Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 Review Exercise Contents: MCQs on properties of \( \iota \), real/imaginary parts, conjugates, modulus; short conceptual questions; simplification of powers of \( \iota \); verification of conjugate properties; solving complex simultaneous equations; finding real/imaginary parts of reciprocal; and solving equations with complex coefficients.
📚 Related Resources – Unit 1: Complex Numbers
Class 10 Math Unit 1 Review Exercise – Complete Revision
The Review Exercise of Unit 1 consolidates all concepts from Complex Numbers – powers of iota (\( \iota \)), real and imaginary parts, conjugates, modulus, and complex equations. This comprehensive review is designed to prepare students for Punjab Board exams with a mix of MCQs, short conceptual questions, and step-by-step solved problems covering every topic from Exercises 1.1 through 1.4.
What You Will Learn
This review exercise tests and reinforces all key concepts: powers of \( \iota \) and their cyclical nature, identifying real and imaginary parts, understanding conjugate properties (\( \overline{z} \)), modulus (\( |z| \)), additive and multiplicative inverses, verifying algebraic laws, and solving simultaneous linear equations with complex coefficients.
Topics Covered in This Review
- MCQs – testing fundamental concepts of complex numbers
- Conceptual Questions – understanding "0" as a complex number, \( z\overline{z} = |z|^2 \), equality condition
- Powers of \( \iota \) – simplifying \( \iota^{37} \), \( \iota^{13} \times \iota^{11} \), \( (-\iota)^{-9} \)
- Complex Arithmetic – multiplication and division of complex numbers
- Inverses – finding additive and multiplicative inverses
- Conjugate Properties – verifying \( \overline{z_1+z_2} = \overline{z_1}+\overline{z_2} \), \( \overline{z_1z_2} = \overline{z_1}\overline{z_2} \), etc.
- Complex Equations – solving simultaneous equations and finding unknown variables
Why Review Exercise Is Important
The Review Exercise is the ultimate preparation tool for board exams. It combines all the skills learned in Exercises 1.1-1.4 and presents them in a format similar to what appears in board papers. Students who master this review exercise are well-prepared for any complex numbers question in their exam.
Exam Tips for Complex Numbers
- Remember the cycle of \( \iota \): \( \iota^1 = \iota \), \( \iota^2 = -1 \), \( \iota^3 = -\iota \), \( \iota^4 = 1 \)
- Always rationalize denominators using conjugates
- For equality of complex numbers, equate real and imaginary parts separately
- \( z\overline{z} = |z|^2 \) is a fundamental identity
📖 Multiple Choice Questions (Unit 1 Review)
(i) \( \iota^2 + \iota^4 = \)
✅ Correct Answer: (B) 0
\( \iota^2 = -1 \) and \( \iota^4 = (\iota^2)^2 = (-1)^2 = 1 \). Therefore, \( \iota^2 + \iota^4 = -1 + 1 = 0 \).
\( \iota^2 = -1 \) and \( \iota^4 = (\iota^2)^2 = (-1)^2 = 1 \). Therefore, \( \iota^2 + \iota^4 = -1 + 1 = 0 \).
(ii) Real part of \( (2 - 3\iota)(2 + 3\iota) \) is:
✅ Correct Answer: (D) 13
\( (2-3\iota)(2+3\iota) = (2)^2 - (3\iota)^2 = 4 - 9\iota^2 = 4 - 9(-1) = 4 + 9 = 13 \). The real part is 13.
\( (2-3\iota)(2+3\iota) = (2)^2 - (3\iota)^2 = 4 - 9\iota^2 = 4 - 9(-1) = 4 + 9 = 13 \). The real part is 13.
(iii) Imaginary part of \( (2 - \iota)(2 + \iota) \) is:
✅ Correct Answer: (A) 0
\( (2-\iota)(2+\iota) = (2)^2 - (\iota)^2 = 4 - (-1) = 5 = 5 + 0\iota \). The imaginary part is 0.
\( (2-\iota)(2+\iota) = (2)^2 - (\iota)^2 = 4 - (-1) = 5 = 5 + 0\iota \). The imaginary part is 0.
(iv) \( x + \iota y \) will be pure imaginary number, when:
✅ Correct Answer: (B) x = 0
A pure imaginary number has real part equal to zero. So \( x = 0 \).
A pure imaginary number has real part equal to zero. So \( x = 0 \).
(v) What is additive inverse of \( 5 - 2\iota \) ?
✅ Correct Answer: (D) \(-5 + 2\iota\)
Additive inverse of \( a + b\iota \) is \( -a - b\iota \). So additive inverse of \( 5 - 2\iota \) is \( -5 + 2\iota \).
Additive inverse of \( a + b\iota \) is \( -a - b\iota \). So additive inverse of \( 5 - 2\iota \) is \( -5 + 2\iota \).
(vi) What is multiplicative inverse of \( z = 1 + \iota \) ?
✅ Correct Answer: (D) \(\frac{1}{2} - \frac{1}{2}\iota\)
\( z^{-1} = \frac{1}{1+\iota} \times \frac{1-\iota}{1-\iota} = \frac{1-\iota}{1-\iota^2} = \frac{1-\iota}{1+1} = \frac{1}{2} - \frac{1}{2}\iota \).
\( z^{-1} = \frac{1}{1+\iota} \times \frac{1-\iota}{1-\iota} = \frac{1-\iota}{1-\iota^2} = \frac{1-\iota}{1+1} = \frac{1}{2} - \frac{1}{2}\iota \).
(vii) If \( z = 4 - 3\iota \), then \( z\overline{z} = \)
✅ Correct Answer: (D) 25
\( z\overline{z} = |z|^2 = (4)^2 + (-3)^2 = 16 + 9 = 25 \).
\( z\overline{z} = |z|^2 = (4)^2 + (-3)^2 = 16 + 9 = 25 \).
(viii) Conjugate of \( 9 - 4\iota \) is:
✅ Correct Answer: (B) \(9 + 4\iota\)
Conjugate of \( a + b\iota \) is \( a - b\iota \). So conjugate of \( 9 - 4\iota \) is \( 9 + 4\iota \).
Conjugate of \( a + b\iota \) is \( a - b\iota \). So conjugate of \( 9 - 4\iota \) is \( 9 + 4\iota \).
(ix) If \( z = 4 + 4\iota \), then \( z + \overline{z} = \)
✅ Correct Answer: (A) 8
\( z + \overline{z} = (4+4\iota) + (4-4\iota) = 8 \).
\( z + \overline{z} = (4+4\iota) + (4-4\iota) = 8 \).
(x) If \( z = 5 + 4\iota \), then \( |z| = \)
✅ Correct Answer: (D) \(\sqrt{41}\)
\( |z| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \).
\( |z| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \).
2 Conceptual Questions
(i) Is "0" a complex number? Explain.
Yes, \(0\) is a complex number because it can be written as \(0 + 0\iota\) with real part \(0\) and imaginary part \(0\).
Yes, \(0\) is a complex number because it can be written as \(0 + 0\iota\) with real part \(0\) and imaginary part \(0\).
(ii) What is the result of multiplying a complex number by its conjugate?
\[ z\overline{z} = |z|^2 \] which is a real non‑negative number.
\[ z\overline{z} = |z|^2 \] which is a real non‑negative number.
(iii) State the condition for two complex numbers to be equal.
\(a + b\iota = c + d\iota \iff a = c\) and \(b = d\).
\(a + b\iota = c + d\iota \iff a = c\) and \(b = d\).
3 Simplify the following powers & expressions
(i) \(\iota^{37}\)
\[ \begin{aligned}
\iota^{37} &= \iota^{36}\cdot\iota = (\iota^2)^{18}\cdot\iota = (-1)^{18}\iota = \iota
\end{aligned} \]
(ii) \(\iota^{13}\times\iota^{11}\)
\[ \begin{aligned}
\iota^{13+11} = \iota^{24} = (\iota^2)^{12} = (-1)^{12}=1
\end{aligned} \]
(iii) \((-\iota)^{-9}\)
\[ \begin{aligned}
(-\iota)^{-9} &= \frac{1}{(-\iota)^9} = \frac{1}{(-1)^9\iota^9} = \frac{1}{-\iota^8\iota} = \frac{1}{-1\cdot\iota} = -\frac{1}{\iota} = \iota
\end{aligned} \]
(iv) \((3-4\iota)(5-6\iota)\)
\[ \begin{aligned}
&=15-18\iota-20\iota+24\iota^2 =15-38\iota-24 = -9-38\iota
\end{aligned} \]
(v) \((3+4\iota)\div(5-7\iota)\)
\[ \begin{aligned}
&=\frac{3+4\iota}{5-7\iota}\times\frac{5+7\iota}{5+7\iota} = \frac{15+21\iota+20\iota+28\iota^2}{25-49\iota^2} = \frac{15+41\iota-28}{25+49} = \frac{-13+41\iota}{74} = -\frac{13}{74}+\frac{41}{74}\iota
\end{aligned} \]
4 Additive and multiplicative inverse of \(z = 8 + 9\iota\)
\[ \begin{aligned}
\text{Additive inverse: } -z &= -8 - 9\iota \\
\text{Multiplicative inverse: } z^{-1} &= \frac{1}{8+9\iota} = \frac{8-9\iota}{64-81\iota^2} = \frac{8-9\iota}{64+81} = \frac{8-9\iota}{145} = \frac{8}{145} - \frac{9}{145}\iota
\end{aligned} \]
5 Verify properties with \(z_1=3+4\iota,\; z_2=2+3\iota\)
(i) \(\overline{z_1+z_2} = \overline{z_1}+\overline{z_2}\)
LHS: \(z_1+z_2 = 5+7\iota \implies \overline{5+7\iota}=5-7\iota\)
RHS: \((3-4\iota)+(2-3\iota)=5-7\iota\). ✓
LHS: \(z_1+z_2 = 5+7\iota \implies \overline{5+7\iota}=5-7\iota\)
RHS: \((3-4\iota)+(2-3\iota)=5-7\iota\). ✓
(ii) \(\overline{z_1z_2} = \overline{z_1}\cdot\overline{z_2}\)
LHS: \(z_1z_2=(3+4\iota)(2+3\iota)=6+9\iota+8\iota+12\iota^2 = -6+17\iota\); conjugate \(-6-17\iota\)
RHS: \((3-4\iota)(2-3\iota)=6-9\iota-8\iota+12\iota^2 = -6-17\iota\). ✓
LHS: \(z_1z_2=(3+4\iota)(2+3\iota)=6+9\iota+8\iota+12\iota^2 = -6+17\iota\); conjugate \(-6-17\iota\)
RHS: \((3-4\iota)(2-3\iota)=6-9\iota-8\iota+12\iota^2 = -6-17\iota\). ✓
(iii) \(\overline{(z_1/z_2)} = \overline{z_1}/\overline{z_2}\)
LHS: \(z_1/z_2 = \frac{3+4\iota}{2+3\iota} = \frac{18-\iota}{13}\), conjugate \(\frac{18+\iota}{13}\)
RHS: \(\frac{3-4\iota}{2-3\iota} = \frac{18+\iota}{13}\). ✓
LHS: \(z_1/z_2 = \frac{3+4\iota}{2+3\iota} = \frac{18-\iota}{13}\), conjugate \(\frac{18+\iota}{13}\)
RHS: \(\frac{3-4\iota}{2-3\iota} = \frac{18+\iota}{13}\). ✓
(iv) \(|z_1| = |-\overline{z_1}|\)
\(|z_1| = \sqrt{3^2+4^2}=5\), \(|-\overline{z_1}| = |-3+4\iota| = \sqrt{9+16}=5\). ✓
\(|z_1| = \sqrt{3^2+4^2}=5\), \(|-\overline{z_1}| = |-3+4\iota| = \sqrt{9+16}=5\). ✓
(v) \(\overline{\overline{z_2}} = z_2\)
\(\overline{z_2}=2-3\iota\), conjugate again \(2+3\iota = z_2\). ✓
\(\overline{z_2}=2-3\iota\), conjugate again \(2+3\iota = z_2\). ✓
(vi) \(z_1\overline{z_1}=|z_1|^2\)
LHS: \((3+4\iota)(3-4\iota)=25\), RHS: \(|z_1|^2=25\). ✓
LHS: \((3+4\iota)(3-4\iota)=25\), RHS: \(|z_1|^2=25\). ✓
6 If \(z_1=5+4\iota,\; z_2=3+2\iota\), find:
(i) \(z_1z_2 = (5+4\iota)(3+2\iota)=15+10\iota+12\iota+8\iota^2 = 7+22\iota\)
(ii) \(\frac{z_1}{z_2} = \frac{5+4\iota}{3+2\iota} = \frac{(5+4\iota)(3-2\iota)}{9-4\iota^2} = \frac{15-10\iota+12\iota-8\iota^2}{13} = \frac{23+2\iota}{13} = \frac{23}{13}+\frac{2}{13}\iota\)
(iii) \(\overline{z_1z_2} = \overline{7+22\iota} = 7-22\iota\)
(iv) \(|z_1z_2| = |7+22\iota| = \sqrt{7^2+22^2} = \sqrt{49+484} = \sqrt{533}\)
7 Real & imaginary parts of \((2+7\iota)^{-1}\)
\[ \begin{aligned}
z = \frac{1}{2+7\iota} &= \frac{2-7\iota}{4 - 49\iota^2} = \frac{2-7\iota}{4+49} = \frac{2-7\iota}{53} \\
\operatorname{Re}(z)=\frac{2}{53},&\quad \operatorname{Im}(z)=-\frac{7}{53}
\end{aligned} \]
8 Solve \(\iota z+(2-\iota)w=4+\iota,\; \iota z+(3+\iota)w=3+3\iota\)
\[ \begin{aligned}
\text{Subtract: } & (\iota z+2w-\iota w) - (\iota z+3w+\iota w) = (4+\iota)-(3+3\iota) \\
& -w -2\iota w = 1-2\iota \;\Rightarrow\; w(-1-2\iota)=1-2\iota \\
w &= \frac{1-2\iota}{-1-2\iota} \times \frac{-1+2\iota}{-1+2\iota} = \frac{-1+2\iota+2\iota-4\iota^2}{1-4\iota^2} = \frac{3+4\iota}{5} = \frac{3}{5}+\frac{4}{5}\iota \\
\text{From first: } &\iota z + 2w -\iota w = 4+\iota \;\Rightarrow\; \iota z = 4+\iota -2w+\iota w \\
&\iota z = 4+\iota -2\left(\frac{3+4\iota}{5}\right)+\iota\left(\frac{3+4\iota}{5}\right) = 2 \;\Rightarrow\; z = \frac{2}{\iota} = -2\iota
\end{aligned} \]
Thus \(z = -2\iota,\; w = \dfrac{3}{5}+\dfrac{4}{5}\iota\).
9 Solve \((3-4\iota)(a+b\iota)=1\)
\[ \begin{aligned}
a+b\iota &= \frac{1}{3-4\iota} = \frac{3+4\iota}{9-16\iota^2} = \frac{3+4\iota}{25} = \frac{3}{25}+\frac{4}{25}\iota \\
\Rightarrow a=\frac{3}{25},\; b=\frac{4}{25}
\end{aligned} \]
10 Solve \((3-2\iota)(x+y\iota)=2(x-2y\iota)+2\iota-1\)
\[ \begin{aligned}
&(3x+2y) + \iota(3y-2x) = (2x-1) + \iota(-4y+2) \\
&\text{Real: } 3x+2y = 2x-1 \;\Rightarrow\; x+2y=-1 \\
&\text{Imag: } 3y-2x = -4y+2 \;\Rightarrow\; -2x+7y=2 \\
&\text{Solve: } 2\times (i): 2x+4y=-2 \;\text{add to (ii)} \Rightarrow 11y=0 \Rightarrow y=0,\; x=-1
\end{aligned} \]
Thus \(x=-1,\; y=0\).
📈 Key Formulas – Complex Numbers Review
- Powers of \(\iota\): \(\iota^2=-1,\ \iota^3=-\iota,\ \iota^4=1\)
- Conjugate: \(\overline{a+b\iota}=a-b\iota\)
- Modulus: \(|z|=\sqrt{a^2+b^2},\ z\bar{z}=|z|^2\)
- Additive inverse: \(-z\), Multiplicative inverse: \(\frac{1}{z}=\frac{\bar{z}}{|z|^2}\)
- Equality: \(a+b\iota = c+d\iota \iff a=c\) and \(b=d\)
❓ Frequently Asked Questions
What is covered in Unit 1 Review Exercise?
The Review Exercise covers all concepts from Unit 1 including MCQs on powers of \( \iota \), real and imaginary parts, conjugates, modulus; short conceptual questions; simplification of powers; verification of conjugate properties; solving complex simultaneous equations; finding real and imaginary parts of reciprocals; and solving equations with complex coefficients.
Is this review exercise important for board exams?
Yes, the review exercise consolidates all key concepts from Unit 1 and is an excellent preparation tool for Punjab Board exams. The MCQs and short questions are representative of what appears in board papers.
Who prepared these review solutions?
These solutions were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.