 # NCERT Exemplar Class 11 Maths Solutions for Chapter 5 - Complex Numbers And Quadratic Equations

NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are provided here for students to practice and prepare for the exam. Students can make use of NCERT exemplar problems with solutions prepared by experts for Complex Numbers and Quadratic Equations to learn the concepts explained in this chapter. These exemplars are prepared in accordance with CBSE syllabus (2018-2019) and curriculum.

## Class 11 Maths NCERT Exemplar Problems on Complex Numbers and Quadratic Equations

In chapter 5, students will learn and solve exemplar problems based on;

• Complex numbers
• Develop the algebra of complex numbers
• Addition, Difference, Multiplication and Division of two complex number
• Power of imaginary number i
• The square roots of a negative real number
• The Modulus and the Conjugate of a Complex Number
• Argand Plane and Polar Representation of the complex number

This chapter defines the concept of Complex numbers and Quadratic Equations. A complex number can simply be defined as a real number, a pure imaginary number or even the sum of both the numbers. The real number can be represented as ‘a’ and the imaginary number can be represented as ‘bi’. Quadratic equations are related to imaginary numbers and complex numbers only when the value of the quadratic formula is negative.

The pdf format of chapter 5, exemplars can be downloaded easily and practised offline as well. It can also be used as a reference tool while preparing for final exams. Also, learn all the chapters of 11th class Maths NCERT exemplar solutions here.

## Download PDF Of NCERT Exemplar Class 10 Maths Chapter 5 – Complex Numbers And Quadratic Equations            ### Access Answers of NCERT Exemplar Class 10 Maths Chapter 5 – Complex Numbers And Quadratic Equations

Exercise Page No: 91

1. For a positive integer n, find the value of (1 – i)n
(1 – 1/i)n.

Solution:

According to the question,

We have, = (1 – i)n (1 + i)n

= (1 – i2)n

= 2n

Therefore, (1 – i)n (1 – 1/i)n = 2n

2. Evaluate where n ϵ N.

Solution:

According to the question,

We have, 3. If , then find (x, y).

Solution:

According to the question,

We have, = i3 – (-i3)

= 2i3

= 0 – 2i

Thus, (x, y) = (0, -2)

4. If , then find the value of x + y.

Solution:

According to the question,

We have, 5. If then find (a, b).

Solution:

According to the question,

We have, = (i4)25

= 1

Hence, (a, b) = (1, 0)

6. If a = cos θ + i sin θ, find the value of Solution:

According to the question,

We have,

a = cos θ + i sin θ 7. If (1 + i)z = (1 – i) z̅, then show that z = i.

Solution:

According to the question,

We have, = -iz̅

Hence proved.

8. If z = x + iy, then show that zz̅ + 2(z + z̅) + b = 0 where bϵR, representing z in the complex plane is a circle.

Solution:

According to the question,

We have,

z = x + iy

⇒ z̅ = x – iy

Now, we also have,

z z̅ + 2 (z + z̅) + b = 0

⇒ (x + iy) (x – iy) + 2 (x + iy + x – iy) + b = 0

⇒ x2 + y2 + 4x + b = 0

The equation obtained represents the equation of a circle.

9. If the real part of ( z̅ + 2)/ ( z̅ – 1) is 4, then show that the locus of the point representing z in the complex plane is a circle.

Solution:

According to the question,

Let z = x + iy

Now, ⇒ x2 + x – 2 + y2 = 4 (x2 – 2x + 1 + y2)

⇒ 3x2 + 3y2 – 9x + 6 = 0

The equation obtained represents the equation of a circle.

Hence, locus of z is a circle.

10. Show that the complex number z, satisfying the condition arg ((z-1)/(z+1)) = π/4 lies on a circle.

Solution:

According to the question,

Let z = x + iy

arg ((z-1)/(z+1)) = π/4

⇒ arg (z – 1) – arg (z + 1) = π/4

⇒ arg (x + iy – 1) – arg (x + iy + 1) = π/4

⇒ arg (x – 1 + iy) – arg (x + 1 + iy) = π/4 ⇒ x2 + y2 – 1 = 2y

⇒ x2 + y2 – 2y – 1 = 0

The equation obtained represents the equation of a circle.

11. Solve that equation |z| = z + 1 + 2i.

Solution:

According to the question,

We have,

|z| = z + 1 + 2i

Substituting z = x + iy, we get,

⇒ |x + iy| = x + iy + 1 + 2i

We know that,

|z| = √(x2 + y2)

√(x2 + y2) = (x + 1) + i(y + 2)

Comparing real and imaginary parts,

We get,

√(x2 + y2) = (x + 1)

And 0 = y + 2

⇒ y = -2

Substituting the value of y in √(x2 + y2) = (x + 1),

We get,

⇒ x2 + (-2)2 = (x + 1)2

⇒ x2 + 4 = x2 + 2x + 1

Hence, x = 3/2

Hence, z = x + iy

= 3/2 – 2i

12. If |z + 1| = z + 2 (1 + i), then find z.

Solution:

According to the question,

We have,

|z + 1| = z + 2 (1 + i)

Substituting z = x + iy, we get,

⇒ |x + iy + 1| = x + iy + 2 (1 + i)

We know,

|z| = √(x2 + y2)

√((x + 1)2 + y2) = (x + 2) + i(y + 1)

Comparing real and imaginary parts,

⇒ √((x + 1)2 + y2) = x + 2

And 0 = y + 2

⇒ y = -2

Substituting the value of y in √((x + 1)2 + y2) = x + 2,

⇒ (x + 1)2 + (-2)2 = (x + 2)2

⇒ x2 + 2x + 1 + 4 = x2 + 4x + 4

⇒ 2x = 1

Hence, x = ½

Hence, z = x + iy

= ½ – 2i

13. If arg (z – 1) = arg (z + 3i), then find x – 1 : y. where z = x + iy

Solution:

According to the question,

Let z = x + iy

Given that,

arg (z – 1) = arg (z + 3i)

⇒ arg (x + iy – 1) = arg (x + iy + 3i)

⇒ arg (x – 1 + iy) = arg (x + I (y) = π/4 ⇒ xy = xy – y + 3x – 3

⇒ 3x – 3 = y

⇒ (x – 1)/y = 1/3

Hence, (x – 1): y = 1: 3

14. Show that |(z – 2) / (z – 3)| = 2 represents a circle. Find its centre and radius.

Solution:

According to the question,

We have,

|(z – 2) / (z – 3)| = 2

Substituting z = x + iy, we get

⇒ |(x + iy – 2) / (x + iy – 3)| = 2

⇒ |x – 2 + iy| = 2 |x – 3 + iy|

⇒ √((x – 2)2 + y2) = 2√((x – 3)2 + y2)

⇒ x2 – 4x + 4 + y2 = 4 (x2 – 6x + 9 + y2)

⇒ 3x2 + 3y2 – 20x + 32 = 0 Therefore, centre of circle is (10/3, 0) and radius is 4/9 or 2/3.

15. If (z – 1)/(z + 1) is a purely imaginary number (z ≠ –1), then find the value of |z|.

Solution:

According to the question,

Let z = x + iy

Now, ⇒ x2 – 1 + y2 = 0

⇒ x2 + y2 = 1

⇒ √(x2 + y2) = 1

Hence, |z| = 1

16. z1 and z2 are two complex numbers such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that z1 = – z̅2.

Solution:

According to the question,

Let z1 = |z1| (cos θ1 + I sin θ1) and z2 = |z2| (cos θ2 + I sin θ2)

Given that |z1| = |z2|

And arg (z1) + arg (z2) = π

⇒ θ1 + θ2 = π

⇒ θ1 = π – θ2

Now, z1 = |z2| (cos (π – θ2) + I sin (π – θ2))

⇒ z1 = |z2| (-cos θ2 + I sin θ2)

⇒ z1 = -|z2| (cos θ2 – I sin θ2)

⇒ z1 = – [|z2| (cos θ2 – I sin θ2)]

Hence, z1 = -z̅2

Hence proved.

17. If |z1| = 1 (z1 ≠ –1) and z2 = (z1 – 1) / (z + 1), then show that the real part of z2 is zero.

Solution:

According to the question,

Let z1 = x + iy Therefore, the real part of z2 is zero.

18. If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find arg(z1/z4) + arg(z2/z3).

Solution:

According to the question,

We have,

z1 and z2 are conjugate complex numbers.

The negative side of the real axis

= r1 (cos θ1 – i sin θ1)

= r1 [cos (-θ1) + I sin (-θ1)]

Similarly, z3 = r2 (cos θ2 – i sin θ2)

⇒ z4 = r2 [cos (-θ2) + I sin (-θ2)] = θ1 – (-θ2) + (-θ1) – θ2

= θ1 + θ2 – θ1 – θ2

= 0

⇒ arg(z1/z4) + arg(z2/z3) = 0

19. If |z1| = |z2| = ….. = |zn| = 1, then show that |z1 + z2 + z3 + …. + zn| = | 1/z1 + 1/z2 + 1/z3 + … + 1/zn|

Solution:

According to the question,

We have,

|z1| = |z2| = … = |zn| = 1

⇒ |z1|2 = |z2|2 = … = |zn|2 = 1

⇒ z1 z̅ 1= z2 z̅ 2= z3 z̅ 3= … = zn z̅ n= 1 Hence proved.

20. If for complex numbers z1 and z2, arg (z1) – arg (z2) = 0, then show that |z1 – z2| = |z1| – |z2|.

Solution:

According to the question,

Let z1 = |z1| (cos θ1 + I sin θ1) and z2 = |z2| (cos θ2 + I sin θ2)

We have,

arg (z1) – arg (z2) = 0

⇒ θ1 – θ2 = 0

⇒ θ1 = θ2

We also have,

z2 = |z2| (cos θ1 + I sin θ1)

⇒ z1 – z2 = ((|z1|cos θ1 – |z2| cos θ1) + i (|z1| sin θ1 – |z2| sin θ1)) Hence, |z1 – z2| = |z1| – |z2|

Hence proved.

21. Solve the system of equations Re (z2) = 0, |z| = 2.

Solution:

According to the question,

We have,

Re (z2) = 0, |z| = 2

Let z = x + iy.

Then, |z| = √(x2 + y2)

Given in the question,

√(x2 + y2)  = 2

⇒ x2 + y2 = 4 … (i)

z2 = x2 + 2ixy – y2

= (x2 – y2) + 2ixy

Now, Re (z2) = 0

⇒ x2 – y2 = 0 … (ii)

Equating (i) and (ii), we get

⇒ x2 = y2 = 2

⇒ x = y = ±√2

Hence, z = x + iy

= ±√2 ± i√2

= √2 + i√2, √2 – i√2, –√2 + i√2 and –√2 – i√2

Hence, we have four complex numbers.

22. Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.

Solution:

According to the question,

We have,

z + √2 |(z + 1)| + i = 0 … (1)

Substituting z = x + iy, we get

⇒ x + iy + √2 |x + iy + 1| + i = 0 ⇒ 2x2 + 4x + 4 = x2

⇒ x2 + 4x + 4 = 0

⇒ (x + 2)2 = 0

⇒ x = -2

Hence, z = x + iy

= – 2 – i

23. Write the complex number in polar from.

Solution:

According to the question,

We have, 24. If z and w are two complex numbers such that |zw| = 1 and arg (z) – arg (w) = π/2, then show that z̅w = – i.

Solution:

Let z = |z| (cos θ1 + I sin θ1) and w = |w| (cos θ2 + I sin θ2)

Given |zw| = |z| |w| = 1

Also arg (z) – arg (w) = π/2

⇒ θ1 – θ2 = π/2

Now, z̅ w = |z| (cos θ1 – I sin θ1) |w| (cos θ2 + I sin θ2)g | = 1

= |z| |w| (cos (-θ1) + I sin (-θ1)) (cos θ2 + I sin θ2)

= 1 [cos (θ2 – θ1) + I sin (θ2 – θ1)]

= [cos (-π/2) + I sin (-π/2)]

= 1 [0 – i]

= – i

Hence proved

 Also Access NCERT Solutions for Class 11 Maths Chapter 5 CBSE Notes for Class 11 Maths Chapter 5

The solved questions in the exemplar of Maths class 11 chapter 5 will help students to know the methods for solving problems based on the Complex Numbers and Quadratic Equation. We are also providing with Maths NCERT solutions for 11th standard, notes and question papers online for free. Solving previous year question papers and sample papers, will let them know the question pattern and marking scheme for chapter 5, Complex Numbers and Quadratic Equations.

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