Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers powers of iota (\( \iota \)), simplification of complex expressions, and solving complex equations. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 1: Complex Numbers
Class 10 Math Unit 1 Exercise 1.1 – Complex Numbers: Complete Guide
Unit 1 of Class 10 Mathematics introduces students to complex numbers, one of the most important foundational topics in the PECTAA 2026 syllabus. Exercise 1.1 is the entry point into this chapter, and it lays the groundwork for understanding the imaginary unit iota (\( \iota \)), its powers, and how to work with complex numbers in equations.
What You Will Learn
By working through Exercise 1.1, students learn how to simplify powers of iota (\( \iota \)), write expressions in terms of \( \iota \), and solve complex equations by equating real and imaginary parts. These are not isolated skills — they are the building blocks used throughout the rest of Unit 1 and reappear in later units when working with quadratic equations and other algebraic expressions.
Topics Covered in This Exercise
- Powers of Iota (\( \iota \)) – simplifying \( \iota^5 \), \( \iota^{16} \), \( (-\iota)^{-19} \), and more
- Complex Expression Simplification – evaluating expressions like \( \iota^{11} + \iota^5 \) and \( (\iota^4 + \iota^3 + \iota^2 + \iota)^2 \)
- Writing in Terms of Iota – expressing radicals of negative numbers using \( \iota \)
- Complex Equations – solving for \( x \) and \( y \) by equating real and imaginary parts
Why Exercise 1.1 Is Important
Exercise 1.1 is the foundation exercise of Unit 1. Punjab Board exam papers frequently draw at least one short question directly from this exercise, particularly on powers of iota (\( \iota \)) and complex equations. Students who master this exercise will find the rest of Unit 1 much easier to understand.
Punjab Board Preparation
For students preparing for board exams under any of the 10 BISE Punjab boards, this exercise should be treated as compulsory practice. The power cycle of iota (\( \iota^{4n} = 1 \), \( \iota^{4n+1} = \iota \), \( \iota^{4n+2} = -1 \), \( \iota^{4n+3} = -\iota \)) is the single most important concept from Exercise 1.1 and appears repeatedly in board exam questions.
PECTAA 2026 Syllabus Alignment
These solved notes are aligned with the Punjab Education & Curriculum Textbook Authority (PECTAA) 2026 syllabus, following the National Curriculum 2023 framework for Class 10 Mathematics. The question numbering, formulas, and terminology match the current Punjab textbook exactly.
Exam Tips for Complex Numbers
- Remember the cycle of powers of iota (\( \iota \)): \( \iota \), \( -1 \), \( -\iota \), \( 1 \), and then repeat.
- When simplifying \( \iota^n \), divide \( n \) by 4 and use the remainder to determine the value.
- For complex equations, always separate real and imaginary parts before equating.
- Check your answers by substituting back into the original equation.
Common Mistakes Students Make
- Forgetting that \( \iota^2 = -1 \) and incorrectly simplifying powers of iota.
- Confusing the real and imaginary parts of a complex equation.
- Not reducing powers of iota to their simplest form.
- Making sign errors when working with negative powers of iota.
Use the table of contents below to jump directly to any question, or scroll through the full solved exercise in order.
1 Simplify the Following
(i) \( \iota^{5} \)
\[
\iota^{5} = \iota^{4} \cdot \iota = (1) \cdot \iota = \iota
\]
Since \( \iota^2 = -1 \), we have \( \iota^4 = (\iota^2)^2 = (-1)^2 = 1 \). Therefore \( \iota^5 = \iota \).
(ii) \( \iota^{16} \)
\[
\iota^{16} = (\iota^{2})^{8} = (-1)^{8} = 1
\]
Using \( \iota^2 = -1 \), we get \( \iota^{16} = (\iota^2)^8 = (-1)^8 = 1 \).
(iii) \( (-\iota)^{-19} \)
\[
(-\iota)^{-19} = \frac{1}{(-\iota)^{19}} = \frac{1}{-\iota^{19}} = -\frac{1}{\iota^{19}}
\]
\[
\iota^{19} = \iota^{16} \cdot \iota^{3} = 1 \cdot (-\iota) = -\iota
\]
\[
-\frac{1}{-\iota} = \frac{1}{\iota} = \frac{\iota}{\iota^2} = \frac{\iota}{-1} = -\iota
\]
Thus \( (-\iota)^{-19} = -\iota \).
(iv) \( \iota^{11} + \iota^{5} \)
\[
\iota^{11} = \iota^{10} \cdot \iota = (\iota^2)^5 \cdot \iota = (-1)^5 \cdot \iota = -\iota
\]
\[
\iota^{5} = \iota
\]
\[
\iota^{11} + \iota^{5} = -\iota + \iota = 0
\]
(v) \( (\iota^{4} + \iota^{3} + \iota^{2} + \iota)^{2} \)
\[
\iota^{4} = 1,\ \iota^{3} = -\iota,\ \iota^{2} = -1,\ \iota = \iota
\]
\[
1 + (-\iota) + (-1) + \iota = (1 - 1) + (-\iota + \iota) = 0
\]
\[
(0)^{2} = 0
\]
The expression simplifies to 0 because the terms cancel out.
(vi) \( \left(\frac{\iota^{8}}{\iota^{5}}\right)^{-5} \)
\[
\frac{\iota^{8}}{\iota^{5}} = \iota^{8-5} = \iota^{3}
\]
\[
(\iota^{3})^{-5} = \iota^{-15} = \frac{1}{\iota^{15}}
\]
\[
\iota^{15} = \iota^{12} \cdot \iota^{3} = 1 \cdot (-\iota) = -\iota
\]
\[
\frac{1}{-\iota} = -\frac{1}{\iota} = -\frac{\iota}{\iota^2} = -\frac{\iota}{-1} = \iota
\]
Thus \( \left(\frac{\iota^{8}}{\iota^{5}}\right)^{-5} = \iota \).
(vii) \( \iota^{13} \times \iota^{29} \)
\[
\iota^{13} \times \iota^{29} = \iota^{13+29} = \iota^{42}
\]
\[
\iota^{42} = (\iota^{2})^{21} = (-1)^{21} = -1
\]
2 Write in Terms of \( \iota \)
(i) \( 2 + \sqrt{-4} \)
\[
2 + \sqrt{-4} = 2 + \sqrt{4 \cdot (-1)} = 2 + 2\sqrt{-1} = 2 + 2\iota
\]
(ii) \( 3 - \sqrt{-7} \)
\[
3 - \sqrt{-7} = 3 - \sqrt{7 \cdot (-1)} = 3 - \sqrt{7}\iota
\]
(iii) \( \frac{2}{5} + \frac{\sqrt{-16}}{5} \)
\[
\frac{2}{5} + \frac{\sqrt{-16}}{5} = \frac{2}{5} + \frac{\sqrt{16 \cdot (-1)}}{5} = \frac{2}{5} + \frac{4\iota}{5} = \frac{2 + 4\iota}{5}
\]
(iv) \( \sqrt{2} - \sqrt{-3} \)
\[
\sqrt{2} - \sqrt{-3} = \sqrt{2} - \sqrt{3 \cdot (-1)} = \sqrt{2} - \sqrt{3}\iota
\]
3 Find the Values of \( x \) and \( y \)
(i) \( (2x+5) + (y-3)\iota = 1 + 2\iota \)
By equating real and imaginary parts:
\[
2x + 5 = 1 \implies 2x = -4 \implies x = -2
\]
\[
y - 3 = 2 \implies y = 5
\]
Therefore, \( x = -2 \) and \( y = 5 \).
(ii) \( (3x+2) - (4-y)\iota = 5 + 3\iota \)
First, rewrite the equation:
\[
(3x+2) + [-(4-y)]\iota = 5 + 3\iota
\]
By equating real and imaginary parts:
\[
3x + 2 = 5 \implies 3x = 3 \implies x = 1
\]
\[
-(4-y) = 3 \implies -4 + y = 3 \implies y = 7
\]
Therefore, \( x = 1 \) and \( y = 7 \).
(iii) \( (2+\iota)x + (1-2\iota)y = 3 + 4\iota \)
Expand the left side:
\[
2x + x\iota + y - 2y\iota = 3 + 4\iota
\]
Group real and imaginary parts:
\[
(2x + y) + (x - 2y)\iota = 3 + 4\iota
\]
By equating real and imaginary parts:
\[
2x + y = 3 \quad \text{...(i)}
\]
\[
x - 2y = 4 \quad \text{...(ii)}
\]
From equation (i): \( y = -2x + 3 \)
Substitute in equation (ii):
\[
x - 2(-2x + 3) = 4 \implies x + 4x - 6 = 4 \implies 5x = 10 \implies x = 2
\]
Now substitute \( x = 2 \) in equation (i):
\[
y = -2(2) + 3 = -4 + 3 = -1
\]
Therefore, \( x = 2 \) and \( y = -1 \).
(iv) \( (1-\iota)x + (2+\iota)y = 4 - \iota \)
Expand the left side:
\[
x - x\iota + 2y + y\iota = 4 - \iota
\]
Group real and imaginary parts:
\[
(x + 2y) + (y - x)\iota = 4 - \iota
\]
By equating real and imaginary parts:
\[
x + 2y = 4 \quad \text{...(i)}
\]
\[
y - x = -1 \quad \text{...(ii)}
\]
From equation (i): \( x = -2y + 4 \)
Substitute in equation (ii):
\[
y - (-2y + 4) = -1 \implies y + 2y - 4 = -1 \implies 3y = 3 \implies y = 1
\]
Now substitute \( y = 1 \) in equation (ii):
\[
1 - x = -1 \implies x = 2
\]
Therefore, \( x = 2 \) and \( y = 1 \).
(v) \( (3x-1) + (2y-3)\iota = 8 + 7\iota \)
By equating real and imaginary parts:
\[
3x - 1 = 8 \implies 3x = 9 \implies x = 3
\]
\[
2y - 3 = 7 \implies 2y = 10 \implies y = 5
\]
Therefore, \( x = 3 \) and \( y = 5 \).
📈 Key Concepts & Quick Revision
📐 Powers of Iota (\( \iota \))
- Basic Powers: \( \iota^2 = -1 \), \( \iota^3 = -\iota \), \( \iota^4 = 1 \)
- General Cycle: \( \iota^{4n} = 1 \), \( \iota^{4n+1} = \iota \), \( \iota^{4n+2} = -1 \), \( \iota^{4n+3} = -\iota \)
- Negative Powers: \( \iota^{-n} = \frac{1}{\iota^n} \)
📐 Complex Numbers
- Standard Form: \( z = a + b\iota \) where \( a, b \in \mathbb{R} \)
- Real Part: \( a = \text{Re}(z) \)
- Imaginary Part: \( b = \text{Im}(z) \)
- Equality: \( a + b\iota = c + d\iota \iff a = c \) and \( b = d \)
📝 Important MCQs for Practice
1. \( \iota^{2026} \) is equal to:
2. \( \iota^{4} + \iota^{3} + \iota^{2} + \iota \) equals:
3. If \( (x+2) + (y-1)\iota = 5 + 3\iota \), then \( x + y \) equals:
4. The value of \( \sqrt{-25} \) in terms of \( \iota \) is:
5. \( \iota^{15} \) is equal to:
🏆 Board Exam Tips & Strategy
- Always reduce powers of iota modulo 4 to find the simplified value.
- For complex equations, write the equation clearly before separating real and imaginary parts.
- Check your answers by substituting back into the original equation.
- Practice solving simultaneous equations — they appear frequently in complex number problems.
❓ Frequently Asked Questions
What is taught in Exercise 1.1 of Class 10 Math Unit 1?
Exercise 1.1 introduces complex numbers and powers of iota (\( \iota \)). It covers simplifying powers of \( \iota \), evaluating complex expressions, and solving complex equations by equating real and imaginary parts.
How many questions are there in Unit 1 Exercise 1.1?
Exercise 1.1 has 3 main sections: simplifying powers of iota (\( \iota \)) (Question 1), writing expressions in terms of \( \iota \) (Question 2), and finding values of \( x \) and \( y \) from complex equations (Question 3 with 5 parts).
What is iota (\( \iota \)) in mathematics?
Iota (\( \iota \)) is the imaginary unit defined as \( \iota = \sqrt{-1} \), with the property that \( \iota^2 = -1 \). It is used to represent complex numbers.
What are the powers of iota (\( \iota \))?
The powers of iota cycle every 4: \( \iota^1 = \iota \), \( \iota^2 = -1 \), \( \iota^3 = -\iota \), \( \iota^4 = 1 \), and then repeat. For any integer \( n \), \( \iota^{4n} = 1 \), \( \iota^{4n+1} = \iota \), \( \iota^{4n+2} = -1 \), \( \iota^{4n+3} = -\iota \).
How do you solve complex equations?
To solve complex equations, separate the equation into real and imaginary parts, then equate the real parts on both sides and the imaginary parts on both sides. This gives two equations that can be solved simultaneously.
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 1.1 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 1.1?
Yes, a complete solved PDF for Exercise 1.1 is embedded on this page and available to download for free.
Can I download the Unit 1 Exercise 1.1 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 1.1 notes to your device.
Is Exercise 1.1 important for Class 10 board exams?
Yes, complex numbers and powers of iota (\( \iota \)) are fundamental topics tested in board exams. Exercise 1.1 builds essential skills for solving complex equations and simplifying complex expressions.
Who prepared these Class 10 Math Unit 1 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.