NCERT Solutions for Exercise 5.1 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

Q: How root(-1) is represented in complex numbers?

root(-1) is represented by the letter i (iota)

Q: How to add two complex numbers?

Consider two complex numbers z1=x+iy and z2=a+ib. Then z1+z2=(x+a)+(b+y)i

Q: What is the multiplicative identity in complex numbers?

If z is a complex number, then multiplicative identity is given as z(1/z)=1

Q: What is the number of a solved example prior to the exercise 5.1 Class 11 Maths?

6 main questions and their sub-questions are solved prior to the Class 11 Maths chapter 5 exercise 5.1

Q: What type of questions are covered in the NCERT solutions for Class 11 Maths chapter 5 exercise 5.1?

The questions from the topics algebra of complex numbers and the modulus and conjugate of complex numbers are covered in the solutions of Class 11 Maths chapter 5 exercise 5.1

NCERT Solutions for Exercise 5.1 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

Edited By Vishal kumar | Updated on Nov 06, 2023 06:51 AM IST

NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Exercise 5.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.1- NCERT solutions for exercise 5.1 Class 11 Maths Chapter 5 introduces the concept of complex numbers through some questions and their solutions. Students are familiar with real numbers. NCERT Class 11 chapter Complex Numbers And Quadratic Equations introduces imaginary numbers. And exercise 5.1 Class 11 Maths introductory exercise includes the concept of imaginary numbers. If the discriminant is negative that is b²-4ac<0, then in a real number system, we don't have a solution.

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NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Exercise 5.1- Download Free PDF
NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.1
Access 11th Class Maths Exercise 5.1 Answers
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Key Features of NCERT Solutions for Class 11 Maths Chapter 5.1 Exercise
NCERT Solutions of Class 11 Subject Wise

NCERT Solutions for Exercise 5.1 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

In NCERT book chapter 5 of Class 11 NCERT Mathematics the concept of solving quadratic equations of the form ax²+bx+c=0 with b²-4ac<0 is also introduced. The Class 11 Maths chapter 5 exercise 5.1 lists a few practice problems on imaginary numbers. The NCERT syllabus Class 11 Maths Chapter 5 Exercise 5.1 covers topics like the concept of complex numbers and the algebra of complex numbers. Also, questions related to modulus and conjugate of complex numbers are discussed in NCERT Solutions for Class 11 Maths chapter 5 exercise 5.1. In addition to Class 11 Maths Chapter 5 Exercise 5.1, the following exercises were written by subject experts at Careers360, providing detailed and easy-to-understand solutions. Students can also download PDF versions of these 11th class maths exercise 5.1 answers free of charge, allowing them to access and use the materials offline at their convenience, anytime and anywhere. This class 11 maths ex 5.1 solution resource is valuable for students seeking comprehensive support in mathematics.

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.1

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Access 11th Class Maths Exercise 5.1 Answers

Question:1 Express the given complex number in the form a+ib: (5i)(-3i/5)?
Answer:

$eginarrayl extGiven a+ib: left(5i ight) imesleft(-frac35i ight)\ left(5i ight) imesleft(-frac35i ight)=-left(5 imesfrac35 ight)left(i^2 ight)\ =-3left(i^2 ight) left[i^2=-1 ight]\ =3.endarray$

Question:2 Express each of the complex number in the form $a+ib$ .

$i^9+i^1^9$

Answer:

We know that $i^4 = 1$
Now, we will reduce $i^9+i^{19}$ into

$i^9+i^1^9$ $= (i^4)^2.i+(i^4)^3.i^3$
$= (1)^2.i+(1)^3.(-i)$ $(\because i^4 = 1 , i^3 = -i\ and \ i^2 = -1)$
$=i-i = 0$
Now, in the form of $a+ib$ we can write it as
$o+io$
Therefore, the answer is $o+io$

Question:3 Express each of the complex number in the form a+ib.

$i^{-39}$

Answer:

We know that $i^4 = 1$
Now, we will reduce $i^{-39}$ into

$i^{-39}$ $= (i^{4})^{-9}.i^{-3}$
$= (1)^{-9}.(-i)^{-1}$ $(\because i^4 = 1 , i^3 = -i)$
$= \frac{1}{-i}$
$= \frac{1}{-i} \times \frac{i}{i}$
$= \frac{i}{-i^2}$ $(\because i^2 = -1)$
$= \frac{i}{-(-1)}$
$=i$
Now, in the form of $a+ib$ we can write it as
$o+i1$
Therefore, the answer is $o+i1$

Question:4 Express each of the complex number in the form a+ib.

$3(7+7i)+i(7+7i)$

Answer:

Given problem is
$3(7+7i)+i(7+7i)$
Now, we will reduce it into

$3(7+7i)+i(7+7i)$ $= 21+21i+7i+7i^2$
$= 21+21i+7i+7(-1)$ $(\because i^2 = -1)$
$= 21+21i+7i-7$
$=14+28i$

Therefore, the answer is $14+i28$

Question:5 Express each of the complex number in the form $a+ib$ .

$(1-i)-(-1+6i)$

Answer:

Given problem is
$(1-i)-(-1+6i)$
Now, we will reduce it into

$(1-i)-(-1+6i)$ $=1-i+1-6i$
$= 2-7i$

Therefore, the answer is $2-7i$

Question:6 Express each of the complex number in the form $a+ib$ .

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )$

Answer:

Given problem is
$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )$
Now, we will reduce it into

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right ) = \frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}$
$= \frac{1-20}{5}+i\frac{(4-25)}{10}$
$= -\frac{19}{5}-i\frac{21}{10}$

Therefore, the answer is $-\frac{19}{5}-i\frac{21}{10}$

Question:7 Express each of the complex number in the form $a+ib$ .

$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right )$

Answer:

Given problem is
$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right )$
Now, we will reduce it into

$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right ) = \frac{1}{3}+i\frac{7}{3} + 4+i\frac{1}{3} + \frac{4}{3}-i$
$=\frac{1+4+12}{3}+i\frac{(7+1-3)}{3}$
$=\frac{17}{3}+i\frac{5}{3}$

Therefore, the answer is $\frac{17}{3}+i\frac{5}{3}$

Question:8 Express each of the complex number in the form $a+ib$ .

$(1-i)^4$

Answer:

The given problem is
$(1-i)^4$
Now, we will reduce it into

$(1-i)^4 = ((1-i)^2)^2$
$= (1^2+i^2-2.1.i)^2$ $(using \ (a-b)^2= a^2+b^2-2ab)$
$=(1-1-2i)^2$ $(\because i^2 = -1)$
$= (-2i)^2$
$= 4i^2$
$= -4$

Therefore, the answer is $-4+i0$

Question:9 Express each of the complex number in the form $a+ib$ .

$\left ( \frac{1}{3}+3i \right )^3$

Answer:

Given problem is
$\left ( \frac{1}{3}+3i \right )^3$
Now, we will reduce it into

$\left ( \frac{1}{3}+3i \right )^3=\left ( \frac{1}{3} \right )^3+(3i)^3+3.\left ( \frac{1}{3} \right )^2.3i+3.\frac{1}{3}.(3i)^2$ $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
$= \frac{1}{27}+27i^3+i + 9i^2$
$= \frac{1}{27}+27(-i)+i + 9(-1)$ $(\because i^3=-i \ and \ i^2 = -1)$
$=\frac{1}{27}-27i+i-9$
$=\frac{1-243}{27}-26i$
$=-\frac{242}{27}-26i$

Therefore, the answer is

$-\frac{242}{27}-26i$

Question:10 Express each of the complex number in the form $a+ib$ .

$\left ( -2-\frac{1}{3}i \right )^3$

Answer:

Given problem is
$\left ( -2-\frac{1}{3}i \right )^3$
Now, we will reduce it into

$\left ( -2-\frac{1}{3}i \right )^3=-\left ( (2)^3+\left ( \frac{1}{3}i \right )^3 +3.(2)^2\frac{1}{3}i+3.\left ( \frac{1}{3}i \right )^2.2 \right )$ $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
$=-\left ( 8+\frac{1}{27}i^3+3.4.\frac{1}{3}i+3.\frac{1}{9}i^2.2 \right )$
$=-\left ( 8+\frac{1}{27}(-i)+4i+\frac{2}{3}(-1) \right )$ $(\because i^3=-i \ and \ i^2 = -1)$
$=-\left ( 8-\frac{1}{27}i+4i-\frac{2}{3} \right )$
$=-\left ( \frac{(-1+108)}{27}i+\frac{24-2}{3} \right )$
$=-\frac{22}{3}-i\frac{107}{27}$

Therefore, the answer is $-\frac{22}{3}-i\frac{107}{27}$

Question:11 Find the multiplicative inverse of each of the complex numbers.

$4-3i$

Answer:

Let $z = 4-3i$
Then,
$\bar z = 4+ 3i$
And
$|z|^2 = 4^2+(-3)^2 = 16+9 =25$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{4+3i}{25}= \frac{4}{25}+i\frac{3}{25}$

Therefore, the multiplicative inverse is

$\frac{4}{25}+i\frac{3}{25}$

Question:12 Find the multiplicative inverse of each of the complex numbers.

$\sqrt{5}+3i$

Answer:

Let $z = \sqrt{5}+3i$
Then,
$\bar z = \sqrt{5}-3i$
And
$|z|^2 = (\sqrt5)^2+(3)^2 = 5+9 =14$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{\sqrt5-3i}{14}= \frac{\sqrt5}{14}-i\frac{3}{14}$

Therefore, the multiplicative inverse is $\frac{\sqrt5}{14}-i\frac{3}{14}$

Question:13 Find the multiplicative inverse of each of the complex numbers.

$-i$

Answer:

Let $z = -i$
Then,
$\bar z = i$
And
$|z|^2 = (0)^2+(1)^2 = 0+1 =1$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{i}{1}= 0+i$

Therefore, the multiplicative inverse is $0+i1$

Question:14 Express the following expression in the form of $a+ib:$

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

Answer:

Given problem is
$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Now, we will reduce it into

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})} = \frac{3^2- (\sqrt5i)^2}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$ $(using \ (a-b)(a+b)=a^2-b^2)$
$=\frac{9-5i^2}{\sqrt3+\sqrt2i-\sqrt3+\sqrt2i}$
$=\frac{9-5(-1)}{2\sqrt2i}$ $(\because i^2 = -1)$
$=\frac{14}{2\sqrt2i}\times \frac{\sqrt2i}{\sqrt2i}$
$=\frac{7\sqrt2i}{2i^2}$
$=-\frac{7\sqrt2i}{2}$
Therefore, the answer is $0-i\frac{7\sqrt2}{2}$

Benefits of NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1

The solved example before the exercise 5.1 Class 11 Maths and the NCERT solutions for Class 11 Maths chapter 5 exercise 5.1 are important as it covers questions from basics of complex numbers
For Class 11 final exams students may get MCQs, short answer or long answer questions from the type covered in the NCERT syllabus Class 11 Maths chapter 5 exercise 5.1

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Key Features of NCERT Solutions for Class 11 Maths Chapter 5.1 Exercise

Step-by-step solutions: Detailed, step-by-step explanations for each ex 5.1 class 11 problem, aiding students in understanding the concepts and problem-solving techniques.
Clarity and accuracy: The class 11 maths ex 5.1 solutions are presented clearly and accurately, ensuring that students can confidently prepare for exams and improve their understanding.
Curriculum alignment: Class 11 ex 5.1 solutions are designed to closely follow the NCERT curriculum, covering topics and concepts as per the official syllabus.
Accessibility: These 11th class maths exercise 5.1 answers are often available for free, making them easily accessible to students.
Format options: PDF versions of the solutions may be provided, allowing students to download and use them conveniently, both online and offline.

NCERT Solutions of Class 11 Subject Wise

Subject wise NCERT Exampler solutions

Frequently Asked Question (FAQs)

1. How root(-1) is represented in complex numbers?

root(-1) is represented by the letter i (iota)

2. What is the value of i^2?

i^2=-1

3. Simplify i^3?

i^3=i^2.i=-1.i=-i

4. What is the value of i^4?

i^4=1

5. How to add two complex numbers?

Consider two complex numbers z1=x+iy and z2=a+ib.

Then z1+z2=(x+a)+(b+y)i

6. What is the multiplicative identity in complex numbers?

If z is a complex number, then multiplicative identity is given as z(1/z)=1

7. What is the number of a solved example prior to the exercise 5.1 Class 11 Maths?

6 main questions and their sub-questions are solved prior to the Class 11 Maths chapter 5 exercise 5.1

8. What type of questions are covered in the NCERT solutions for Class 11 Maths chapter 5 exercise 5.1?

The questions from the topics algebra of complex numbers and the modulus and conjugate of complex numbers are covered in the solutions of Class 11 Maths chapter 5 exercise 5.1

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Exercise 5.1 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Exercise 5.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.1

Access 11th Class Maths Exercise 5.1 Answers

More About NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1

Benefits of NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1

Key Features of NCERT Solutions for Class 11 Maths Chapter 5.1 Exercise

NCERT Solutions of Class 11 Subject Wise

Subject wise NCERT Exampler solutions

Frequently Asked Question (FAQs)

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