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 focuses on finding real and imaginary parts of complex numbers after algebraic manipulation, and solving simultaneous linear equations with complex coefficients. Step-by-step solutions using conjugates, rationalization, and substitution methods. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 1: Complex Numbers
Class 10 Math Unit 1 Exercise 1.4 – Real & Imaginary Parts: Complete Guide
Exercise 1.4 of Unit 1 focuses on two important skills in complex numbers: finding real and imaginary parts of complex expressions and solving simultaneous linear equations with complex coefficients. This exercise consolidates all the skills learned in previous exercises — operations with complex numbers, conjugates, and rationalization — and applies them to more advanced problems.
What You Will Learn
By working through Exercise 1.4, students master the following: simplifying complex expressions using algebraic manipulation, rationalizing denominators using conjugates, identifying real parts (\( \operatorname{Re}(z) \)) and imaginary parts (\( \operatorname{Im}(z) \)), and solving systems of linear equations where coefficients are complex numbers.
Topics Covered in This Exercise
- Real and Imaginary Parts – finding \( \operatorname{Re}(z) \) and \( \operatorname{Im}(z) \) from expressions like \( (8-3\iota)^2 \), \( (5+3\iota)^{-1} \), and more complex forms
- Rationalization – using conjugates to simplify division of complex numbers
- Complex Simultaneous Equations – solving for \( z \) and \( w \) using substitution and elimination
Why Exercise 1.4 Is Important
Exercise 1.4 is the culminating exercise of Unit 1. Board exam papers frequently feature questions on finding real and imaginary parts and solving complex equations. This exercise tests students' ability to apply all the skills learned in Exercises 1.1, 1.2, and 1.3 in a integrated manner.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Finding real and imaginary parts is a common short question type, and complex simultaneous equations appear in multi-part questions. Practising every part of Q1 and Q2 by hand is highly recommended.
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 Real & Imaginary Parts
- Always simplify \( \iota^2 = -1 \) immediately when expanding expressions.
- For reciprocal expressions, multiply numerator and denominator by the conjugate.
- When solving complex equations, treat \( \iota \) as a constant and group real and imaginary terms separately.
- Check your answers by substituting back into the original equations.
Common Mistakes Students Make
- Forgetting that \( \iota^2 = -1 \) when expanding powers.
- Not rationalizing denominators properly.
- Confusing the real part with the coefficient of \( \iota \).
- Making sign errors when solving simultaneous equations.
📖 Exercise 1.4 – Solved Problems
1
Find the real and imaginary parts of the following complex numbers.
(i) \((8 - 3\iota)^2\)
\[
\begin{aligned}
z &= (8 - 3\iota)^2 \\
&= 8^2 + (3\iota)^2 - 2(8)(3\iota) \\
&= 64 + 9\iota^2 - 48\iota \\
&= 64 + 9(-1) - 48\iota \\
&= 64 - 9 - 48\iota \\
&= 55 - 48\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=55,\quad \operatorname{Im}(z)=-48\).
(ii) \((5 + 3\iota)^{-1}\)
\[
\begin{aligned}
z &= (5 + 3\iota)^{-1} = \frac{1}{5 + 3\iota} \\
&= \frac{1}{5 + 3\iota} \times \frac{5 - 3\iota}{5 - 3\iota} \\
&= \frac{5 - 3\iota}{(5)^2 - (3\iota)^2} \\
&= \frac{5 - 3\iota}{25 - 9\iota^2} = \frac{5 - 3\iota}{25 - 9(-1)} \\
&= \frac{5 - 3\iota}{25 + 9} = \frac{5 - 3\iota}{34} \\
&= \frac{5}{34} - \frac{3}{34}\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=\dfrac{5}{34},\quad \operatorname{Im}(z)=-\dfrac{3}{34}\).
(iii) \((4 - 5\iota)^{-1}\)
\[
\begin{aligned}
z &= \frac{1}{4 - 5\iota} = \frac{1}{4 - 5\iota} \times \frac{4 + 5\iota}{4 + 5\iota} \\
&= \frac{4 + 5\iota}{4^2 - (5\iota)^2} = \frac{4 + 5\iota}{16 - 25\iota^2} \\
&= \frac{4 + 5\iota}{16 - 25(-1)} = \frac{4 + 5\iota}{16 + 25} \\
&= \frac{4 + 5\iota}{41} = \frac{4}{41} + \frac{5}{41}\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=\dfrac{4}{41},\quad \operatorname{Im}(z)=\dfrac{5}{41}\).
(iv) \((4 - 3\iota)^{-2}\)
\[
\begin{aligned}
z &= \frac{1}{(4 - 3\iota)^2} = \frac{1}{16 + 9\iota^2 - 24\iota} \\
&= \frac{1}{16 + 9(-1) - 24\iota} = \frac{1}{7 - 24\iota} \\
&= \frac{1}{7 - 24\iota} \times \frac{7 + 24\iota}{7 + 24\iota} = \frac{7 + 24\iota}{49 - (24\iota)^2} \\
&= \frac{7 + 24\iota}{49 - 576\iota^2} = \frac{7 + 24\iota}{49 - 576(-1)} \\
&= \frac{7 + 24\iota}{49 + 576} = \frac{7 + 24\iota}{625} = \frac{7}{625} + \frac{24}{625}\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=\dfrac{7}{625},\quad \operatorname{Im}(z)=\dfrac{24}{625}\).
(v) \(\dfrac{3 + 2\iota}{4 + 3\iota}\)
\[
\begin{aligned}
z &= \frac{3 + 2\iota}{4 + 3\iota} \times \frac{4 - 3\iota}{4 - 3\iota} \\
&= \frac{(3)(4) + 3(-3\iota) + 2\iota(4) + 2\iota(-3\iota)}{(4)^2 - (3\iota)^2} \\
&= \frac{12 - 9\iota + 8\iota - 6\iota^2}{16 - 9\iota^2} \\
&= \frac{12 - \iota -6(-1)}{16 - 9(-1)} = \frac{12 - \iota + 6}{16 + 9} \\
&= \frac{18 - \iota}{25} = \frac{18}{25} - \frac{1}{25}\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=\dfrac{18}{25},\quad \operatorname{Im}(z)=-\dfrac{1}{25}\).
(vi) \(\left( \dfrac{2-\iota}{2+\iota} \right)^{-2}\)
\[
\begin{aligned}
z &= \left( \frac{2+\iota}{2-\iota} \right)^{2} = \frac{(2+\iota)^2}{(2-\iota)^2} \\
&= \frac{4 + \iota^2 + 4\iota}{4 + \iota^2 - 4\iota} = \frac{4 - 1 + 4\iota}{4 - 1 - 4\iota} \\
&= \frac{3 + 4\iota}{3 - 4\iota} \times \frac{3 + 4\iota}{3 + 4\iota} = \frac{9 + 12\iota + 12\iota + 16\iota^2}{9 - 16\iota^2} \\
&= \frac{9 + 24\iota + 16(-1)}{9 - 16(-1)} = \frac{9 + 24\iota - 16}{9 + 16} \\
&= \frac{-7 + 24\iota}{25} = -\frac{7}{25} + \frac{24}{25}\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=-\dfrac{7}{25},\quad \operatorname{Im}(z)=\dfrac{24}{25}\).
(vii) \(\left( \dfrac{1-2\iota}{1+\iota} \right)^{2}\)
\[
\begin{aligned}
z &= \frac{(1-2\iota)^2}{(1+\iota)^2} = \frac{1 + 4\iota^2 - 4\iota}{1 + \iota^2 + 2\iota} \\
&= \frac{1 + 4(-1) - 4\iota}{1 + (-1) + 2\iota} = \frac{1 - 4 - 4\iota}{2\iota} \\
&= \frac{-3 - 4\iota}{2\iota} \times \frac{2\iota}{2\iota} = \frac{-6\iota - 8\iota^2}{4\iota^2} \\
&= \frac{-6\iota - 8(-1)}{4(-1)} = \frac{-6\iota + 8}{-4} \\
&= \frac{8 - 6\iota}{-4} = -2 + \frac{3}{2}\iota
\end{aligned}
\]
Hence \(\operatorname{Re}(z)=-2,\quad \operatorname{Im}(z)=\dfrac{3}{2}\).
2
Solve the simultaneous linear equations with complex coefficients for \(w\) and \(z\).
(i)
\[
\begin{cases}
3z + (2+\iota)w = 11 - \iota \\
(2-\iota)z - w = -1 + \iota
\end{cases}
\]
\[
\begin{aligned}
&\text{From }(2-\iota)z - w = -1+\iota \implies w = (2-\iota)z + 1 - \iota. \\
&\text{Substitute into first equation:}\\
3z + (2+\iota)\big[(2-\iota)z + 1 - \iota\big] &= 11 - \iota \\
3z + (2+\iota)(2-\iota)z + (2+\iota)(1 - \iota) &= 11 - \iota \\
3z + (4 - \iota^2)z + \big[2 - 2\iota + \iota - \iota^2\big] &= 11 - \iota \\
3z + (4 + 1)z + \big[2 - \iota - (-1)\big] &= 11 - \iota \\
3z + 5z + (2 - \iota + 1) &= 11 - \iota \\
8z + (3 - \iota) &= 11 - \iota \\
8z &= 8 \implies z = 1.
\end{aligned}
\]
Then \(w = (2-\iota)(1) + 1 - \iota = 2 - \iota + 1 - \iota = 3 - 2\iota\).
\(\boxed{z=1,\; w=3-2\iota}\).
\(\boxed{z=1,\; w=3-2\iota}\).
(ii)
\[
\begin{cases}
2z + (3+\iota)w = 9 - \iota \\
-\iota z - \iota w = -1 + \iota
\end{cases}
\]
\[
\begin{aligned}
&\text{From } -\iota(z+w) = -1+\iota \;\Rightarrow\; z+w = \frac{-1+\iota}{-\iota} = \frac{-1+\iota}{-\iota}\cdot \frac{\iota}{\iota} \\
&= \frac{-\iota + \iota^2}{-\iota^2} = \frac{-\iota -1}{1} = -1 - \iota \\
&\Rightarrow z = -w -1 - \iota.
\end{aligned}
\]
Substitute into first equation:
\[
2(-w -1 - \iota) + (3+\iota)w = 9 - \iota \\
-2w -2 -2\iota + 3w + \iota w = 9 - \iota \\
w + \iota w -2 -2\iota = 9 - \iota \\
w(1+\iota) = 9 - \iota + 2 + 2\iota = 11 + \iota.
\]
\[
w = \frac{11+\iota}{1+\iota} = \frac{11+\iota}{1+\iota}\cdot\frac{1-\iota}{1-\iota} = \frac{11-11\iota + \iota -\iota^2}{1 - \iota^2} = \frac{11 -10\iota +1}{1+1} = \frac{12 -10\iota}{2} = 6 -5\iota.
\]
Then \(z = -(6-5\iota) -1 - \iota = -6+5\iota -1 -\iota = -7 + 4\iota\).
\(\boxed{z = -7+4\iota,\; w = 6-5\iota}\).
\(\boxed{z = -7+4\iota,\; w = 6-5\iota}\).
(iii)
\[
\begin{cases}
z - 4w = 3\iota \\
2z + 3w = 11 - 5\iota
\end{cases}
\]
From first: \(z = 3\iota + 4w\). Substitute into second:
\[
2(3\iota + 4w) + 3w = 11 - 5\iota \implies 6\iota + 8w + 3w = 11 - 5\iota \implies 11w + 6\iota = 11 - 5\iota.
\]
\[
11w = 11 - 5\iota - 6\iota = 11 - 11\iota \implies w = 1 - \iota.
\]
Then \(z = 3\iota + 4(1 - \iota) = 3\iota + 4 - 4\iota = 4 - \iota\).
\(\boxed{z = 4 - \iota,\; w = 1 - \iota}\).
\(\boxed{z = 4 - \iota,\; w = 1 - \iota}\).
(iv)
\[
\begin{cases}
z + w = 3\iota \\
2z + 3w = 2
\end{cases}
\]
From first: \(z = 3\iota - w\). Insert into second:
\[
2(3\iota - w) + 3w = 2 \implies 6\iota - 2w + 3w = 2 \implies 6\iota + w = 2 \implies w = 2 - 6\iota.
\]
Then \(z = 3\iota - (2 - 6\iota) = 3\iota - 2 + 6\iota = -2 + 9\iota\).
\(\boxed{z = -2 + 9\iota,\; w = 2 - 6\iota}\).
\(\boxed{z = -2 + 9\iota,\; w = 2 - 6\iota}\).
(v)
\[
\begin{cases}
2z + (3+\iota)w = 1 \\
-z - (1-\iota)w = 2
\end{cases}
\]
From second: \(-z = 2 + (1-\iota)w \;\Rightarrow\; z = -2 - (1-\iota)w = -2 - w + \iota w\).
Substitute into first:
\[
2\big[-2 - w + \iota w\big] + (3+\iota)w = 1 \\
-4 - 2w + 2\iota w + 3w + \iota w = 1 \\
-4 + w + 3\iota w = 1 \implies w + 3\iota w = 5 \\
w(1+3\iota) = 5 \implies w = \frac{5}{1+3\iota}.
\]
\[
w = \frac{5}{1+3\iota} \cdot \frac{1-3\iota}{1-3\iota} = \frac{5 - 15\iota}{1 - 9\iota^2} = \frac{5 - 15\iota}{1 + 9} = \frac{5 - 15\iota}{10} = \frac{1}{2} - \frac{3}{2}\iota.
\]
Then \(z = -2 - \left(\frac12 - \frac32\iota\right) + \iota\left(\frac12 - \frac32\iota\right)\):
\[
z = -2 - \frac12 + \frac32\iota + \frac12\iota - \frac32\iota^2 = -\frac52 + 2\iota - \frac32(-1) = -\frac52 + 2\iota + \frac32 = -1 + 2\iota.
\]
\(\boxed{z = -1 + 2\iota,\; w = \dfrac{1}{2} - \dfrac{3}{2}\iota}\).
📈 Key Concepts & Quick Revision
📐 Operations with Complex Numbers
- Reciprocal: \( \frac{1}{a+b\iota} = \frac{a-b\iota}{a^2+b^2} \)
- Square of binomial: \( (a\pm b\iota)^2 = a^2 - b^2 \pm 2ab\iota \)
- Division: \( \frac{a+b\iota}{c+d\iota} = \frac{(a+b\iota)(c-d\iota)}{c^2+d^2} \)
- Real Part: \( \operatorname{Re}(a+b\iota) = a \)
- Imaginary Part: \( \operatorname{Im}(a+b\iota) = b \)
📝 Important MCQs for Practice
1. The real part of \((2+3\iota)^2\) is:
2. The imaginary part of \(\frac{1}{2+3\iota}\) is:
3. If \(3z + (2+\iota)w = 11-\iota\) and \(z=1\), then \(w\) equals:
🏆 Board Exam Tips & Strategy
- Always show the intermediate steps when finding real and imaginary parts.
- For simultaneous equations, clearly write the substitution step.
- Check your answers by substituting back into the original equations.
- Practice all parts of Q2 — complex equations appear frequently in board exams.
❓ Frequently Asked Questions
What is taught in Exercise 1.4 of Class 10 Math Unit 1?
Exercise 1.4 covers finding real and imaginary parts (\( \operatorname{Re}(z) \), \( \operatorname{Im}(z) \)) of complex numbers after algebraic manipulation, and solving simultaneous linear equations with complex coefficients using substitution and elimination methods.
How many questions are there in Unit 1 Exercise 1.4?
Exercise 1.4 has 2 main questions: Question 1 has 7 parts covering finding real and imaginary parts, and Question 2 has 5 parts covering solving simultaneous linear equations with complex coefficients.
How do you find the real and imaginary parts of a complex number?
To find real and imaginary parts, simplify the complex expression using algebraic operations, rationalize denominators using conjugates, and write in the form \( a + b\iota \). The real part is \( \operatorname{Re}(z) = a \) and the imaginary part is \( \operatorname{Im}(z) = b \).
How do you solve simultaneous equations with complex coefficients?
To solve simultaneous equations with complex coefficients, use substitution or elimination methods just like real equations. Treat \( \iota \) as a constant and simplify using \( \iota^2 = -1 \). The solutions will be complex numbers.
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 1.4 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 1.4?
Yes, a complete solved PDF for Exercise 1.4 is embedded on this page and available to download for free.
Can I download the Unit 1 Exercise 1.4 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 1.4 notes to your device.
Is Exercise 1.4 important for Class 10 board exams?
Yes, finding real and imaginary parts and solving complex equations are frequently tested in board exams. Exercise 1.4 builds essential skills in complex number manipulation and algebraic problem-solving.
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.