📘 Unit 1: Complex Numbers – Review Exercise (Complete Solutions)
Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 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
📘 Multiple Choice Questions (Unit 1 Review)
| Q.No | Question | A | B | C | D |
|---|---|---|---|---|---|
| (i) | \(\iota^2 + \iota^4 =\) | \(-1\) | \(0\) | \(1\) | \(2\) |
| (ii) | Real part of \((2 - 3\iota)(2 + 3\iota)\) is: | \(-3\) | \(1\) | \(4\) | \(13\) |
| (iii) | Imaginary part of \((2 - \iota)(2 + \iota)\) is: | \(0\) | \(1\) | \(7\) | \(9\) |
| (iv) | \(x + \iota y\) will be pure imaginary number, when: | \(y = 0\) | \(x = 0\) | \(\iota = 0\) | \(x = 0, y = 0\) |
| (v) | Additive inverse of \(5 - 2\iota\) is: | \(5 + 2\iota\) | \(-5 - 2\iota\) | \(5 - 2\iota\) | \(-5 + 2\iota\) |
| (vi) | Multiplicative inverse of \(z = 1 + \iota\) is: | \(1 - \iota\) | \(\iota\) | \(\frac12 - 2\iota\) | \(\frac12 - \frac12\iota\) |
| (vii) | If \(z = 4 - 3\iota\), then \(z\overline{z} =\) | \(3\) | \(9\) | \(16\) | \(25\) |
| (viii) | Conjugate of \(9 - 4\iota\) is: | \(-9 - 4\iota\) | \(9 + 4\iota\) | \(9 + 9\iota\) | \(4 - 9\iota\) |
| (ix) | If \(z = 4 + 4\iota\), then \(z + \overline{z} =\) | \(8\) | \(8 + 8\iota\) | \(8\iota\) | \(0\) |
| (x) | If \(z = 5 + 4\iota\), then \(|z| =\) | \(9\) | \(25\) | \(41\) | \(\sqrt{41}\) |
✅ Answer Key
(i) B (0)
(ii) D (13)
(iii) A (0)
(iv) B (x = 0)
(v) D (\(-5 + 2\iota\))
(vi) D (\(\frac{1}{2} - \frac{1}{2}\iota\))
(vii) D (25)
(viii) B (\(9 + 4\iota\))
(ix) A (8)
(x) D (\(\sqrt{41}\))
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) Multiplying a complex number by its conjugate gives:
\[ z\overline{z} = |z|^2 \] which is a real non‑negative number.
\[ z\overline{z} = |z|^2 \] which is a real non‑negative number.
(iii) Condition for equality: \(a + b\iota = c + d\iota \iff a = c\) 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 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 \(\overline{5+7\iota}=5-7\iota\), RHS \((3-4\iota)+(2-3\iota)=5-7\iota\). ✓
(ii) \(\overline{z_1z_2} = \overline{z_1}\cdot\overline{z_2}\): \(z_1z_2=(3+4\iota)(2+3\iota)=6+9\iota+8\iota+12\iota^2 = -6+17\iota\); conjugate \(-6-17\iota\). RHS \((3-4\iota)(2-3\iota)=6-9\iota-8\iota+12\iota^2 = -6-17\iota\). ✓
(iii) \(\overline{(z_1/z_2)} = \overline{z_1}/\overline{z_2}\): \(z_1/z_2 = \frac{3+4\iota}{2+3\iota} = \frac{18-\iota}{13}\), conjugate \(\frac{18+\iota}{13}\). RHS \(\frac{3-4\iota}{2-3\iota} = \frac{18+\iota}{13}\). ✓
(iv) \(|z_1| = |-\overline{z_1}|\): \(|z_1|=5\), \(|-\overline{z_1}|=| -3+4\iota| = \sqrt{9+16}=5\). ✓
(v) \(\overline{\overline{z_2}} = z_2\): \(\overline{z_2}=2-3\iota\), conjugate again \(2+3\iota = z_2\). ✓
(vi) \(z_1\overline{z_1}=|z_1|^2\): LHS \((3+4\iota)(3-4\iota)=25\), \(|z_1|^2=25\). ✓
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: \(\frac{1}{z}=\frac{\bar{z}}{|z|^2}\)
Unit 1✨ Complex Numbers (Current Chapter)
Unit 2Quadratic Equations and Inequalities
Unit 3Matrices and Determinants
Unit 4Functions and Graphs
Unit 5Algebraic Fractions
Unit 6Vectors in Plane
Unit 7Trigonometry
Unit 8Chords and Arcs of a Circle
Unit 9Tangent and Angles of a Circle
Unit 10Practical Geometry of Circles
Unit 11Information Handling
Unit 12Probability
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus