.

For example, 5(12i3)=5×125×i3

5(12i3)=60i15

Q.4. How do you solve complex numbers with multiplication?
Ans:
 While multiplying two complex numbers, remember to multiply each part of the first complex number with each part of the second complex number. Alternatively, remember the pneumonic ‘FOIL’ that expands to First, Outer, Inner, and Last that denotes the order in which pairs of numbers are multiplied.

Example: (p+iq)(x+iy)

Here,

  • First: p×x
  • Outer: p×iy
  • Inner: iq×x
  • Last: iq×iy

Q.5. Do complex numbers commute?
Ans:
 Complex numbers are commutative under addition and multiplication. This means that while adding or multiplying complex numbers, changing the order of the numbers does not alter the result.

Learn About Complex Numbers Here

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Practice Complex Numbers Questions with Hints & Solutions