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

# 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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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.

## Access 11th Class Maths Exercise 5.1 Answers

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

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 $o+io$

$i^{-39}$

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 $o+i1$

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

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 $14+i28$

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

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 $2-7i$

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

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 $-\frac{19}{5}-i\frac{21}{10}$

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

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 $\frac{17}{3}+i\frac{5}{3}$

$(1-i)^4$

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 $-4+i0$

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

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

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

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

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 $-\frac{22}{3}-i\frac{107}{27}$

$4-3i$

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

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

$\sqrt{5}+3i$

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 $\frac{\sqrt5}{14}-i\frac{3}{14}$

$-i$

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 $0+i1$

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 $0-i\frac{7\sqrt2}{2}$

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

The problems from the concepts of addition, multiplication and division of complex numbers and their properties, power of i, and the square root of negative real numbers are covered in exercise 5.1 Class 11 Maths. The first 10 questions of NCERT solutions for ex 5.1 class 11 is to represent the complex numbers of the form p+iq. And question 11 to 13 of Class 11 Maths chapter 5 exercise 5.1 is to find the multiplicative inverse and the 14th question discussed in Class 11th Maths chapter 5 exercise 5.1 is to simplify a given expression of the form p+iq.

## 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

1. 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.

2. 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.

3. 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.

4. Accessibility: These 11th class maths exercise 5.1 answers are often available for free, making them easily accessible to students.

5. 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

### Frequently Asked Questions (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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