What are Imaginary Numbers and iota (i), Powers of iota, Solved Examples

Imaginary Numbers and iota are fundamental concepts in Complex Numbers and Quadratic Equations for CBSE Class 11 Maths. Imaginary numbers are introduced to solve equations that do not have real solutions. This article covers imaginary numbers, iota definition, positive and negative integral powers, the cyclic pattern of i, solved examples including evaluating large powers and proving identities — all with complete step-by-step working Understanding integral powers of iota is essential for simplifying complex expressions and solving algebraic problems efficiently. This topic forms the foundation for advanced mathematics and is important for CBSE board exams, JEE, and other competitive examinations.

What are Imaginary Numbers ?

Linear and quadratic equations are fundamental in math, but some, like $x^2+1=0$, have no solution within the domain of real numbers. Because real numbers cannot solve equations involving the square root of a negative value, it became necessary to extend the real number system into a larger system.

The mathematician Euler was the first to introduce the symbol $i$ (iota) to represent $\sqrt{-1}$.

The concept of complex numbers was primarily driven by the need to find solutions for algebraic equations that were previously considered “unsolvable”. By defining $i=\sqrt{-1}$, mathematicians were able to create a consistent framework for these calculations.

Learn more about Venn Diagrams of SETS : Laws, Theorems, Proofs, Examples“

---

What is the definition of Imaginary Numbers?

An imaginary number is defined as the square root of a negative number.

Examples : Numbers such as $\sqrt{-2}$, $\sqrt{-3}$, and $\sqrt{-17}$ are all classified as imaginary numbers.

$\sqrt{-1}$ is called an imaginary unit iota, which is denoted by $i$.

The core property of the imaginary unit is that $i^2=-1$.

“Learn Class 11 Maths Full Chapter Notes”

---

What is iota (i) ?

The iota is referred to by the alphabet ‘i’. iota is helpful to represent the imaginary part of the complex number. It is also very helpful to find the square root of negative numbers as the value of i² = -1, which is used to find the value of the square root of negative numbers,

Example: $\sqrt{-9}$ =$\sqrt{i^2{3}^2}$= ±3i

“Learn Full Chapter Notes of Complex Number for Class 11“

---

What are the Integral Powers of iota (i) ?

The value of i² = -1 is the fundamental aspect of a complex number. 

The powers of (i) follow a repeating cycle of four :

$i=\sqrt{-1}$,

$i^2=i\cdot i=\sqrt{-1}\cdot \sqrt{-1}=-1,$

$i^3=i^2\cdot i=(-1)\cdot i=-i,$

$i^4=(i^2{)}^2=(-1{)}^2=1;$

After this, the powers repeat continuously:

$i^5=i,i^6=-1,i^7=-i,i^8=1$

What is the Cyclic Pattern of Powers of iota (i) ?

$i^{4n}=1$

$i^{4n+1}=i$

$i^{4n+2}=-1$

$i^{4n+3}=-i$

where (n) is any positive integer.

How to Find Large Powers of iota (i) ?

To simplify large powers of (i), divide the exponent by 4 and find the remainder.

• Remainder 0 $(\to 1)$
• Remainder 1 $(\to i)$
• Remainder 2 $(\to -1)$
• Remainder 3 $(\to -i)$

Example Problems

Example.1 : Find $(i^{17})$

Since : $17÷4=4\text{ remainder }1$

Therefore : $i^{17}=i$

Example 2 : Find $(i^{26}).$

Since : $26÷4=6\text{ remainder }2$

Therefore : $i^{26}=-1$

Example 3 : Find $(i^{43}).$

Since : $43÷4=10\text{ remainder }3$

Therefore : $i^{43}=-i$

What are the Positive Integral Powers of iota (i) ?

$i=\sqrt{-1}$,

$i^2=i\cdot i=\sqrt{-1}\cdot \sqrt{-1}=-1,$

$i^3=i^2\cdot i=(-1)\cdot i=-i,$

$i^4=(i^2{)}^2=(-1{)}^2=1;$

$i^{4n}=1$

$i^{4n+1}=i$

$i^{4n+2}=-1$

$i^{4n+3}=-i$ and so on.

where (n) is any positive integer.

What are the Negative Integral Powers of iota (i) ?

$i^{-1}={\displaystyle \frac{1}{i}}={\displaystyle \frac{i^3}{i^4}}={\displaystyle \frac{i^3}{1}}=-i,$

$i^{-2}={\displaystyle \frac{1}{i^2}}={\displaystyle \frac{1}{-1}}=-1$

$i^{-3}={\displaystyle \frac{1}{i^3}}={\displaystyle \frac{i}{i^4}}={\displaystyle \frac{i}{1}}=i,$

$i^{-4}={\displaystyle \frac{1}{i^4}}={\displaystyle \frac{1}{1}}=1$ and so on.

What is the Zero power of iota (i) ?

$i^0=1.$

Key Point
$i^n=-1,1,i$ or $-i$ $\forall$ n $\in Z.$

“Access Complex Numbers and Quadratic Equations NCERT Solutions of Exercise 4.1“

---

Why $\sqrt{a}\times \sqrt{b}\ne \sqrt{ab}$ for Negative Numbers?

For any two real numbers a and b, $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ is true only when at least one of a and b is either zero or positive.

Actually, $\sqrt{-a}\times \sqrt{-b}=(i\sqrt{a})(i\sqrt{b})=i^2\sqrt{ab}=-\sqrt{ab},$ where a and b are both positive real numbers.

So, $\sqrt{-3}\times \sqrt{-5}=\sqrt{(-3)\times (-5)}=\sqrt{15}$ is wrong.

Thus the correct result is :

$\sqrt{-3}\times \sqrt{-5}=(i\sqrt{3})\times (i\sqrt{5})=i^2(\sqrt{3}\times \sqrt{5})=-\sqrt{15}.$

Key Point
$\sqrt{a}\times \sqrt{b}\ne \sqrt{ab}$ if a, b are both negative.

---

Solved Examples Based on Imaginary Numbers for CBSE Class 11 Maths Complex Numbers and Quadratic Equations Chapter

Example 1. Evaluate the following : 

(i) $i^9$  (ii) $i^{342}$  (iii) $i^{998}$ 

(iv) $i^{-63}$  (v) $(i^3+{\displaystyle \frac{1}{i^3}})$

Solution.

(i)

$i^9=i^8\times i=(i^4{)}^2\times i=(1{)}^2\times i$ $=(1)\times i=i.$ $[\because i^4=1]$

(ii)

On dividing 342 by 4, we get 85 as quotient and 2 as remainder.

$i^{342}=(i^4{)}^{85}\times i^2=(1{)}^{85}\times i^2=i^2=-1.$ $[\because i^4=1]$

(iii)

On dividing 998 by 4, we get 249 as quotient and 2 as remainder.

$i^{998}=(i^4{)}^{249}\times i^2=(1{)}^{249}\times i^2=i^2=-1.$

(iv)

$i^{-63}={\displaystyle \frac{1}{i^{63}}}={\displaystyle \frac{1}{i^{63}}}\times {\displaystyle \frac{i}{i}}={\displaystyle \frac{i}{i^{64}}}={\displaystyle \frac{i}{(i^4{)}^{16}}}$ $={\displaystyle \frac{i}{1}}=i.$

(v)

$i^3=i^2\times i=(-1)\times i=-i$ $[\because i^2=-1]$

$(i^3+{\displaystyle \frac{1}{i^3}})=-i+{\displaystyle \frac{1}{-i}}=-(i+{\displaystyle \frac{1}{i}})$ $=-{\displaystyle \frac{i^2+1}{i}}=-{\displaystyle \frac{-1+1}{i}}=0$

“Learn more about Set Operations Laws and Theorems, DeMorgan’s Law“

---

Example 2. Evaluate the following : 

(i) $\sqrt{-25}\times \sqrt{-81}$ 

(ii) $\sqrt{-36}\times \sqrt{16}$ 

(iii) $4\sqrt{-4}+5\sqrt{-9}-3\sqrt{-16}$.

Solution.

(i)

$\sqrt{-25}\times \sqrt{-81}=(5i)\times (9i)=45i^2$ $=-45.$

(ii)

$\sqrt{-36}\times \sqrt{16}=(6i)\times 4=24i.$

(iii)

$4\sqrt{-4}+5\sqrt{-9}-3\sqrt{-16}$
= $4(2i)+5(3i)-3(4i)$ $=8i+15i-12i=11i$

“Learn set theory from basic Set Theory : Representation of Sets, Roster or Tabular Form, Set builder Form“

---

Example 3. Prove that: 

(i) $1+i^{10}+i^{100}-i^{1000}=0$ 

(ii) $i^{107}+i^{112}+i^{117}+i^{122}=0$ 

(iii) $(1+i^{14}+i^{18}+i^{22})$ is a real number.

Solution.

(i) $(1+i^{10}+i^{100}-i^{1000})$

$=1+(i^4{)}^2\times i^2+(i^4{)}^{25}-(i^4{)}^{250}$

$[\because i^4=1\text{ and }{i}^2=-1]$

$=1+(i^2{)}^2\times (-1)+(1{)}^{25}-(1{)}^{250}$

$=1-1+1-1=0.$

(ii) $(i^{107}+i^{112}+i^{117}+i^{122})$

$=i^{107}(1+i^5+i^{10}+i^{15})$

$[\because i^5=i^4\times i=i;\text{ etc.}]$

$=i^{107}(1+i+i^2+i^3)$

$[\because i^2=-1\text{ and }{i}^3=-i]$

$=i^{107}(1+i-1-i)$

$=i^{107}(0)=0.$

(iii) $(1+i^{14}+i^{18}+i^{22})$

$=1+(i^4{)}^3\times i^2+(i^4{)}^4\times i^2+(i^4{)}^5\times i^2$

$[\because i^4=1\text{ and }{i}^2=-1]$

$=1+(1{)}^3\times (-1)+(1{)}^4\times (-1)+(1{)}^5\times (-1)$

$=1-1-1-1=-2,$

which is a real number.

---

Example 4. Prove that ${(i^{17}-{\left({\displaystyle \frac{1}{i}}\right)}^{34})}^2=2i.$

Solution.

${(i^{17}-{\left({\displaystyle \frac{1}{i}}\right)}^{34})}^2={(i^{17}-{\displaystyle \frac{1}{i^{34}}})}^2$

$={(i-{\displaystyle \frac{1}{i^2}})}^2={(i-{\displaystyle \frac{1}{-1}})}^2=(i+1{)}^2$

$=i^2+2i+1=-1+2i+1=2i,$

which is true.

---

Example 5. For any positive integer n, prove that :

$i^n+i^{n+1}+i^{n+2}+i^{n+3}+i^{n+4}+i^{n+5}+i^{n+6}+i^{n+7}=0.$

Solution.

$\text{LHS}=i^n+i^{n+1}+i^{n+2}+i^{n+3}+i^{n+4}+i^{n+5}+i^{n+6}+i^{n+7}$

$=i^n(1+i+i^2+i^3+i^4+i^5+i^6+i^7)$

$=i^n(1+i+i^2+i^3+1+i+i^2+i^3)[\because i^4=1]$

$=2i^n(1+i+i^2+i^3)$

$=2i^n(1+i-1-i)[\because i^2=-1\text{ and }{i}^3=-i]$

$=2i^n(0)=0=\text{RHS.}$

Learn more about JEE Previous Years Questions and Solved Papers”

---

FAQs about Imaginary Numbers and Iota (i)

What are Imaginary Numbers in Class 11 Maths?

Imaginary numbers are numbers involving the imaginary unit (i), used to solve equations that do not have real solutions.

Why are imaginary numbers important?

Imaginary numbers are important in algebra, engineering, physics, higher mathematics, and competitive examinations like JEE.

How do you find large powers of (i)?

Large powers of (i) can be simplified by dividing the exponent by 4 and checking the remainder.

Which chapter contains Imaginary Numbers and iota in CBSE Class 11 Maths?

Imaginary Numbers and iota are included in the chapter Complex Numbers and Quadratic Equations.

Are Integral Powers of iota important for JEE?

Yes, questions based on powers of (i), simplification, and complex numbers are frequently asked in JEE Main and Advanced.

What are the applications of complex numbers?

Complex numbers are used in electrical engineering, signal processing, physics, quantum mechanics, and advanced mathematics.

Where can students find solved examples on Integral Powers of iota?

Students can practice solved examples, NCERT questions, exemplar problems, and JEE PYQs to strengthen conceptual understanding.

---

Important Chapter Links For Set Theory

Sets Exercise 1.4 of Class 11 NCERT Mathematics covers important concepts of set operations including union, intersection, difference, and complement. This exercise helps students understand how to perform operations on sets and solve problems using Venn diagrams. It also introduces key properties and laws of sets that simplify calculations. With detailed solutions and examples, this exercise strengthens problem-solving skills and is highly useful for CBSE exams and competitive exams of India.

Exercise-wise NCERT Solutions of Previous Chapter SETS

Exercise 1.1

• Basic definition of sets
• Writing sets in roster and set-builder form

Exercise 1.2

• Types of sets
• Finite and infinite sets

Exercise 1.3

• Subsets and proper subsets
• Number of subsets

Exercise 1.4

• Set operations (union, intersection, complement)

Exercise 1.5

• Advanced problems on set operations and Venn diagrams

Miscellaneous Exercise

• Mixed problems covering all concepts of the chapter