case study questions class 11 maths complex numbers

CBSE Case Study Questions for Class 11 Maths Complex Numbers Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Complex Numbers in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Complex Numbers PDF

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Case Study Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

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[PDF] Download Case Study Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Here we are providing case study questions for class 11 maths. In this article, we are sharing Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations case study questions. All case study questions of class 11 maths are solved so that students can check their solutions after attempting questions.

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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Unit 4: Complex numbers and quadratic equations

What are the imaginary numbers.

Intro to the imaginary numbers (Opens a modal)
Powers of the imaginary unit (Opens a modal)
Simplifying roots of negative numbers (Opens a modal)
i as the principal root of -1 (Opens a modal)
Powers of the imaginary unit Get 3 of 4 questions to level up!
Simplify roots of negative numbers Get 3 of 4 questions to level up!

What are the complex numbers?

Intro to complex numbers (Opens a modal)
Classifying complex numbers (Opens a modal)
Plotting numbers on the complex plane (Opens a modal)
The complex plane (Opens a modal)
Parts of complex numbers Get 3 of 4 questions to level up!
Classify complex numbers Get 3 of 4 questions to level up!
Plot numbers on the complex plane Get 3 of 4 questions to level up!

Adding and subtracting complex numbers

Adding complex numbers (Opens a modal)
Subtracting complex numbers (Opens a modal)
Add & subtract complex numbers Get 3 of 4 questions to level up!
Graphically add & subtract complex numbers Get 3 of 4 questions to level up!

Multiplying complex numbers

Multiplying complex numbers (Opens a modal)
Complex number operations review (Opens a modal)
Multiply complex numbers (basic) Get 3 of 4 questions to level up!
Multiply complex numbers Get 3 of 4 questions to level up!

Dividing complex numbers

Intro to complex number conjugates (Opens a modal)
Complex number conjugates (Opens a modal)
Dividing complex numbers (Opens a modal)
Dividing complex numbers review (Opens a modal)
Complex number conjugates Get 3 of 4 questions to level up!
Divide complex numbers Get 3 of 4 questions to level up!

Absolute value and angle of complex numbers

Absolute value of complex numbers (Opens a modal)
Absolute value & angle of complex numbers (Opens a modal)
Complex number absolute value & angle review (Opens a modal)
Modulus (absolute value) of complex numbers Get 3 of 4 questions to level up!
Angle of complex numbers Get 3 of 4 questions to level up!

Polar form of complex numbers

Polar & rectangular forms of complex numbers (Opens a modal)
Complex number forms review (Opens a modal)
Complex numbers from absolute value & angle Get 3 of 4 questions to level up!
Polar & rectangular forms of complex numbers Get 3 of 4 questions to level up!

CBSE Class 11 Maths – Chapter 5 Complex Numbers and Quadratic Equations- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

Concepts of Complex Numbers and Quadratic Equations
Complex Numbers and Quadratic Equations Master File
Complex Numbers and Quadratic Equations Revision Notes
R D Sharma Solution of Complex Numbers
R D Sharma Solution of Quadratic Equations
NCERT Solution Complex Numbers and Quadratic Equations
NCERT Exemplar Solution Complex Numbers and Quadratic Equations
Complex Numbers and Quadratic Equations : Solved Example 1

Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers

Question 1. The value of √(-16) is (a) -4i (b) 4i (c) -2i (d) 2i

Answer: (b) 4i Hint: Given, √(-16) = √(16) × √(-1) = 4i {since i = √(-1) }

Question 2. The value of √(-144) is (a) 12i (b) -12i (c) ±12i (d) None of these

Answer: (a) 12i Hint: Given, √(-144) = √{(-1) × 144} = √(-1) × √(144) = i × 12 {Since √(-1) = i} = 12i So, √(-144) = 12i

Question 3: The value of √(-25) + 3√(-4) + 2√(-9) is (a) 13i (b) -13i (c) 17i (d) -17i

Answer: (c) 17i Hint: Given, √(-25) + 3√(-4) + 2√(-9) = √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9} = √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9} = 5i + 3×2i + 2×3i {since √(-1) = i} = 5i + 6i + 6i = 17i So, √(-25) + 3√(-4) + 2√(-9) = 17i

Question 4. if z lies on |z| = 1, then 2/z lies on (a) a circle (b) an ellipse (c) a straight line (d) a parabola

Answer: (a) a circle Hint: Let w = 2/z Now, |w| = |2/z| => |w| = 2/|z| => |w| = 2 This shows that w lies on a circle with center at the origin and radius 2 units.

Question 5. If ω is an imaginary cube root of unity, then (1 + ω – ω²) 7 equals (a) 128 ω (b) -128 ω (c) 128 ω² (d) -128 ω²

Answer: (d) -128 ω² Hint: Given ω is an imaginary cube root of unity. So 1 + ω + ω² = 0 and ω³ = 1 Now, (1 + ω – ω²) 7 = (-ω² – ω²) 7 ⇒ (1 + ω – ω 2 ) 7 = (-2ω 2 ) 7 ⇒ (1 + ω – ω 2 ) 7 = -128 ω 14 ⇒ (1 + ω – ω 2 ) 7 = -128 ω 12 × ω 2 ⇒ (1 + ω – ω 2 ) 7 = -128 (ω 3 ) 4 ω 2 ⇒ (1 + ω – ω 2 ) 7 = -128 ω 2

Question 6. The least value of n for which {(1 + i)/(1 – i)} n is real, is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b) 2 Hint: Given, {(1 + i)/(1 – i)} n = [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}] n = [{(1 + i)²}/{(1 – i²)}] n = [(1 + i² + 2i)/{1 – (-1)}] n = [(1 – 1 + 2i)/{1 + 1}] n = [2i/2] n = i n Now, in is real when n = 2 {since i2 = -1 } So, the least value of n is 2

Question 7. Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z = (a) -2√3 + 2i (b) 2√3 + 2i (c) 2√3 – 2i (d) -√3 + i

Answer: (a) -2√3 + 2i Hint: Let z = r(cos θ + i × sin θ) Then r = 4 and θ = 5π/6 So, z = 4(cos 5π/6 + i × sin 5π/6) ⇒ z = 4(-√3/2 + i/2) ⇒ z = -2√3 + 2i

Question 8: The value of i -999 is (a) 1 (b) -1 (c) i (d) -i

Answer: (c) i Hint: Given, i -999 = 1/i 999 = 1/(i 996 × i³) = 1/{(i 4 ) 249 × i 3 } = 1/{1 249 × i 3 } {since i 4 = 1} = 1/i 3 = i 4 /i 3 {since i 4 = 1} = i So, i -999 = i

Question 9. Let z 1 and z 2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z 1 and z 1 form an equilateral triangle. Then (a) a² = b (b) a² = 2b (c) a² = 3b (d) a² = 4b

Answer: (c) a² = 3b Hint: Given, z 1 and z 2 be two roots of the equation z² + az + b = 0 Now, z 1 + z 2 = -a and z 1 × z 2 = b Since z 1 and z 2 and z 3 from an equilateral triangle. ⇒ z 1 2 + z 2 2 + z 3 2 = z 1 × z 2 + z 2 × z 3 + z 1 × z 3 ⇒ z 1 2 + z 2 2 = z 1 × z 2 {since z 3 = 0} ⇒ (z 1 + z 2 )² – 2z 1 × z 2 = z 1 × z 2 ⇒ (z 1 + z 2 )² = 2z 1 × z 2 + z 1 × z 2 ⇒ (z 1 + z 2 )² = 3z 1 × z 2 ⇒ (-a)² = 3b ⇒ a² = 3b

Question 10: The complex numbers sin x + i cos 2x are conjugate to each other for (a) x = nπ (b) x = 0 (c) x = (n + 1/2) π (d) no value of x

Answer: (d) no value of x Hint: Given complex number = sin x + i cos 2x Conjugate of this number = sin x – i cos 2x Now, sin x + i cos 2x = sin x – i cos 2x ⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part} ⇒ tan x = 1 and tan 2x = 1 Now both of them are not possible for the same value of x. So, there exist no value of x

Question 11. The curve represented by Im(z²) = k, where k is a non-zero real number, is (a) a pair of striaght line (b) an ellipse (c) a parabola (d) a hyperbola

Answer: (d) a hyperbola Hint: Let z = x + iy Now, z² = (x + iy)² ⇒ z² = x² – y² + 2xy Given, Im(z²) = k ⇒ 2xy = k ⇒ xy = k/2 which is a hyperbola.

Question 12. The value of x and y if (3y – 2) + i(7 – 2x) = 0 (a) x = 7/2, y = 2/3 (b) x = 2/7, y = 2/3 (c) x = 7/2, y = 3/2 (d) x = 2/7, y = 3/2

Answer: (a) x = 7/2, y = 2/3 Hint: Given, (3y – 2) + i(7 – 2x) = 0 Compare real and imaginary part, we get 3y – 2 = 0 ⇒ y = 2/3 and 7 – 2x = 0 ⇒ x = 7/2 So, the value of x = 7/2 and y = 2/3

Question 13. Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary (a) θ = nπ ± π/2 where n is an integer (b) θ = nπ ± π/3 where n is an integer (c) θ = nπ ± π/4 where n is an integer (d) None of these

Answer: (b) θ = nπ ± π/3 where n is an integer Hint: Given, (3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ) (3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1 Now, equation 1 is imaginary if 3 – 4sin² θ = 0 ⇒ 4sin² θ = 3 ⇒ sin² θ = 3/4 ⇒ sin θ = ±√3/2 ⇒ θ = nπ ± π/3 where n is an integer

Question 14. If {(1 + i)/(1 – i)} n = 1 then the least value of n is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (d) 4 Hint: Given, {(1 + i)/(1 – i)} n = 1 ⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}] n = 1 ⇒ [{(1 + i)²}/{(1 – i²)}] n = 1 ⇒ [(1 + i² + 2i)/{1 – (-1)}] n = 1 ⇒ [(1 – 1 + 2i)/{1 + 1}] n = 1 ⇒ [2i/2] n = 1 ⇒ i n = 1 Now, i n is 1 when n = 4 So, the least value of n is 4

Question 15. If arg (z) < 0, then arg (-z) – arg (z) = (a) π (b) -π (c) -π/2 (d) π/2

Answer: (a) π Hint: Given, arg (z) < 0 Now, arg (-z) – arg (z) = arg(-z/z) ⇒ arg (-z) – arg (z) = arg(-1) ⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}

Question 16. if x + 1/x = 1 find the value of x 2000 + 1/x 2000 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Given x + 1/x = 1 ⇒ (x² + 1) = x ⇒ x² – x + 1 = 0 ⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1) ⇒ x = {1 ± √(1 – 4)}/2 ⇒ x = {1 ± √(-3)}/2 ⇒ x = {1 ± √(-1)×√3}/2 ⇒ x = {1 ± i√3}/2 {since i = √(-1)} ⇒ x = -w, -w² Now, put x = -w, we get x 2000 + 1/x 2000 = (-w) 2000 + 1/(-w) 2000 = w 2000 + 1/w 2000 = w 2000 + 1/w 2000 = {(w³) 666 × w²} + 1/{(w³) 666 × w²} = w² + 1/w² {since w³ = 1} = w² + w³ /w² = w² + w = -1 {since 1 + w + w² = 0} So, x 2000 + 1/x 2000 = -1

Question 17. The value of √(-144) is (a) 12i (b) -12i (c) ±12i (d) None of these

Answer: (a) 12i Hint: Given, √(-144) = √{(-1)×144} = √(-1) × √(144) = i × 12 {Since √(-1) = i} = 12i So, √(-144) = 12i

Question 18. If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are (a) -1, -1 + 2ω, – 1 – 2ω² (b) – 1, -1, – 1 (c) – 1, 1 – 2ω, 1 – 2ω² (d) – 1, 1 + 2ω, 1 + 2ω²

Answer: (c) – 1, 1 – 2ω, 1 – 2ω² Hint: Note that since 1, ω, and ω² are the cube roots of unity (the three cube roots of 1), they are the three solutions to x³ = 1 (note: ω and ω² are the two complex solutions to this) If we let u = x – 1, then the equation becomes u³ + 8 = (u + 2)(u² – 2u + 4) = 0. So, the solutions occur when u = -2 (giving -2 = x – 1 ⇒ x = -1), or when: u² – 2u + 4 = 0, which has roots, by the Quadratic Formula, to be u = 1 ± i√3 So, x – 1 = 1 ± i√3 ⇒ x = 2 ± i√3 Now, x³ = 1 when x³ – 1 = (x – 1)(x² + x + 1) = 0, giving x = 1 and x² + x + 1 = 0 ⇒ x = (-1 ± i√3)/2 If we let ω = (-1 – i√3)/2 and ω₂ = (-1 + i√3)/2 then 1 – 2ω and 1 – 2ω² yield the two complex solutions to (x – 1)³ + 8 = 0 So, the roots of (x – 1)³ + 8 are -1, 1 – 2ω, and 1 – 2ω²

Question 19. (1 – w + w²)×(1 – w² + w 4 )×(1 – w 4 + w 8 ) × …………… to 2n factors is equal to (a) 2 n (b) 2 2n (c) 2 3n (d) 2 4n

Answer: (b) 2 2n Hint: Given, (1 – w + w²)×(1 – w² + w 4 )×(1 – w 4 + w 8 ) × …………… to 2n factors = (1 – w + w 2 )×(1 – w 2 + w )×(1 – w + w 2 ) × …………… to 2n factors {Since w 4 = w, w 8 = w 2 } = (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors = (2² w³)×(2² w³)×(2² w³) …………… to 2n factors = (2²) n {since w³ = 1} = 2 2n

Question 20. The modulus of 5 + 4i is (a) 41 (b) -41 (c) √41 (d) -√41

Answer: (c) √41 Hint: Let Z = 5 + 4i Now modulus of Z is calculated as |Z| = √(5² + 4²) ⇒ |Z| = √(25 + 16) ⇒ |Z| = √41 So, the modulus of 5 + 4i is √41

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Complex Numbers and Quadratic Equations Class 11 Notes CBSE Maths Chapter 5 (Free PDF Download)
Revision Notes

Revision Notes for CBSE Class 11 Maths Chapter 5 (Complex Numbers and Quadratic Equations) - Free PDF Download

NCERT Class 11 revision notes maths ch 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at Vedantu. This class 11 revision notes maths ch 5 help the students in solving the problems quickly, accurately, and efficiently. Also, Vedantu provides step by step solutions for all NCERT problems, thereby ensuring students understand them and clear their exams with flying colors.

This class 11 revision notes maths ch 5 for Complex Numbers and Quadratic Equations comprises different critical mathematical theorems and formulae. The NCERT textbook has many practice problems to cover all these concepts, which would help students to easily understand higher concepts in the future. Vedantu provides solutions for all these problems by explaining all the steps with proper explanations. This particular NCERT ch 5 class 11 maths revision notes of Vedantu will help students who aim to clear their maths exams even with the last-minute preparations.

Download CBSE Class 11 Maths Revision Notes 2023-24 PDF

Also, check CBSE Class 11 Maths revision notes for all chapters:

Complex Numbers and Quadratic Equations Chapter-Related Important Study Materials It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Complex Numbers and Quadratic Equations Class 11 Notes Maths - Basic Subjective Questions

Section–a (1 mark questions).

1. If a+2i and 3+bi are conjugate of each other then find value of a and b.

Ans. Let $z=a+2 i$

$\bar{z}=a-2 i$

But we are given that $\bar{z}=3+b i$

$$ \therefore a-2 i=3+b i $$

Comparing the real and imaginary parts, we get $a=3$ and $b=-2$.

2. The conjugate of the complex number $\dfrac{1-i}{1+i}$ is ______.

Ans. $$\text { 2. } \begin{aligned} & \frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i} \\ & =\frac{1+i^2-2 i}{1-i^2} \\ & =\frac{1-1-2 i}{1+1}=-i \end{aligned} $$

Hence, conjugate of $\frac{1-i}{1+i}$ is $0+i$.

3. If z=2+3i , then value of $\left| z-1+4i \right |$ is______.

Ans. Given that: $z=2+3 i$

Then, $|z-1+4 i|=|2+3 i-1+4 i|$

$$ \begin{aligned} & =|1+7 i|=\sqrt{1^2+7^2} \\ & =\sqrt{50}=5 \sqrt{2} \end{aligned} $$

4. The sum of the series $i+i^{2}+i^{3}+\cdots$ up to 1000 terms is equal to ……….

Ans. $i+i^2+i^3+\ldots .$. upto 1000 terms

$$=i+i^2+i^3+\ldots+i^{1000}=0$$

Hence, the value of the filler is 0 .

5. Find the amplitude of $\sqrt{3}+i$ .

Ans. Let $z=\sqrt{3}+i$

Here, $z$ lies in first quadrant.

Therefore

$\tan\theta=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|=\left|\frac{1}{\sqrt{3}}\right|=\frac{1}{\sqrt{3}}$

$$ \therefore \theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \text { i.e., } \theta=\frac{\pi}{6} \text {. } $$

Section–B (2 Marks Questions)

6. Show that $(z+3)(\vec{z}+3)$ is equivalent to $\left| z+3 \right |^{2}$ .

Ans. Let $z=x+i y$

$$ \begin{aligned} & \operatorname{So}(z+3)(\bar{z}+3)=(x+i y+3)(x-i y+3) \\ & =[(x+3)+i y][(x+3)-i y] \\ & =(x+3)^2-i^2 y^2 \\ & =(x+3)^2+y^2 \\ & =|x+3+i y|^2=|z+3|^2 . \end{aligned} $$

7. If $\left ( \dfrac{1+i}{1-i} \right )^{x}=1$ , then show that $x=4n$ , where $n\epsilon N $.

Ans. Given that: $\left(\frac{1+i}{1-i}\right)^x=1$

$$ \begin{aligned} & \Rightarrow\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^x=1 \\ & \Rightarrow\left(\frac{1+i^2+2 i}{1-i^2}\right)^x=1 \\ & \Rightarrow\left(\frac{1-1+2 i}{1+1}\right)^x=1 \\ & \Rightarrow\left(\frac{2 i}{2}\right)^x=1 \\ & \Rightarrow(i)^x=(i)^{4 n} \\ & \Rightarrow x=4 n, n \in N \end{aligned} $$

8. Find the value of p such that the difference of the roots of the equation $x^{2}-px+8=0$ is 2.

Ans. Let $\alpha, \beta$ be the roots of the equation

$$ x^2-p x+8=0 $$

Therefore $\alpha+\beta=p$ and $\alpha \beta=8$

Now, $\alpha-\beta= \pm \sqrt{(\alpha+\beta)^2-4 \alpha \beta}$

Therefore, $2= \pm \sqrt{p^2-32}$

$$ \begin{aligned} & \Rightarrow p^2-32=4 \\ & \Rightarrow p= \pm 6 . \end{aligned} $$

9. If $x+iy=\dfrac{a+ib}{a-ib}$, prove that $x^{2}+y^{2}=1$.

Ans. We have,

$$ \begin{aligned} & x+i y=\frac{(a+i b)(a+i b)}{(a-i b)(a+i b)} \\ & =\frac{a^2-b^2+2 a b i}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+\frac{2 a b}{a^2+b^2} i \end{aligned} $$

So that, $x-i y=\frac{a^2-b^2}{a^2+b^2}-\frac{2 a b}{a^2+b^2} i$

$$ \begin{aligned} & x^2+y^2=(x+i y)(x-i y) \\ & =\frac{\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)^2}+\frac{4 a^2 b^2}{\left(a^2+b^2\right)^2}=\frac{\left(a^2+b^2\right)^2}{\left(a^2+b^2\right)^2}=1 \end{aligned} $$

10. Express $\frac{5+\sqrt{2} i}{1-\sqrt{2} i}$ in the form of.

Ans. $$ \text { 10. } \begin{aligned} & \frac{5+\sqrt{2} i}{1-\sqrt{2} i} \\ & =\frac{5+\sqrt{2} i}{1-\sqrt{2} i} \times \frac{1+\sqrt{2} i}{1+\sqrt{2} i} \\ & =\frac{5+5 \sqrt{2} i+\sqrt{2} i-2}{1-(\sqrt{2} i)^2}=\frac{3+6 \sqrt{2} i}{1+2} \\ & =\frac{3(1+2 \sqrt{2} i)}{3}=1+2 \sqrt{2} i . \end{aligned} $$

11. What is the conjugate of $\dfrac{\sqrt{5+12 i}+\sqrt{5-12 i}}{\sqrt{5+12 i}-\sqrt{5-12 i}}$

Ans. Let, $z=\frac{\sqrt{5+12 i}+\sqrt{5-12 i}}{\sqrt{5+12 i}-\sqrt{5-12 i}}$

$=\frac{\sqrt{5+12 i}+\sqrt{5-12 i}}{\sqrt{5+12 i}-\sqrt{5-12 i}} \times \frac{\sqrt{5+12 i}+\sqrt{5-12 i}}{\sqrt{5+12 i}+\sqrt{5-12 i}}$

$=\frac{5+12 i+5-12 i+2 \sqrt{25+144}}{5+12 i-5+12 i}$

$=\frac{10+2 \times 13}{24 i}=\frac{36}{24 i}$

$=-\frac{3 i}{2}$

$=0-\frac{3}{2} i$

Therefore, the conjugate of $z=0+\frac{3}{2} i$.

12. Express $z=\frac{1}{(2+i)^2}$ in standard form.

Ans. $$ \begin{aligned} & \text { 12. } \frac{1}{(2+i)^2}=\frac{1}{4+i^2+4 i} \\ & =\frac{1}{4-1+4 i}=\frac{1}{3+4 i} \\ & =\frac{1}{3+4 i} \times \frac{3-4 i}{3-4 i} \\ & =\frac{3-4 i}{9-16 i^2}=\frac{3-4 i}{25} \end{aligned} $$

13. If $\frac{(1+i)^2}{2-i}=x+i y$, find x+y.

Ans. Given: $\frac{(1+i)^2}{2-i}=x+i y$

Upon expansion we get,

$$ \begin{aligned} & \frac{1^2+i^2+2(1)(i)}{2-i}=x+i y \\ & \frac{1+(-1)+2 i}{2-i}=x+i y \\ & \frac{2 i}{2-i}=x+i y \end{aligned} $$

Now, let us multiply and divide by $(2+i)$, we get

$$ \begin{aligned} & \frac{2 i}{2-i} \times \frac{2+i}{2+i}=x+i y \\ & \frac{4 i+2 i^2}{2^2-i^2}=x+i y \\ & \frac{2(-1)+4 i}{4-(-1)}=x+i y \\ & \frac{-2+4 i}{5}=x+i y \end{aligned} $$

Let us equate real and imaginary parts on both sides we get,

$x=-\frac{2}{5}$ and $y=\frac{4}{5}$

So, $x+y=-\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$.

PDF Summary - Class 11 Maths Complex Numbers and Quadratic Equations Notes (Chapter 5)

1. definition.

When a given number is in the form of $a + ib$, where $a,b \in R$ and $i = \sqrt { - 1} $ it is called a complex number and such number is denoted by ‘$z$’.

$z = a + ib$

$a$= real part of complex number and,

$b$= imaginary part of complex number.

1.1 Conjugate of a Complex Number

Consider a complex number $z = a + ib$,

Then its conjugate is written as '$\bar z$'.

Whose value is defined as $\bar z = a - ib$.

2. Algebra of Complex Numbers

Let ${z_1} = a + ib$ and ${z_2} = c + id$ be two complex numbers where ${\text{a}},{\text{b}},{\text{c}},{\text{d}} \in {\text{R}}$ and ${\text{i}} = \sqrt { - 1} $.

1. Addition :

$ z_{1}+z_{2} =(a+b i)+(c+d i) $

$ =(a+c)+(b+d) i $

2. Subtraction :

$ z_{1}-z_{2} =(a+b i)-(c+d i) $

$ =(a-c)+(b-d) $

3. Multiplication :

$ z_{1} \cdot z_{2} =(a+b i)(c+d i) $

$ =a(c+d i)+b i(c+d i) $

$ =a c+a d i+b c i+b d i^{2} $

$ =a c-b d+(a d+b c) i $ $\left(\because i^{2}=-1\right)$

1. $\begin{array}{r} a+i b=c+i d \\ \Leftrightarrow a=c \& b=d \end{array}$

2. $i^{4 k+r}=\left\{\begin{array}{cl}1 ; & r=0 \\ i ; & r=1 \\ -1 ; & r=2 \\ -i ; & r=3\end{array}\right.$

3. $\sqrt b \sqrt a = \sqrt {ba} $ is only possible if atleast one of either a or b is non-negative.

3. Argand Plane

Any complex number $z = a + ib$ can be represented by a unique point $P(a,\,\,b)$ in the argand plane.

(Image will be updated soon)

$P(a,\,\,b)$ represents the complex number $z = a + ib$.

3.1 Modulus and Argument of Complex Number

Consider a complex number $z = a + ib$.

(i) Distance of $z$ from the origin is referred to as the modulus of complex number $z$.

It is represented by ${\text{r}} = |{\text{z}}| = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} $

(ii) Here, $\theta $ i.e.The angle made by ray OP with positive direction of the real axis is called the argument of ${\mathbf{z}}$.

${{\text{z}}_1} > {{\text{z}}_2}$ or ${{\text{z}}_1} < {{\text{z}}_2}$ has no meaning but $\left| {{{\text{z}}_1}} \right| > \left| {{{\text{z}}_2}} \right|$ or $\left| {{{\text{z}}_1}} \right| < \left| {{{\text{z}}_2}} \right|$ holds meaning.

3.2 Principal Argument

The argument $'\theta '$ of complex numbers $z = a + ib$ is called the principal argument of $z$if $ - \pi < \theta \leqslant \pi $.

Consider$\tan \alpha = \left| {\dfrac{b}{a}} \right|$, and $\theta $ be the $\arg \left( z \right)$.

In (iii) and (iv) the principal argument is given by $ - \pi + \alpha $ and $ - \alpha $respectively.

4. Polar Form

$\begin{array}{l} a=r \cos \theta \quad \& b=r \sin \theta \\ \text { where } r=|z| \text { and } \theta=\arg (z) \\ \therefore \quad z \quad=a+i b \\ \quad=r(\cos \theta+\operatorname{isin} \theta) \end{array}$

A complex number $z$ can also be represented as $z = r{e^{i\theta }}$, it is known as Euler’s form.

$\mathrm{r}=|\mathrm{Z}| \& \theta=\arg (\mathrm{Z})$

5. Some Important Properties

1. $\overline {(\bar z)} = z$

2. $z + \bar z = 2\operatorname{Re} (z)$

3. $z - \bar z = 2i\operatorname{Im} (z)$

4. $\overline {{{\text{z}}_1} + {{\text{z}}_2}} = {\overline {\text{z}} _1} + {\overline {\text{z}} _2}$

5. $\overline {{{\text{z}}_1}{{\text{z}}_2}} = {\overline {\text{z}} _1}.{\overline {\text{z}} _2}$

6. $|{\text{z}}| = 0 \Rightarrow {\text{z}} = 0$

7. $z.\,\bar z = |z{|^2}$

8. $\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|;\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}$

9. $|\overline {\text{z}} | = |{\text{z}}| = | - {\text{z}}|$

10. ${\left| {{z_1} \pm {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} \pm 2\operatorname{Re} \left( {{z_1}{{\bar z}_2}} \right)$

11. $\left| {{z_1} + {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|$ (Triangle Inequality)

12. $\left| {\,{{\text{z}}_1} - {{\text{z}}_2}} \right|\,\, \geqslant \,\,\,\left| {\left| {\,{{\text{z}}_1}} \right| - \left| {\,{{\text{z}}_2}} \right|} \right|$

13. ${\left| {{\text{a}}{{\text{z}}_1} - {\text{b}}{{\text{z}}_2}} \right|^2} + {\left| {{\text{b}}{{\text{z}}_1} + {\text{a}}{{\text{z}}_2}} \right|^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2}} \right)\left( {{{\left| {{{\text{z}}_1}} \right|}^2} + {{\left| {{{\text{z}}_2}} \right|}^2}} \right)$

14. $\operatorname{amp} \left( {{{\text{z}}_1} \cdot {{\text{z}}_2}} \right) = \operatorname{amp} {{\text{z}}_1} + \operatorname{amp} {{\text{z}}_2} + 2{\text{k}}\pi ;\,\,\,\,{\text{k}} \in {\text{I}}$

15. $\operatorname{amp} \left( {\dfrac{{{{\text{y}}_0}}}{{{{\text{y}}_1}}}} \right) = \operatorname{amp} {{\text{z}}_1} - \operatorname{amp} {{\text{z}}_2} + 2{\text{k}}\pi ;\,\,\,\,\,{\text{k}} \in {\text{I}}$

16. $\operatorname{amp} \left( {{z^n}} \right) = n\operatorname{amp} (z) + 2k\pi ;\,\,\,k \in I$

6. De-Moivre's Theorem

Statement: $\cos n\theta + i\sin n\theta $ is the value or one of the values of ${(\cos \theta + i\sin \theta )^n}$ according as if 'n' is integer or a rational number. The theorem is very useful in determining the roots of any complex quantity.

7. Cube Root of Unity

Roots of the equation ${{\text{x}}^3} = 1$ are called cube roots of unity.$

Roots of the equation $x^{3}=1$ are called cube roots of unity.

$\begin{array}{l} \quad x^{3}-1=0 \\ (x-1)\left(x^{2}+x+1\right)=0 \\ x=1 \quad \text { or } \quad x^{2}+x+1=0 \end{array}$

i.e $x=\underbrace{\dfrac{-1+\sqrt{3} \mathrm{i}}{2}}_{\mathrm{w}}$ or $\mathrm{x}=\underbrace{\dfrac{-1-\sqrt{3} \mathrm{i}}{2}}_{w^{2}}$

(i) The cube roots of unity are $1,\dfrac{{ - 1 + i\sqrt 3 }}{2},\dfrac{{ - 1 - i\sqrt 3 }}{2}$.

(ii) ${\omega ^3} = 1$

(iii) If ${\text{w}}$ is one of the imaginary cube roots of unity then $1 + \omega + {\omega ^2} = 0$.

(iv) In general $1 + {\omega ^{\text{r}}} + {\omega ^{2{\text{r}}}} = 0$; where ${\text{r}} \in {\text{I}}$ but is not the multiple of 3 .

(v) In polar form the cube roots of unity are:

$\cos 0 + i\sin 0;\cos \dfrac{{2\pi }}{3} + i\sin \dfrac{{2\pi }}{3},\cos \dfrac{{4\pi }}{3} + i\sin \dfrac{{4\pi }}{3}$

(vi) The three cube roots of unity when plotted on the argand plane constitute the vertices of an equilateral triangle.

(vii) The following factorisation should be remembered:

$\begin{array}{l} a^{3}-b^{3}=(a-b)(a-\omega b)\left(a-\omega^{2} b\right) \\ x^{2}+x+1=(x-\omega)\left(x-\omega^{2}\right) \\ a^{3}+b^{3}=(a+b)(a+\omega b)\left(a+\omega^{2} b\right) \\ a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a+\omega b+\omega^{2} c\right)\left(a+\omega^{2} b+\omega c\right) \end{array}$

8. ‘n’ nth Roots of Unity

Solution of equation ${x^n} = 1$ is given by,

$\begin{array}{*{20}{l}} {{\text{x}} = \cos \dfrac{{2{\text{k}}\pi }}{{\text{n}}} + {\text{i}}\sin \dfrac{{2{\text{k}}\pi }}{{\text{n}}}}&{;{\text{k}} = 0,1,2, \ldots ,{\text{n}} - 1} \\ { = {{\text{e}}^{{\text{i}}\left( {\dfrac{{2{\text{k}}\pi }}{{\text{n}}}} \right)}}}&{;{\text{k}} = 0,1, \ldots .,{\text{n}} - 1} \end{array}$

1. We may take any n consecutive integral values of ${\text{k}}$ to get ' ${\text{n}}$ ' ${{\text{n}}^{{\text{th }}}}$ roots of unity.

2. Sum of ' $'n'\,\,{n^{{\text{th }}}}$ roots of unity is zero, $n \in N$

3. The points represented by '${\text{n}}$', nth roots of unity are located at the vertices of a regular polygon of $n$ sides inscribed in a unit circle, centered at the origin and one vertex being one positive real axis.

Properties:

If $1,{\alpha _1},{\alpha _2},{\alpha _3} \ldots .{\alpha _{n - 1}}$ are the ${\text{n}},{{\text{n}}^{{\text{th }}}}$ root of unity then:

(i) They are in G.P. with common ratio ${{\text{e}}^{{\text{i}}(2\pi /{\text{n}})}}$

(ii) ${1^p} + \alpha _0^0 + \alpha _1^\circ + \ldots . + \alpha _{m - 0}^\circ = \left[ {\begin{array}{*{20}{l}} {0,{\text{ if }}p \ne {k^n}} \\ {n,{\text{ if }}p = kn} \end{array}} \right.$ where $k \in Z$

(iii) $\left( {1 - {\alpha _1}} \right)\left( {1 - {\alpha _2}} \right) \ldots \ldots \left( {1 - {\alpha _{n - 1}}} \right) = {\text{n}}$

(iv) $\left( {1 + {\alpha _1}} \right)\left( {1 + {\alpha _2}} \right) \ldots \ldots \left( {1 + {\alpha _{n - 1}}} \right) = \left[ {\begin{array}{*{20}{l}} {0,{\text{ if n is even }}} \\ {1,{\text{ if n is odd }}} \end{array}} \right.$

(v) $1.{\alpha _1} \cdot {\alpha _2} \cdot {\alpha _3} \ldots \ldots \ldots {\alpha _{n - 1}} = \left[ {\begin{array}{*{20}{c}} { - 1,{\text{ if }}n{\text{ is even }}} \\ {1,{\text{ if }}n{\text{ is odd }}} \end{array}} \right.$

(i) $\cos \theta + \cos 2\theta + \cos 3\theta + \ldots . + \cos n\theta = \dfrac{{\sin (n\theta /2)}}{{\sin (\theta /2)}}\cos \left( {\dfrac{{n + 1}}{2}} \right)\theta $

(ii) $\sin \theta + \sin 2\theta + \sin 3\theta + \ldots . + \sin n\theta = \dfrac{{\sin (n\theta /2)}}{{\sin (\theta /2)}}\sin \left( {\dfrac{{n + 1}}{2}} \right)\theta $.

9. Square Root of Complex Number

Let ${\text{x}} + {\text{iy}} = \sqrt {{\text{a}} + {\text{ib}}} $, Squaring both sides, we get

${(x + iy)^2} = a + ib$

i.e., ${x^2} - {y^2} = a,{\mkern 1mu} {\mkern 1mu} 2xy = b$

Solving these equations, we get square roots of $z$.

10. LOCI in Complex Plane

(i) $\left| {z - {z_0}} \right| = $ a represents the circumference of a circle, centred at ${{\text{z}}_{\text{o}}}$, radius a .

(ii) $\left| {{\text{z}} - {{\text{z}}_{\text{0}}}} \right| < $ a represents the interior of the circle.

(iii) $\left| {z - {z_0}} \right| > $ a represents the exterior of this circle.

(iv) $\left| {z - {z_1}} \right| = \left| {z - {z_2}} \right|$ represents $ \bot $ bisector of segment with endpoints ${z_1}{\text{ and }}{z_2}$.

(v) $\left| {\dfrac{{ - {z_1}}}{{ - {z_2}}}} \right| = k$ represents: $\left\{ {\begin{array}{*{20}{l}} {{\text{ circle, k}} \ne 1} \\ { \bot {\text{ bisector, k}} = 1} \end{array}} \right\}$

(vi) $\arg ({\text{z}}) = \theta $ is a ray starting from the origin (excluded) inclined at an $\angle \theta $ with a real axis.

(vii) Circle described on line segment joining ${z_1}{\text{ and }}{z_2}$ as diameter is:

$\left( {z - {z_1}} \right)\left( {\bar z - {{\bar z}_2}} \right) + \left( {z - {z_2}} \right)\left( {\bar z - {{\bar z}_1}} \right) = 0$

(viii) If $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}$ are the vertices of an equilateral triangle where $\mathrm{z}_{0}$ is its circumcentre then

(a) $\dfrac{1}{\mathrm{z}_{2}-\mathrm{z}_{3}}+\dfrac{1}{\mathrm{z}_{3}-\mathrm{z}_{1}}+\dfrac{1}{\mathrm{z}_{1}-\mathrm{z}_{2}}=0$

(b) $z_{0}^{1}+z_{1}^{1}+z_{2}^{1}-z_{1} z_{2}-z_{2} z_{3}-z_{3} z_{1}=0$

(ix) If $\mathrm{A}, \mathrm{B}, \mathrm{C} \& \mathrm{D}$ are four points representing the complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3} \& \mathrm{z}_{4}$ then

$\mathrm{AB}|| \mathrm{CD}$ if $\dfrac{\mathrm{Z}_{4}-\mathrm{Z}_{3}}{\mathrm{z}_{2}-\mathrm{z}_{1}}$ is purely real ;

$\mathrm{AB} \perp \mathrm{CD}$ if $\dfrac{\mathrm{z}_{4}-\mathrm{z}_{3}}{\mathrm{z}_{2}-\mathrm{z}_{1}}$ is purely imaginary $]$

11. Vectorial Representation of a Complex

Every complex number can be considered as if it is the position vector of that point. If the point P represents the complex number z then,

$\overrightarrow {OP} \, = \,\,z{\text{ and }}\left| {\overrightarrow {OP} } \right| = \left| {\,z\,} \right|.$

(“Image will be updated soon)

(i) If $\overrightarrow {{\text{OP}}} = {\text{z}} = {\text{r}}{{\text{e}}^{{\text{i}}\theta }}$ then $\overrightarrow {{\text{OQ}}} = {{\text{z}}_1} = {\text{r}}{{\text{e}}^{{\text{i}}(\theta + \,\phi )}} = {\text{z}}{\text{.}}{{\text{e}}^{{\text{i}}\phi }}$.

If $\overrightarrow {{\text{OP}}} $ and $\overrightarrow {{\text{OQ}}} $ are of unequal magnitude, then

$\widehat {{\text{OQ}}} = \widehat {{\text{OP}}}{{\text{e}}^{{\text{i}}\phi }}$

(ii) If ${z_1},\,\,{z_2}{\text{ and }}{{\text{z}}_3}$ are three vertices of a triangle ABC described in the counterclockwise sense, then

$\dfrac{{{z_3} - z}}{{{z_2} - z}} = \dfrac{{AC}}{{AB}}(\cos \alpha + i\sin \alpha ) = \dfrac{{AC}}{{AB}} \cdot {e^{i\alpha }} = \dfrac{{\left| {{z_3} - {z_1}} \right|}}{{\left| {{z_2} - {z_1}} \right|}} \cdot {e^{i\alpha }}$

12. Some Important Results

(i) If ${{\text{z}}_1}$ and ${{\text{z}}_2}$ are two complex numbers, then the distance between ${z_1}$ and ${z_2}$ is $\left| {{z_2} - {z_1}} \right|$.

(ii) Segment Joining points ${\text{A}}\left( {{{\text{z}}_1}} \right)$ and ${\text{B}}\left( {{{\text{z}}_2}} \right)$ is divided by point ${\text{P}}({\text{z}})$ in the ratio ${{\text{m}}_1}:{{\text{m}}_2}$ then ${\text{z}} = \dfrac{{{{\text{m}}_1}{{\text{z}}_2} + {{\text{m}}_2}{\text{z}}}}{{{{\text{m}}_1} + {{\text{m}}_2}}},\;{{\text{m}}_1}$ and ${{\text{m}}_2}$ are real.

(iii) The equation of the line joining ${z_1}$ and ${z_2}$ is given by

$\left| {\begin{array}{*{20}{c}} z&{\bar z} \\ z&{\bar z} \\ {{z_2}}&{{{\bar z}_2}} \end{array}} \right| = 0{\text{ (non parametric form) }}$

$\dfrac{{z - z}}{{\bar z - \bar z}} = \dfrac{{z - {z_2}}}{{\bar z - {{\bar z}_2}}}$

(iv) $\bar az + a\bar z\, + b = 0$ represents a general form of line.

(v) The general eqn. of circle is:

$z\bar z + a\bar z + \bar az + b = 0$ (where ${\text{b}}$ is real no.).

Centre : $( - a)$ and radius,

$\sqrt {|a{|^2} - b} = \sqrt {a\bar a - b} $.

(vi) Circle described on line segment joining ${z_1}$ and ${z_2}$ as diameter is:

(vii) Four pts. ${z_1},{z_2},{z_3},{z_4}$ in anticlockwise order will be concyclic, if and only if

$\theta = \arg \cdot \left( {\dfrac{{{z_2} - {z_4}}}{{{z_1} - {z_4}}}} \right) = \arg \left( {\dfrac{{{z_2} - {z_3}}}{{{z_1} - {z_3}}}} \right)$

$\Rightarrow \arg \left( {\dfrac{{{z_2} - {z_4}}}{{{z_1} - {z_4}}}} \right) - \arg \cdot \left( {\dfrac{{{z_2} - {z_3}}}{{{z_1} - {z_3}}}} \right) = 2n\pi ;(n \in I)$

$\Rightarrow \arg \left[ {\left( {\dfrac{{{z_2} - {z_4}}}{{{z_1} - {z_4}}}} \right)\left( {\dfrac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}}} \right)} \right] = 2n\pi$

$ \Rightarrow \quad \left( {\dfrac{{{z_2} - {z_4}}}{{{z_1} - {z_4}}}} \right) \times \left( {\dfrac{{{z_1} - {z_3}}}{{{z_2} - {z_3}}}} \right)$ is real and positive.

(viii) If ${z_1},{z_2},{z_3}$ are the vertices of an equilateral triangle where ${z_0}$ is its circumcentre then

(a) $\dfrac{1}{{{z_2} - {z_3}}} + \dfrac{1}{{{z_3} - {z_1}}} + \dfrac{1}{{{z_1} - {z_2}}} = 0$

(b) $z_0^1 + z_1^1 + z_2^1 - {z_1}{z_2} - {z_2}{z_3} - {z_3}{z_1} = 0$

(ix) If $A,B,C{\text{ and }}D$ are four points representing the complex numbers ${z_1},{z_2},{z_3}{\text{ and }}{z_4}$ then

${\text{AB}}||{\text{CD}}$ if $\dfrac{{{z_4} - {z_3}}}{{{z_2} - {z_1}}}$ is purely real;

$AB \bot CD{\text{ if }}\dfrac{{{z_4} - {z_3}}}{{{z_2} - {z_1}}}$ is purely imaginary.

(x) Two points $P\left( {{z_1}} \right)$ and $Q\left( {{z_2}} \right)$ lie on the same side or opposite side of the line $\bar az\, + \,a\bar z\, + b$ accordingly as $\bar a{z_1}\, + \,a{\bar z_1}\, + b{\text{ and }}\bar a{z_2}\, + \,a{\bar z_2}\, + b$ have same sign or opposite sign.

Important Identities

(i) ${{\text{x}}^2} + {\text{x}} + 1 = ({\text{x}} - \omega )\left( {{\text{x}} - {\omega ^2}} \right)$

(ii) ${{\text{x}}^2} - {\text{x}} + 1 = ({\text{x}} + \omega )\left( {{\text{x}} + {\omega ^2}} \right)$

(iii) ${x^2} + xy + {y^2} = (x - y\omega )\left( {x - y{\omega ^2}} \right)$

(iv) ${{\text{x}}^2} - {\text{xy}} + {{\text{y}}^2} = ({\text{x}} + \omega {\text{y}})\left( {{\text{x}} + {\text{y}}{\omega ^2}} \right)$

(v) ${x^2} + {y^2} = (x + iy)(x - iy)$

(vi) ${x^3} + {y^3} = (x + y)(x + y\omega )\left( {x + y{\omega ^2}} \right)$

(vii) ${{\text{x}}^3} - {{\text{y}}^3} = ({\text{x}} - {\text{y}})({\text{x}} - {\text{y}}\omega )\left( {{\text{x}} - {\text{y}}{\omega ^2}} \right)$

(viii) ${x^2} + {y^2} + {z^2} - xy - yz - zx = \left( {x + y\omega + z{\omega ^2}} \right)\left( {x + y{\omega ^2} + z\omega } \right)$

or $\quad \left( {{\text{x}}\omega + {\text{y}}{\omega ^2} + {\text{z}}} \right)\left( {{\text{x}}{\omega ^2} + {\text{y}}\omega + {\text{z}}} \right)$

or $\quad \left( {{\text{x}}\omega + {\text{y}} + {\text{z}}{\omega ^2}} \right)\left( {{\text{x}}{\omega ^2} + {\text{y}} + {\text{z}}\omega } \right)$

(ix) ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)\left( {x + \omega y + {\omega ^2}z} \right)$$\left( {{\text{x}} + {\omega ^2}{\text{y}} + \omega {\text{z}}} \right)$

1. Quadratic Expression

The standard form of a quadratic expression in ${\text{x}}$ is, $f({\text{x}}) = a{x^2} + bx + c$, where $a,b,c \in R{\text{ and }}a \ne 0$. General form of a quadratic equation in ${\text{x}}$ is, $a{x^2} + bx + c = 0$, where $a,b,c \in R{\text{ and }}a \ne 0$.

2. Roots of Quadratic Equation

(a) The solution of the quadratic equation,

$a{x^2} + bx + c = 0$ is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

The expression ${\text{D}} = {{\text{b}}^2} - 4{\text{ac}}$ is called the discriminant of the quadratic equation.

(b) If $\alpha {\text{ and }}\beta $ are the roots of the quadratic equation $a{x^2} + bx + c = 0$, then

(i) $\alpha + \beta = \dfrac{{ - {\text{b}}}}{a}$

(ii) $\alpha \beta = \dfrac{{\text{c}}}{a}$

(iii) $|\alpha - \beta | = \dfrac{{\sqrt {\text{D}} }}{{|{\text{a}}|}}$

${{\text{x}}^2} - (\alpha + \beta ){\text{x}} + \alpha \beta = 0\quad $ i.e.,

${{\text{x}}^2} - ($ sum of roots $){\text{x}} + $ product of roots $ = 0$

$ y=\left(a x^{2}+b x+c\right) \equiv a(x-\alpha)(x-\beta) $

$ =a\left(x+\dfrac{b}{2 a}\right)^{2}-\dfrac{D}{4 a} $

3. Nature of Roots

(a) Consider the quadratic equation $a{x^2} + bx + c = 0$ where a, $a,b,c \in R{\text{ and }}a \ne 0$ then;

(i) ${\text{D}} > 0 \Leftrightarrow $ roots are real and distinct (unequal).

(ii) ${\text{D}} = 0 \Leftrightarrow $ roots are real and coincident (equal).

(iii) ${\text{D}} < 0 \Leftrightarrow $ roots are imaginary.

(iv) If ${\text{p}} + {\text{iq}}$ is one root of a quadratic equation, then the other must be the conjugate $p - iq$ and vice versa. $({\text{p}},{\text{q}} \in {\text{R and i}} = \sqrt { - 1} )$.

(b) Consider the quadratic equation $a{x^2} + bx + c = 0$ where $a,b,c \in Q{\text{ and }}a \ne 0$ then;

(i) If ${\text{D}} > 0$ and is a perfect square, then roots are rational and unequal.

(ii) If $\alpha = p + \sqrt q $ is one root in this case, (where $p$ is rational and $\sqrt q $ is a surd) then the other root must be the conjugate of it i.e., $\beta = p - \sqrt q $ and vice versa.

Remember that a quadratic equation cannot have three different roots and if it has, it becomes an identity.

4. Graph of Quadratic Equation

Consider the quadratic expression, ${\text{y}} = {\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}}$, $a \ne 0{\text{ and }}a,b,c \in R$ then;

(i) The graph between ${\text{x}},{\text{y}}$ is always a parabola. If ${\text{a}} > 0$ then the shape of the parabola is concave upwards and if ${\text{a}} < 0$ then the shape of the parabola is concave downwards.

(ii) ${\text{y}} > 0\forall {\text{x}} \in {\text{R}}$, only if ${\text{a}} > 0{\text{ and D}} < 0$

(iii) ${\text{y}} < 0\forall {\text{x}} \in {\text{R}}$, only if ${\text{a}} < 0{\text{ and D}} < 0$

5. Solution of Quadratic Inequalities

$a x^{2}+b x+c>0(a \neq 0)$

(i) If $D>0$, then the equation $a x^{2}+b x+c=0$ has two different roots $\left(\mathrm{x}_{1}<\mathrm{x}_{2}\right)$

Then $\mathrm{a}>0 \quad \Rightarrow \quad \mathrm{x} \in\left(-\infty, \mathrm{x}_{1}\right) \cup\left(\mathrm{x}_{2}, \infty\right)$

$\mathrm{a}<0 \quad \Rightarrow \quad \mathrm{x} \in\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right)$

(ii) Inequalities of the form $\dfrac{\mathrm{P}(\mathrm{x})}{\mathrm{Q}(\mathrm{x})} \gtrless 0$ can be quickly solved using the method of intervals (wavy curve).

6. Maximum and Minimum Value of Quadratic Equation

Maximum and minimum value of $y = a{x^2} + bx + c$ occurs at $x = - \left( {\dfrac{b}{{2a}}} \right)$ according as:

For $a > 0,$ we have:

$y \in \left[ {\dfrac{{4ac - {b^2}}}{{4a}},\,\infty } \right)$

${y_{\min }} = \dfrac{{ - D}}{{4a}}{\text{ at }}x = \dfrac{{ - b}}{{2a}},{\text{ and }}{{\text{y}}_{\max }} \to \infty $

For $a < 0,$ we have:

$y \in \left( { - \infty ,\,\,\dfrac{{4ac - {b^2}}}{{4a}}} \right]$

${y_{\max }} = \dfrac{{ - D}}{{4a}}{\text{ at }}x = \dfrac{{ - b}}{{2a}},{\text{ and }}{{\text{y}}_{\min }} \to \infty $

7. Theory of Equations

If ${\alpha _1},{\alpha _2},{\alpha _3}, \ldots \ldots ,{\alpha _n}$ are the roots of the ${{\text{n}}^{{\text{th }}}}$ degree polynomial equation:

$f({\text{x}}) = {{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{{\text{n}} - 1}} + {{\text{a}}_2}{{\text{x}}^{{\text{n}} - 2}} + \ldots \ldots + {{\text{a}}_{{\text{n}} - 1}}{\text{x}} + {{\text{a}}_{\text{n}}} = 0$

where ${a_0},{a_1}, \ldots \ldots .{a_n}$ are all real and ${a_0} \ne 0$

$\sum {{\alpha _1}} = - \dfrac{{{{\text{a}}_1}}}{{{{\text{a}}_0}}}$

$\sum {{\alpha _1}} {\alpha _2} = \dfrac{{{{\text{a}}_2}}}{{{{\text{a}}_0}}}$

$\sum {{\alpha _1}} {\alpha _2}{\alpha _3} = - \dfrac{{{a_3}}}{{{a_0}}}$;

${\alpha _1}{\alpha _2}{\alpha _3} \ldots .{\alpha _{\text{n}}} = {( - 1)^{\text{n}}}\dfrac{{{{\text{a}}_{\text{n}}}}}{{{{\text{a}}_0}}}$

8. Location of Roots

Let $f(x) = a{x^2} + bx + c$, where $a > 0{\text{ and }}a,b,c \in R$

(i) Conditions for both the roots of $f(x) = 0$ to be greater than a specified number ‘k’ are:

${\text{D}} \geqslant 0{\text{ and }}f({\text{k}}) > 0{\text{ and }}( - {\text{b}}/2{\text{a}}) > {\text{k}}$

(ii) Conditions for both roots of $f({\text{x}}) = 0$ to lie on either side of the number ‘k’ (in other words the number ‘k’ lies between the roots of $f({\text{x}}) = 0$ is:

$a\,f({\text{k}}) < 0$

(iii) Conditions for exactly one root of $f\left( x \right) = 0$ to lie in the interval $\left( {{k_1},\,\,{k_2}} \right){\text{ i}}{\text{.e}}{\text{., }}{k_1} < x < {k_2}$ are:

$D > 0{\text{ and }}f\left( {{k_1}} \right).f\left( {{k_2}} \right) < 0$

(iv) Conditions that both the roots of $f\left( x \right) = 0$ to be confined between the numbers ${k_1}{\text{ and }}{{\text{k}}_2}$ are $\left( {{k_1} < {k_2}} \right)$:

$D \geqslant 0{\text{ and }}f\left( {{k_1}} \right) > 0{\text{ and }}f\left( {{k_2}} \right) > 0{\text{ and }}{k_1} < \left( {\dfrac{{ - b}}{{2a}}} \right) < {k_2}$

9. Maximum and Minimum Values of Rational Numbers

Here we shall find the values attained by a rational expresion of the form $\dfrac{{{a_1}{x^2} + {b_1}x + {c_1}}}{{{a_2}{x^2} + {b_2}x + {c_2}}}$ for real values of $x$.

10. Common Factors

(a) only one common root.

Let $\alpha $ be the common root of $a{x^2} + bx + c = 0$ and ${a^\prime }{x^2} + {b^\prime }x + {c^\prime } = 0$, such that $a,{a^\prime } \ne 0$ and a ${b^\prime } \ne {a^\prime }b$. Then, the condition for one common root is:

${\left( {c{a^\prime } - {c^\prime }a} \right)^2} = \left( {a{b^\prime } - {a^\prime }b} \right)\left( {b{c^\prime } - {b^\prime }c} \right)$

(b) Two common roots

Let $\alpha ,\,\beta $ be the two common roots of

$a{x^2} + bx + c = 0{\text{ and }}{a^\prime }{x^2} + {b^\prime }x + {c^\prime } = 0$ such that ${\text{a}},{{\text{a}}^\prime } \ne 0$.

Then, the condition for two common roots is: $\dfrac{a}{{{a^\prime }}} = \dfrac{b}{{{b^\prime }}} = \dfrac{c}{{{c^\prime }}}$

11. Resolution Into Two Linear Factors

The condition that a quadratic function $f(x,y) = a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c$ may be resolved into two linear factors is that:

$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$ or,

$\left| {\begin{array}{*{20}{l}} a&h&g \\ h&b&f \\ g&f&c \end{array}} \right| = 0$

12. Formation of a Polynomial Equations

If ${\alpha _1},{\alpha _2},{\alpha _3}, \ldots \ldots ,{\alpha _n}$ are the roots of the ${n^{{\text{th }}}}$ degree polynomial equation, then the equation is

${{\text{x}}^{\text{n}}} - {{\text{S}}_1}{{\text{x}}^{{\text{n}} - 1}} + {{\text{S}}_2}{{\text{x}}^{{\text{n}} - 2}} + {{\text{S}}_3}{{\text{x}}^{{\text{n}} - 3}} + \ldots \ldots + {( - 1)^{\text{n}}}{{\text{S}}_{\text{n}}} = 0$

where ${{\text{S}}_{\text{k}}}$ denotes the sum of the products of roots taken ${\text{k}}$ at a time.

Particular Cases

(a) Quadratic Equation: If $\alpha, \beta$ be the roots the quadratic equation, then the equation is :

$\mathrm{x}^{2}-\mathrm{S}_{1} \mathrm{x}+\mathrm{S}_{2}=0 \quad \text { i.e. } \quad \mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta=0$

(b) Cubic Equation: If $\alpha, \beta, \gamma$ be the roots the cubic equation, then the equation is :

$\mathrm{x}^{3}-\mathrm{S}_{1} \mathrm{x}^{2}+\mathrm{S}_{2} \mathrm{x}-\mathrm{S}_{3}=0 \quad \text { i.e }$

$x^{3}-(\alpha+\beta+\gamma) x^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma=0$

(i) If $\alpha$ is a root of equation $\mathrm{f}(\mathrm{x})=0$, the polynomial $\mathrm{f}(\mathrm{x})$ is exactly divisible by $(\mathrm{x}-\alpha)$. In other words, $(x-\alpha)$ is a factor of $f(x)$ and conversely.

(ii) Every equation of nth degree $(n \geq 1)$ has exactly n roots $\&$ if the equation has more than $\mathrm{n}$ roots, it is an identity.

13. Transformation of Equations

(i) To obtain an equation whose roots are reciprocals of the roots of a given equation, it is obtained by replacing ${\text{x}}$ by $1/{\text{x}}$ in the given equation.

(ii) Transformation of an equation to another equation whose roots are negative of the roots of a given equation-replace ${\text{x}}$ by $ - {\text{x}}$.

(iii) Transformation of an equation to another equation whose roots are square of the roots of a given equation-replace ${\text{x}}$ by $\sqrt {\text{x}} $.

(iv) Transformation of an equation to another equation whose roots are cubes of the roots of a given equation-replace $x$ by ${x^{1/3}}$.

Class 11 Maths Notes of Complex Numbers

What is a complex number.

A complex number can be noted as a + ib, here “a” is a real number and “b” is an imaginary number. The complex number comprised of the symbol “i” which assures the condition i2 = −1. Complex numbers are mentioned as the addition of one-dimensional number lines. In a complex plane, a complex number can be denoted by a + bi and is usually represented in the form of the point (a, b).

We have to see that a complex number with no real part, such as – i, -5i, etc, is called as entirely imaginary. Also, a complex number with absolutely no imaginary part is also known as a real number.

A number of form x + iy, where x and y are real numbers, is called a complex number, x is called the real part, and y is called the imaginary part of the complex number i.e. Re(Z) = x and Im(Z) = y.

Properties of Complex Numbers

The properties of a complex number are listed below:

a and b are two real numbers and if, a + ib = 0 then a = 0, b = 0

When the real numbers are a, b and c; and a + ib = c + id ,then a = c and b = d.

A set of three complex numbers namely z1, z2, and z3 satisfy the commutative, associative, as well as distributive laws.

If both the sum as well as the product of any two given complex numbers are real then the complex numbers are conjugate to each other.

For two complex numbers z1 and z2 : |z1 + z2| ≤ |z1| + |z2|

The sum of any two conjugate complex numbers is always real.

The product of any two conjugate complex numbers is always real.

Importance of Ch 5 Class 11 Maths Revision Notes for Complex Numbers and Quadratic Equations

All the class 11 maths chapter 5 revision notes have been made by Vedantu in accordance with the latest CBSE syllabus so that there is no mistake if changes have been created by the CBSE board. The ch 5 class 11 maths revision notes for Complex Numbers and Quadratic Equations have been made by teachers who have vast experience and know exactly what is needed.

Further, students will be able to see chapter-wise Mathematics Revision Notes with short keynotes that could raise their preparations. The CBSE revision notes class 11 maths chapter 5 are prepared in a concise manner from the latest edition of the books.

The CBSE class 11 Maths Chapter 5 revision notes for Complex Numbers and Quadratic Equations are available in a PDF format so that students can simply refer to it whenever required thorough Vedantu. The teachers who prepare these class 11 maths chapter 5 revision notes have done so after rigorously going through the last ten year's question papers and then taking them down.

Because of the large amount of content present in most of the books, it becomes very difficult for the students to keep up with all of it. The class 11 revision notes maths ch 5 can play a role in helping easier studying methods. Vedantu is one of India’s leading education platforms for students all over the country.

Algebra of Complex Numbers - In this section of Notes of Class 11 Revision Notes Chapter 5, you would learn about different operations that can be performed on complex numbers. Let us say there are two complex numbers:

C1 = x + iy and C2 = a + ib

Addition of Complex Numbers - Two complex numbers are added by adding each part (real and imaginary) separately. Therefore C1 + C2 = (x + iy) + (a + ib) = (x + a) + (y + b)i.

Subtraction of Complex Numbers - Two complex numbers are subtracted by subtracting each part (real and imaginary) separately. Therefore C1 - C2 = (x + iy) - (a + ib) = (x - a) + (y - b)i.

Multiplication of Complex Numbers - Two complex numbers are multiplied by multiplying each part (real and imaginary) of the first complex number with each part of the second complex number. Hence:

C1 * C2 = (x + iy) * (a + ib) = (x * a) + (a * ib) + (y * a * i) + (y * b * i 2. )

Division of Complex Numbers - We use the concept of conjugates to divide one complex number by another. Both the numerator and denominator are multiplied by the conjugate of the denominator in the division process. Hence:

C1/C2 = (x + iy) / (a + ib) = (x + iy) * (a - ib)/(a + ib) * (a - ib) = ((xa + yb)/ (a 2 + b 2 )) + ((ya - xb)/(a 2 + b 2 )) i

Modulus and Argument of a Complex Number - A complex number C = x + iy can be represented on an argand plane by a point P(x, y). In that plane:

The distance of point P from the origin is called the modulus of the complex number C and is denoted by r = |C| = √(x 2 + y 2 ).

The angle which a straight line joining the origin and point P makes with the positive direction of the X-axis is known as the argument of the complex number C.

If the argument (θ) of C lies between -π and +π i.e. -π < θ < +π, then θ is called the principal argument of C.

Why Should Students go Through Vedantu Revision Notes?

The studying process is usually student-specific, that is some students prefer kinesthetic studying, some are auditory students, and others may find visual studying to be more efficient. But these different processes are just a part of the studying experience of every student, the other important aspect is the revision. Revision is very essential. Students have to strengthen their studying, hence revisions are a way to begin about this phase as it will help them grasp the concept easily.

Enables the student to strengthen their studying.

Students become more confident during examinations.

Saves valuable time during examinations.

The correctness of answers are higher.

FAQs on Complex Numbers and Quadratic Equations Class 11 Notes CBSE Maths Chapter 5 (Free PDF Download)

1. What is the Equality of Complex Numbers in Maths?

Two complex numbers z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 are equal, if x 1 = x 2 and y 1 = y 2 i.e. Re( z 1 ) = Re( z 2 ) and Im( z 1 ) = Im( z 2 ) Note: Order relation of “less than” and “greater than’’ are not defined for complex numbers.

2. Why are Revision Notes Important?

Revision notes are important as they help:

Exam stress and anxiety are decreased.

Reduced chances of making easy, but conspicuous mistakes.

3. What are Purely Real and Purely Imaginary Complex Numbers?

A complex number suppose Z = x + iy is purely real if its imaginary part is equal to 0, i.e. Im(z) is equal to 0 and purely imaginary if its real part is equal to 0 i.e. Re (z) is equal to 0.

4. What are Some Properties of Complex Numbers?

The properties of complex numbers are:

A set of three complex numbers namely z 1 , z 2 , and z 3 satisfy the commutative, associative, as well as distributive laws.

5. How should I prepare for Math Class 11 Chapter 5?

To prepare Maths Class 11 Chapter 5 Complex Numbers and Quadratic Equations thoroughly, check out the tips given below.

Try to understand the chapter carefully, make a list of concepts that you haven’t understood. Practise several questions of those concepts that are clear to you.

To strengthen your weak concepts, take help from your teachers or friends. You can also read this chapter on Vedantu.

Make notes of different concepts mentioned in the chapter. This will help you to comprehend things better. Also, with the help of notes, you can do your home assignments on time.

6. How to download Chapter 5 Revision Notes of Class 11 Maths?

If you want to download Chapter 5 Complex Numbers and Quadratic Equations Revision notes of Class 11 Maths textbook then follow the steps given.

Visit the page Class 11 Maths Revision Notes for Chapter 5.

After that, you will be redirected to the official website of Vedantu.

Thereupon, you will see a webpage having the option of downloading the PDF.

Click on that option and your PDF will be downloaded.

In the PDF you will get notes and summaries of different concepts mentioned in the chapter. Plus, you’ll also get multiple sample questions to practise these concepts. You can have this content for free.

You can also download the Vedantu app to access these study materials for free.

7. What are real numbers?

Real numbers are rational numbers that are present across the number line. Hence, there are millions and billions of numbers on the number line. These numbers include both rational and irrational numbers. They can be negative and positive. Real numbers are denoted by the symbol ‘R’. At the same time, they also include natural numbers, whole numbers and integers. Example; numbers starting from 1- Natural Numbers, numbers starting from 0-whole numbers.

8. What are the properties of real numbers?

There are four different properties of real numbers. To understand these properties, consider x, y, z as three real numbers.

Commutative property- In this property, the general of the real numbers are described as x+y=y+x for addition and x*y=y*x for multiplication.

Associative property- The numbers are represented as x+(y+z)=(x+y)+z for addition and (xy)*r=x*(yz) for multiplication.

Distributive property- In this property, the numbers are represented as x(y+z)=xy+xz and (x+y)z=xz+yz

Identity property- For addition identity- x+0=x. Here 0 is the additive identity

For multiplication- x*1=1*x=x. Here 1 is the multiplicative identity.

9. What are imaginary numbers?

Imaginary numbers are complex numbers that are written in the format of real numbers. They are multiplied with an unreal or imaginary unit that is denoted as i. Imaginary numbers are added to real numbers to make complex numbers in the form of x+yi where x and y are real numbers. However, x is the real part of the complex number and yi is the imaginary part of the complex number.

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Complex Numbers and Quadratic Equations Class 11 Notes PDF (Short & Handwritten)

Revision notes make you aware of those topics that you might have missed during your regular classes. Thus, Complex Numbers and Quadratic Equations is an extremely important chapter of class 11 Maths and so, all students who have opted for Maths in their intermediate should refer to the Complex Numbers and Quadratic Equations Class 11 Notes. The revision notes work as a reference that help students like you to revise the concepts and formulas which you have studied earlier from your Mathematics textbook.

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Complex Numbers and Quadratic Equations Notes PDF

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Chapter End Questions of Complex Numbers and Quadratic Equations: The Complex Numbers and Quadratic Equations notes PDF that we provide here are prepared referring to the NCERT Class 11 Maths Book so, those who want to use other study resources than revision notes of Complex Numbers and Quadratic Equations can use the NCERT Books. The book contains exercises that enable the learners to solve various questions to cross-check how well they understood the topic.
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On a weekly basis after completing the chapter Complex Numbers and Quadratic Equations: There’s a few students who dedicate themselves in learning Maths during regular classes; however, experts suggests that revision just after completing the chapter Complex Numbers and Quadratic Equations can help students to deepen their understanding in the concepts that were explained in the lesson. Thus, the right time to use the Complex Numbers and Quadratic Equations Class 11 Notes can be on a weekly basis after finishing the chapter’s basics subtopics.
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The Complex Numbers and Quadratic Equations is an important part of higher level Maths, because in standard like 11, students have to study various topics at the same time, it becomes a challenge to handle so many topics and subjects at the same time. Thus, here we have listed what you definitely need to revise Class 11 Complex Numbers and Quadratic Equations for the ease of your study and exam preparation.

A Short Notes of Complex Numbers and Quadratic Equations: You wouldn’t like to use the revision Notes of Complex Numbers and Quadratic Equations that look like bulky textbooks, cluttered with so much information in an unorganised manner, right? So, you need a short yet precise Short Notes of Complex Numbers and Quadratic Equations that can help you recall your learnings from the chapter within a few minutes. You can find such Notes of Complex Numbers and Quadratic Equations here on Selfstudys website.
Easy To Understand Definitions: Unclear and lengthy definitions never work, and it becomes very challenging to retain such definitions for a longer time. Therefore, you need revision notes that can help you easily understand the definitions of Complex Numbers and Quadratic Equations. Easy to understand definitions help you retain them for a longer time and give you an ability to recall them wherever you need them to solve a particular question related to Complex Numbers and Quadratic Equations.
Some Questions of Complex Numbers and Quadratic Equations to Practise: Without solving the questions of Complex Numbers and Quadratic Equations a revision is incomplete. If you really want to revise the topics, you must solve the questions based on Complex Numbers and Quadratic Equations. Such questions can be found from Notes of Complex Numbers and Quadratic Equations that we provide here or from the NCERT Class 11 Maths textbooks. Our Maths experts have mentioned the questions of Complex Numbers and Quadratic Equations especially in Class 11 Maths notes so that a learner can use them to practise.

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Maths assignment class viii | quadrilateral, complex numbers chapter 5 class 11.

Complex Numbers Chapter-5 Class-11

Following concepts will be discussed in this Blog

Sign convention while converting a complex number from Cartesian form to polar form.

Dividend = Divisor X Quotient + Remainder

P(x) = ( x 2 + 10x + 41)(x 2 – x + 4) – 160

= 0 X ( x 2 – x + 4) – 160 = - 160 Ans

Lesson Plan Math Class 10 (Ch-1) | Real Numbers

Lesson Plan Maths Class XII | For Maths Teacher

Assignment 10 15
Assignment 11 12
Assignment 12 14
Assignment 8 8
Assignment 9 5
Lesson plan 10 15
Lesson Plan 12 13
Lesson Plan 8 10
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Maths 11 21
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Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations MCQs

Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations MCQs are provided here for students to help them in their preparation for the exams. The MCQs of Class 11 Maths Chapter 5 given here contain the right answers and explanations. These objective type questions cover all the concepts of the NCERT curriculum, which are important from the examination point of view.

Get MCQs for all chapters of Class 11 Maths here.

Students can solve the MCQs provided here for Chapter 5 of Class 11 Maths to improve their skills in identifying complex numbers and solving quadratic equations, whose roots are complex numbers.

Download PDF – Complex Numbers and Quadratic Equations MCQs

Also check:

Complex numbers Class 11 Notes
Quadratic equations for Class 11 Notes
Important Questions for Class 11 Maths Chapter 5

MCQs for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers

1. The value of 1 + i 2 + i 4 + i 6 + … + i 2n is

(a) positive

(b) negative

(d) cannot be evaluated

Correct option: (d) cannot be evaluated

1 + i 2 + i 4 + i 6 + … + i 2n = 1 – 1 + 1 – 1 + … (–1) n

This cannot be evaluated unless the value of n is known.

2. If a + ib = c + id, then

(a) a 2 + c 2 = 0

(b) b 2 + c 2 = 0

(d) a 2 + b 2 = c 2 + d 2

Correct option: (d) a 2 + b 2 = c 2 + d 2

a + ib = c + id

⇒ |a + ib| = |c + id|

⇒ √(a 2 + b 2 ) = √(c 2 + d 2 )

Squaring on both sides, we get;

a 2 + b 2 = c 2 + d 2

3. If a complex number z lies in the interior or on the boundary of a circle of radius 3 units and centre (– 4, 0), the greatest value of |z +1| is

Correct option: (b) 6

The distance of the point representing z from the centre of the circle is |z – (-4 + i0)| = |z + 4|

According to the given,

|z + 4| ≤ 3

|z + 1| = |z + 4 – 3| ≤ |z + 4| + |-3| ≤ 3 + 3 ≤ 6

Hence, the greatest value of |z + 1| is 6.

4. The value of arg (x) when x < 0 is

(d) none of these

Correct option: (c) π

Let z = x + 0i and x < 0

Since the point (-x, 0) lies on the negative side of the real axis,

|z| = |x + oi| = √[(-1) 2 + 0)] = 1

∴ Principal argument (z) = π

Alternative method:

Let x = cos θ + i sin θ

For θ = π, x should be negative.

Thus, x < 0 for θ = π.

5. If 1 – i, is a root of the equation x 2 + ax + b = 0, where a, b ∈ R, then the value of a – b is

Correct option: (a) -4

Given that 1 – i is the root of x 2 + ax + b = 0.

Thus, 1 + i is also the root of the given equation since non-real complex roots occur in conjugate pairs.

Sum of roots = −a/1 = (1 – i) + (1 + i)

Product of roots, b/1 = (1 – i)(1 + i)

b = 1 – i 2

b = 1 + 1 {since i 2 = -1}

Now, a – b = -2 – 2 = -4

6. Number of solutions of the equation z 2 + |z| 2 = 0 is

(d) infinitely many

Correct option: (d) infinitely many

z 2 + |z| 2 = 0, z ≠ 0

⇒ (x + iy) 2 + [√(x 2 + y 2 )] 2 = 0

⇒ x 2 – y 2 + i2xy + x 2 + y 2 = 0

⇒ 2x 2 + i2xy = 0

⇒2x (x + iy) = 0

⇒ x = 0 or x + iy = 0 (not possible)

Therefore, x = 0 and z ≠ 0.

Thus, y can have any real value.

Hence, there exist infinitely many solutions.

7. If [(1 + i)/(1 – i)] x = 1, then

(a) x = 2n + 1, where n ∈ N

(b) x = 4n, where n ∈ N

(d) x = 4n + 1, where n ∈ N

Correct option: (b) x = 4n, where n ∈ N

By rationalising the denominator,

Thus, i x = i 4n , where n is any positive integer.

8. If the complex number z = x + iy satisfies the condition |z + 1| = 1, then z lies on

(b) circle with centre (1, 0) and radius 1

Correct option: (c) circle with centre (–1, 0) and radius 1

|z + 1| = 1

|x + iy + 1| = 1

⇒ |(x + 1) + iy| = 1

⇒ √[(x +1) 2 + y 2 ] = 1

Squaring on both sides,

(x + 1) 2 + y 2 = 1

This is the equation of a circle with centre (–1, 0) and radius 1.

9. The simplified value of (1 – i) 3 /(1 – i 3 ) is

Correct option: (b) -2

(1 – i) 3 /(1 – i 3 )

= (1 – i) 3 /(1 3 – i 3 )

= (1 – i) 2 /(1 + i – 1)

= (1 – i) 2 /i

= (1 + i 2 – 2i)/i

= (1 – 1 – 2i)/i

10. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for:

(b) x = [n + (1/2)] (π/2)

(d) No value of x

Correct option: (d) No value of x

Consider sin x + i cos 2x and cos x – i sin 2x are conjugate to each other.

So, sin x – i cos 2x = cos x – i sin 2x

On comparing real and imaginary parts of both sides, we get

⇒ sin x = cos x and cos 2x = sin 2x

⇒ sin x/cos x = 1 and (cos 2x/sin 2x) = 1

⇒ tan x = 1 and tan 2x = 1

Now, consider tan 2x = 1

Using the formula tan 2A = 2 tan A/(1 – tan 2 A),

(2 tan x)/(1 – tan 2 x) = 1

However, this is not possible for tan x = 1.

Therefore, for no value of x, sinx + i cos 2x and cos x – i sin 2x are conjugate to each other

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NCERT Solutions Class 11 Maths Complex Numbers and Quadratic Equations
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Class 11 Mathematics Case Study Questions
Class 11th Maths Chapter 05 ( complex numbers) Exercise 5.6
Class 11 Important Questions for Maths
NCERT Solutions Class 11 Maths Chapter 5 Miscellaneous Exercise

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CLASS 11 MATHS || COMPLEX NUMBERS || PYQS
CBSE NCERT CLASS 11 MATHS COMPLEX NUMBERS AND QUADRATIC EQUATIONS EX 4.1 SUM 3 (SOLUTION IN TAMIL)
CBSE NCERT CLASS 11 MATHS COMPLEX NUMBERS AND QUADRATIC EQUATIONS EX 4.1 SUM 4 (SOLUTION IN TAMIL)
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Review Exercise 1 unit 1 complex numbers class 11 New mathematics book
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CBSE Case Study Questions For Class 11 Maths Complex Numbers Free PDF
Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Complex Numbers in order to fully complete your preparation.They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!. I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams.
Case Study Questions for Class 11 Maths Chapter 5 Complex Numbers and
Importance of Solving Case Study Questions for Class 11 Maths. Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills.
Class 11 Mathematics Case Study Questions
Graph the following inequality on the number line: x > −32. Answer Key: (b) 9.8 km and 13.8 km (a) ... Class 11 Mathematics case study question 3. Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the ...
Complex numbers and quadratic equations
Class 11 14 units · 180 skills. Unit 1 Sets. Unit 2 Relations and functions. Unit 3 Trigonometric functions. Unit 4 Complex numbers and quadratic equations. Unit 5 Linear inequalities. Unit 6 Permutations and Combinations. Unit 7 Binomial theorem. Unit 8 Sequence and series.
Case Studies on the topic Complex Numbers
Case Studies on the topic Complex Numbers - Class 11 MathematicsThis video lecture discusses 2 case study questions on the topic Complex Numbers for class 1...
CBSE Class 11 Maths
Square Root of a Complex Number. Solution of a Quadratic Equation The equation ax 2 + bx + c = 0, where a, b and c are numbers (real or complex, a ≠ 0) is called the general quadratic equation in variable x. The values of the variable satisfying the given equation are called roots of the equation. The quadratic equation ax 2 + bx + c = 0 with ...
Important Questions for Class 11 Maths Chapter 5 with Solutions
Practice Problems for Class 11 Maths Chapter 5. Solve the below-given problems from class 11 Maths Chapter 5: If |z 2 -1|= |z| 2 +1, prove that z lies on the imaginary axis. Compute the value of p, such that the difference of the roots of the equation is x 2 +px+8=0 is 2. Express each of the following complex numbers in the form a+ib.
Case Study Questions for Class 11 Maths
Thanks for Watching this Video..Pls SUBSCRIBE the Channel and Like & SHARE the Video..🔗 PDF Link - Coming soon Case Study Questions (Class 11) https://www....
Important Questions for CBSE Class 11 Maths Chapter 5
Here, a and b are real numbers and i is iota. The value of iota is √-1. Therefore, z (complex number) = a + ib where a is the real part, and ib is the imaginary part. a = Re (z), b= Im (z). Download important questions for class 11 maths chapter 5 and ace your upcoming exams. Share this with your friends.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and
Chapter 5 of Class 11 Complex Numbers and Quadratic Equations has 3 exercises and a miscellaneous exercise to help the students practise the required number of problems to understand all the concepts. The topics and sub-topics discussed in the PDF of NCERT Solutions for Class 11 of this chapter include. 5.1 Introduction.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers ...
Quadratic equation is an equation of degree 2. It implies that the highest component of this function is 2. The standard form of quadratic equation is given as: ax2 + bx + c = 0. Here, a,b,and c R, a 0, b2 - 4ac < 0 are given by -b 4ac - b2 i2a. Some Examples of Complex Number Class 11 Solutions: Question 1 to 10.
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and
Free download NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1, Ex 5.2, Ex 5.3 and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE ...
NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic
NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations. Short Answer Type Questions. Q6. If a = cos θ + i sin θ, then find the value of (1+a/1-a) Sol: a = cos θ + i sin θ. Q10. Show that the complex number z, satisfying the condition arg lies on arg (z-1/z+1) = π/4 lies on a circle. Sol: Let z = x + iy.
Complex Numbers and Quadratic Equations Class 11 Notes CBSE Maths
FAQs on Complex Numbers and Quadratic Equations Class 11 Notes CBSE Maths Chapter 5 (Free PDF Download) Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal, if x1 = x2 and y1 = y2 i.e. Re (z1) = Re (z2) and Im (z1) = Im (z2) Note: Order relation of "less than" and "greater than'' are not defined for complex numbers.
Complex Numbers Class 11 Questions
Practice Questions. Solve the following complex numbers class 11 questions. Write the expression i -39 in the form of a + ib. Solve the equation: √2x 2 + x + √2 = 0. Write the given complex number in the polar form: (1 + 2i) / (1 - 3i).
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This set of Class 11 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on "Complex Numbers and Quadratic Equations". These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation. 1. Value of i (iota) is ____________. a) -1. b) 1.
Complex Numbers and Quadratic Equations Class 11 Notes PDF ...
To do this, click on NCERT Notes within the NCERT Books and Solutions menu. Choose Class: There is a list of classes, but you have to choose class 11 in order to download the Complex Numbers and Quadratic Equations Class 11 Notes PDF. Now, Select Maths and Access Complex Numbers and Quadratic Equations Notes for Free: At this moment, you are ...
Complex Numbers Class 11 Notes with Examples
For a complex number z = p + iq, p is known as the real part, represented by Re z and q is known as the imaginary part, it is represented by Im z of complex number z. Complex Numbers Class 11 Concepts. The topics and subtopics covered in Complex numbers class 11 are: Introduction; Complex Numbers; Algebra of Complex Numbers Addition of two ...
Complex Numbers Chapter 5 Class 11
Complex Numbers Chapter-5 Class-11 Basic points and complete explanation of the complex numbers with examples and diagrams. ... Interesting maths riddles and maths magic, Class-wise mathematics study material for students from 8th to 12. ... Math Assignment Class VIII Understanding Quadrilateral Important extra questions on Quadrilateral class ...
NCERT Exemplar Solutions for Class 11 Maths Chapter 5 Complex Numbers
NCERT Exemplar Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Here, we bring to you detailed solutions to exercises of the NCERT Exemplar Class 11 Maths Chapter 5. Subject matter experts who created the solutions have compiled these questions for you to revise Chapter 5 of the NCERT Exemplar Textbook.
Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations MCQs
Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations MCQs are provided here for students to help them in their preparation for the exams. The MCQs of Class 11 Maths Chapter 5 given here contain the right answers and explanations. These objective type questions cover all the concepts of the NCERT curriculum, which are important from ...