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 modulus of complex numbers (\( |z| \)), conjugate properties (\( \overline{z} \)), and verification of identities. Step-by-step solutions for finding modulus, verifying \( \overline{z_1+z_2} = \overline{z_1}+\overline{z_2} \), \( \overline{z_1z_2} = \overline{z_1}\cdot\overline{z_2} \), and other fundamental properties. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 1: Complex Numbers
Class 10 Math Unit 1 Exercise 1.3 – Modulus and Conjugates: Complete Guide
Exercise 1.3 of Unit 1 introduces students to the modulus and conjugate of complex numbers. This exercise is where students learn to find the modulus (\( |z| \)), understand the geometric interpretation as distance from origin, work with conjugates (\( \overline{z} \)), and verify important algebraic properties. These concepts are essential for the remainder of Unit 1 and form the basis of many problems in later units.
What You Will Learn
By working through Exercise 1.3, students master the following concepts: finding the modulus of complex numbers using \( |z| = \sqrt{a^2+b^2} \), understanding conjugates as reflections across the real axis, verifying properties like \( \overline{z_1+z_2} = \overline{z_1}+\overline{z_2} \), \( \overline{z_1z_2} = \overline{z_1}\overline{z_2} \), and the fundamental identity \( z\overline{z} = |z|^2 \).
Topics Covered in This Exercise
- Modulus of Complex Numbers – finding \( |z| = \sqrt{a^2+b^2} \) for various complex numbers
- Conjugate Properties – verifying \( \overline{z_1+z_2} = \overline{z_1}+\overline{z_2} \)
- Product of Conjugates – verifying \( \overline{z_1z_2} = \overline{z_1}\overline{z_2} \)
- Division of Conjugates – verifying \( \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}} \)
- Modulus Properties – \( |z| = |\overline{z}| = |-z| = |-\overline{z}| \)
- Fundamental Identity – \( z\overline{z} = |z|^2 \)
Why Exercise 1.3 Is Important
Exercise 1.3 introduces two of the most important concepts in complex numbers: modulus and conjugate. Board exam papers frequently feature questions on finding modulus, verifying conjugate properties, and using the identity \( z\overline{z} = |z|^2 \). These concepts are essential for understanding the geometric interpretation of complex numbers.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Finding modulus and verifying conjugate properties are among the most common question types in the complex numbers chapter. Practising every part of Q1, Q2, and Q3 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 Modulus and Conjugates
- Remember that modulus represents distance from origin: \( |z| = \sqrt{a^2+b^2} \).
- Conjugate flips the sign of the imaginary part: \( \overline{a+b\iota} = a-b\iota \).
- The identity \( z\overline{z} = |z|^2 \) is extremely useful and appears frequently.
- For division, always rationalize using the conjugate of the denominator.
Common Mistakes Students Make
- Forgetting to square negative components when finding modulus.
- Confusing \( \overline{z_1z_2} \) with \( \overline{z_1}\overline{z_2} \) (they are equal).
- Not simplifying surds in modulus answers.
- Forgetting that \( |z|^2 = z\overline{z} \).
📖 Exercise 1.3 – Solved Problems (Modulus & Conjugate Properties)
1
Find the modulus of the following complex numbers:
(i) \( 4 + 3\iota \)
\[ \begin{aligned} |4+3\iota| &= \sqrt{(4)^{2}+(3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \end{aligned} \]
\[ \begin{aligned} |4+3\iota| &= \sqrt{(4)^{2}+(3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \end{aligned} \]
(ii) \( -5 - 4\iota \)
\[ \begin{aligned} |-5-4\iota| &= \sqrt{(-5)^{2}+(-4)^{2}} \\ &= \sqrt{25+16} \\ &= \sqrt{41} \end{aligned} \]
\[ \begin{aligned} |-5-4\iota| &= \sqrt{(-5)^{2}+(-4)^{2}} \\ &= \sqrt{25+16} \\ &= \sqrt{41} \end{aligned} \]
(iii) \( \frac{3}{5} - \frac{4}{5}\iota \)
\[ \begin{aligned} \left|\frac{3}{5}-\frac{4}{5}\iota\right| &= \sqrt{\left(\frac{3}{5}\right)^{2}+\left(-\frac{4}{5}\right)^{2}} \\ &= \sqrt{\frac{9}{25}+\frac{16}{25}} \\ &= \sqrt{\frac{9+16}{25}} \\ &= \sqrt{\frac{25}{25}} \\ &= \sqrt{1} \\ &= 1 \end{aligned} \]
\[ \begin{aligned} \left|\frac{3}{5}-\frac{4}{5}\iota\right| &= \sqrt{\left(\frac{3}{5}\right)^{2}+\left(-\frac{4}{5}\right)^{2}} \\ &= \sqrt{\frac{9}{25}+\frac{16}{25}} \\ &= \sqrt{\frac{9+16}{25}} \\ &= \sqrt{\frac{25}{25}} \\ &= \sqrt{1} \\ &= 1 \end{aligned} \]
(iv) \( -\sqrt{2} - \sqrt{3}\iota \)
\[ \begin{aligned} |-\sqrt{2}-\sqrt{3}\iota| &= \sqrt{(-\sqrt{2})^{2}+(-\sqrt{3})^{2}} \\ &= \sqrt{2+3} \\ &= \sqrt{5} \end{aligned} \]
\[ \begin{aligned} |-\sqrt{2}-\sqrt{3}\iota| &= \sqrt{(-\sqrt{2})^{2}+(-\sqrt{3})^{2}} \\ &= \sqrt{2+3} \\ &= \sqrt{5} \end{aligned} \]
2
If \( z_{1}=2+7\iota \) and \( z_{2}=4-3\iota \), verify the following:
(i) \( \overline{z_{1}+z_{2}} = \overline{z_{1}}+\overline{z_{2}} \)
L.H.S
\[ \begin{aligned} z_{1}+z_{2} &= (2+7\iota)+(4-3\iota) \\ &= 2+7\iota+4-3\iota \\ &= 2+4+7\iota-3\iota \\ &= 6+4\iota \end{aligned} \] \[ \text{L.H.S} = \overline{z_{1}+z_{2}} = \overline{6+4\iota} = 6-4\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)
\[ \begin{aligned} \text{R.H.S} = \overline{z_{1}}+\overline{z_{2}} &= (2-7\iota)+(4+3\iota) \\ &= 2-7\iota+4+3\iota \\ &= 2+4-7\iota+3\iota \\ &= 6-4\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)
L.H.S
\[ \begin{aligned} z_{1}+z_{2} &= (2+7\iota)+(4-3\iota) \\ &= 2+7\iota+4-3\iota \\ &= 2+4+7\iota-3\iota \\ &= 6+4\iota \end{aligned} \] \[ \text{L.H.S} = \overline{z_{1}+z_{2}} = \overline{6+4\iota} = 6-4\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)
\[ \begin{aligned} \text{R.H.S} = \overline{z_{1}}+\overline{z_{2}} &= (2-7\iota)+(4+3\iota) \\ &= 2-7\iota+4+3\iota \\ &= 2+4-7\iota+3\iota \\ &= 6-4\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)
(ii) \( \overline{z_{1}z_{2}} = \overline{z_{1}}\overline{z_{2}} \)
L.H.S
\[ \begin{aligned} z_{1}z_{2} &= (2+7\iota)(4-3\iota) \\ &= 2(4-3\iota)+7\iota(4-3\iota) \\ &= 8-6\iota+28\iota-21\iota^{2} \\ &= 8+22\iota-21(-1) \\ &= 8+22\iota+21 \\ &= 29+22\iota \end{aligned} \] \[ \text{L.H.S} = \overline{z_{1}z_{2}} = \overline{29+22\iota} = 29-22\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)
\[ \begin{aligned} \text{R.H.S} = \overline{z_{1}}\overline{z_{2}} &= (2-7\iota)(4+3\iota) \\ &= 2(4+3\iota)-7\iota(4+3\iota) \\ &= 8+6\iota-28\iota-21\iota^{2} \\ &= 8-22\iota-21(-1) \\ &= 8-22\iota+21 \\ &= 29-22\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)
L.H.S
\[ \begin{aligned} z_{1}z_{2} &= (2+7\iota)(4-3\iota) \\ &= 2(4-3\iota)+7\iota(4-3\iota) \\ &= 8-6\iota+28\iota-21\iota^{2} \\ &= 8+22\iota-21(-1) \\ &= 8+22\iota+21 \\ &= 29+22\iota \end{aligned} \] \[ \text{L.H.S} = \overline{z_{1}z_{2}} = \overline{29+22\iota} = 29-22\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)
\[ \begin{aligned} \text{R.H.S} = \overline{z_{1}}\overline{z_{2}} &= (2-7\iota)(4+3\iota) \\ &= 2(4+3\iota)-7\iota(4+3\iota) \\ &= 8+6\iota-28\iota-21\iota^{2} \\ &= 8-22\iota-21(-1) \\ &= 8-22\iota+21 \\ &= 29-22\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)
(iii) \( \overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{\overline{z_{1}}}{\overline{z_{2}}} \)
L.H.S
\[ \begin{aligned} \frac{z_{1}}{z_{2}} &= \frac{2+7\iota}{4-3\iota} \\ &= \frac{2+7\iota}{4-3\iota} \times \frac{4+3\iota}{4+3\iota} \\ &= \frac{2(4+3\iota)+7\iota(4+3\iota)}{(4)^{2}-(3\iota)^{2}} \\ &= \frac{8+6\iota+28\iota+21\iota^{2}}{16-9\iota^{2}} \\ &= \frac{8+34\iota+21(-1)}{16-9(-1)} \\ &= \frac{8+34\iota-21}{16+9} \\ &= \frac{-13+34\iota}{25} \\ &= -\frac{13}{25}+\frac{34}{25}\iota \end{aligned} \] \[ \text{L.H.S} = \overline{\left(\frac{z_{1}}{z_{2}}\right)} = \overline{-\frac{13}{25}+\frac{34}{25}\iota} = -\frac{13}{25}-\frac{34}{25}\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)
\[ \begin{aligned} \text{R.H.S} = \frac{\overline{z_{1}}}{\overline{z_{2}}} &= \frac{2-7\iota}{4+3\iota} \\ &= \frac{2-7\iota}{4+3\iota} \times \frac{4-3\iota}{4-3\iota} \\ &= \frac{2(4-3\iota)-7\iota(4-3\iota)}{(4)^{2}-(3\iota)^{2}} \\ &= \frac{8-6\iota-28\iota+21\iota^{2}}{16-9\iota^{2}} \\ &= \frac{8-34\iota+21(-1)}{16-9(-1)} \\ &= \frac{8-34\iota-21}{16+9} \\ &= \frac{-13-34\iota}{25} \\ &= -\frac{13}{25}-\frac{34}{25}\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)
L.H.S
\[ \begin{aligned} \frac{z_{1}}{z_{2}} &= \frac{2+7\iota}{4-3\iota} \\ &= \frac{2+7\iota}{4-3\iota} \times \frac{4+3\iota}{4+3\iota} \\ &= \frac{2(4+3\iota)+7\iota(4+3\iota)}{(4)^{2}-(3\iota)^{2}} \\ &= \frac{8+6\iota+28\iota+21\iota^{2}}{16-9\iota^{2}} \\ &= \frac{8+34\iota+21(-1)}{16-9(-1)} \\ &= \frac{8+34\iota-21}{16+9} \\ &= \frac{-13+34\iota}{25} \\ &= -\frac{13}{25}+\frac{34}{25}\iota \end{aligned} \] \[ \text{L.H.S} = \overline{\left(\frac{z_{1}}{z_{2}}\right)} = \overline{-\frac{13}{25}+\frac{34}{25}\iota} = -\frac{13}{25}-\frac{34}{25}\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)
\[ \begin{aligned} \text{R.H.S} = \frac{\overline{z_{1}}}{\overline{z_{2}}} &= \frac{2-7\iota}{4+3\iota} \\ &= \frac{2-7\iota}{4+3\iota} \times \frac{4-3\iota}{4-3\iota} \\ &= \frac{2(4-3\iota)-7\iota(4-3\iota)}{(4)^{2}-(3\iota)^{2}} \\ &= \frac{8-6\iota-28\iota+21\iota^{2}}{16-9\iota^{2}} \\ &= \frac{8-34\iota+21(-1)}{16-9(-1)} \\ &= \frac{8-34\iota-21}{16+9} \\ &= \frac{-13-34\iota}{25} \\ &= -\frac{13}{25}-\frac{34}{25}\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)
3
If \( z=5-2\iota \), verify the following:
(i) \( \overline{\overline{z}} = z \)
\[ \begin{aligned} z &= 5-2\iota \\ \overline{z} &= 5+2\iota \\ \overline{\overline{z}} &= 5-2\iota \end{aligned} \] Since \( 5-2\iota = z \), hence proved \( \overline{\overline{z}} = z \).
\[ \begin{aligned} z &= 5-2\iota \\ \overline{z} &= 5+2\iota \\ \overline{\overline{z}} &= 5-2\iota \end{aligned} \] Since \( 5-2\iota = z \), hence proved \( \overline{\overline{z}} = z \).
(ii) \( |z| = |\overline{z}| \)
\[ \begin{aligned} z = 5-2\iota \implies |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ \begin{aligned} \overline{z} = 5+2\iota \implies |\overline{z}| &= \sqrt{(5)^{2}+(2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] Hence proved \( |z| = |\overline{z}| \).
\[ \begin{aligned} z = 5-2\iota \implies |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ \begin{aligned} \overline{z} = 5+2\iota \implies |\overline{z}| &= \sqrt{(5)^{2}+(2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] Hence proved \( |z| = |\overline{z}| \).
(iii) \( |z| = |-\overline{z}| \)
\[ \begin{aligned} z = 5-2\iota \implies |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ \begin{aligned} \overline{z} = 5+2\iota \implies -\overline{z} &= -5-2\iota \\ |-\overline{z}| &= \sqrt{(-5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] Hence proved \( |z| = |-\overline{z}| \).
\[ \begin{aligned} z = 5-2\iota \implies |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ \begin{aligned} \overline{z} = 5+2\iota \implies -\overline{z} &= -5-2\iota \\ |-\overline{z}| &= \sqrt{(-5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] Hence proved \( |z| = |-\overline{z}| \).
(iv) \( z\overline{z} = |z|^{2} \)
L.H.S
\[ \begin{aligned} z\overline{z} &= (5-2\iota)(5+2\iota) \\ &= (5)^{2}-(2\iota)^{2} \\ &= 25-4\iota^{2} \\ &= 25-4(-1) \\ &= 25+4 \\ &= 29 \end{aligned} \] R.H.S
\[ \begin{aligned} |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ |z|^{2} = (\sqrt{29})^{2} = 29 \] Hence proved \( z\overline{z} = |z|^{2} \).
L.H.S
\[ \begin{aligned} z\overline{z} &= (5-2\iota)(5+2\iota) \\ &= (5)^{2}-(2\iota)^{2} \\ &= 25-4\iota^{2} \\ &= 25-4(-1) \\ &= 25+4 \\ &= 29 \end{aligned} \] R.H.S
\[ \begin{aligned} |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ |z|^{2} = (\sqrt{29})^{2} = 29 \] Hence proved \( z\overline{z} = |z|^{2} \).
4
If \( z=4-3\iota \), verify that \( |z|=|-z|=|\overline{\overline{z}}|=|-\overline{z}| \)
\[ \begin{aligned}
z = 4-3\iota \implies |z| &= \sqrt{(4)^{2}+(-3)^{2}} \\
&= \sqrt{16+9} \\
&= \sqrt{25} \\
&= 5 \quad \dots \text{(i)}
\end{aligned} \]
\[ \begin{aligned} -z = -4+3\iota \implies |-z| &= \sqrt{(-4)^{2}+(3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(ii)} \end{aligned} \]
\[ \begin{aligned} \overline{z} = 4+3\iota \implies \overline{\overline{z}} &= 4-3\iota \\ |\overline{\overline{z}}| &= \sqrt{(4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(iii)} \end{aligned} \]
\[ \begin{aligned} -\overline{z} = -4-3\iota \implies |-\overline{z}| &= \sqrt{(-4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(iv)} \end{aligned} \]
From equation (i), (ii), (iii) and (iv):
\[ |z|=|-z|=|\overline{\overline{z}}|=|-\overline{z}| \]
\[ \begin{aligned} -z = -4+3\iota \implies |-z| &= \sqrt{(-4)^{2}+(3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(ii)} \end{aligned} \]
\[ \begin{aligned} \overline{z} = 4+3\iota \implies \overline{\overline{z}} &= 4-3\iota \\ |\overline{\overline{z}}| &= \sqrt{(4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(iii)} \end{aligned} \]
\[ \begin{aligned} -\overline{z} = -4-3\iota \implies |-\overline{z}| &= \sqrt{(-4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(iv)} \end{aligned} \]
From equation (i), (ii), (iii) and (iv):
\[ |z|=|-z|=|\overline{\overline{z}}|=|-\overline{z}| \]
5
If \( z_{1}=2+3\iota \) and \( z_{2}=-1+\iota \), verify the following:
First, find \( z_{1}z_{2} \):
\[ \begin{aligned} z_{1}z_{2} &= (2+3\iota)(-1+\iota) \\ &= 2(-1+\iota)+3\iota(-1+\iota) \\ &= -2+2\iota-3\iota+3\iota^{2} \\ &= -2-\iota+3(-1) \\ &= -2-\iota-3 \\ &= -5-\iota \end{aligned} \]
Now, evaluate parts (i) and (ii):
(i) \( \text{Re}(z_{1}z_{2}) = -5 \)
(ii) \( \text{Im}(z_{1}z_{2}) = -1 \)
\[ \begin{aligned} z_{1}z_{2} &= (2+3\iota)(-1+\iota) \\ &= 2(-1+\iota)+3\iota(-1+\iota) \\ &= -2+2\iota-3\iota+3\iota^{2} \\ &= -2-\iota+3(-1) \\ &= -2-\iota-3 \\ &= -5-\iota \end{aligned} \]
Now, evaluate parts (i) and (ii):
(i) \( \text{Re}(z_{1}z_{2}) = -5 \)
(ii) \( \text{Im}(z_{1}z_{2}) = -1 \)
📈 Key Concepts & Quick Revision
📐 Modulus and Conjugates
- Modulus: \( |a+b\iota| = \sqrt{a^2 + b^2} \)
- Conjugate: \( \overline{a+b\iota} = a - b\iota \)
- Double Conjugate: \( \overline{\overline{z}} = z \)
- Modulus Equality: \( |z| = |\overline{z}| = |-z| = |-\overline{z}| \)
- Fundamental Identity: \( z\overline{z} = |z|^2 \)
📐 Conjugate Properties
- Sum: \( \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \)
- Product: \( \overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2} \)
- Division: \( \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}} \)
📝 Important MCQs for Practice
1. The modulus of \( 3-4\iota \) is:
2. The conjugate of \( 2+5\iota \) is:
3. \( z\overline{z} \) is equal to:
🏆 Board Exam Tips & Strategy
- Always show the formula \( |a+b\iota| = \sqrt{a^2+b^2} \) when finding modulus.
- For conjugate properties, clearly show L.H.S and R.H.S separately.
- Remember that \( \overline{\overline{z}} = z \) is a useful check.
- Practice all parts of Q2 — they appear frequently in board exams.
❓ Frequently Asked Questions
What is taught in Exercise 1.3 of Class 10 Math Unit 1?
Exercise 1.3 covers modulus of complex numbers (\( |z| \)), conjugate of complex numbers (\( \overline{z} \)), and verification of properties like \( \overline{z_1+z_2} = \overline{z_1}+\overline{z_2} \), \( \overline{z_1z_2} = \overline{z_1}\overline{z_2} \), and \( z\overline{z} = |z|^2 \).
How many questions are there in Unit 1 Exercise 1.3?
Exercise 1.3 has 5 main questions covering finding modulus (4 parts), verification of conjugate properties (3 parts), verification of modulus properties (4 parts), proving equality of moduli, and finding real and imaginary parts.
What is the modulus of a complex number?
The modulus of a complex number \( z = a + b\iota \) is defined as \( |z| = \sqrt{a^2 + b^2} \). It represents the distance of the point \( (a,b) \) from the origin in the complex plane.
What is the conjugate of a complex number?
The conjugate of a complex number \( z = a + b\iota \) is \( \overline{z} = a - b\iota \). It is obtained by changing the sign of the imaginary part.
What is \( z\overline{z} \) equal to?
For any complex number \( z \), \( z\overline{z} = |z|^2 \). This is a fundamental property that relates a complex number, its conjugate, and its modulus.
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 1.3 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 1.3?
Yes, a complete solved PDF for Exercise 1.3 is embedded on this page and available to download for free.
Can I download the Unit 1 Exercise 1.3 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 1.3 notes to your device.
Is Exercise 1.3 important for Class 10 board exams?
Yes, modulus and conjugate properties are frequently tested in board exams. Exercise 1.3 builds essential skills in understanding the geometric interpretation of complex numbers and their algebraic properties.
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.