NCERT Solutions: Complex Numbers

NCERT Solutions Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

Exercise 4.1:

Q1: Express the given complex number in the form a  + ib:
Ans:

Q2: Express the given complex number in the form a + ib: i9+ i19
Ans:

Q3: Express the given complex number in the form a  + ib: i–39
Ans:

Q4: Express the given complex number in the form a  + ib: 3(7 +  i7) +  i(7 +  i7)
Ans:

Q5: Express the given complex number in the form a +  ib: (1 – i) – (–1  i6)
Ans:

Q6: Express the given complex number in the form a  + ib:
Ans:

Q7: Express the given complex number in the form a +  ib:
Ans:

Q8: Express the given complex number in the form a  + ib:  (1 – i)4
Ans:

Q9: Express the given complex number in the form a  + ib:
Ans:

Q10: Express the given complex number in the form a +  ib:
Ans:

Q11: Find the multiplicative inverse of the complex number 4 – 3i
Ans:
Let z = 4 – 3i
Then,  = 4 +3and
Therefore, the multiplicative inverse of 4 – 3i is given by

Q12: Find the multiplicative inverse of the complex number
Ans:
Let z =

Therefore, the multiplicative inverse of  is given by

Q13: Find the multiplicative inverse of the complex number –i
Ans: Let z = –i

Therefore, the multiplicative inverse of –i is given by

Q14: Express the following expression in the form of a  + ib.

Ans:

Exercise Miscellaneous

Question 1: Evaluate:

Question 2: For any two complex numbers z1 and z2, prove that

Re (z1z2) = Re zRe z2 – Im z1 Im z2

Question 3: Reduce  to the standard form.

[ On Multiplying numerator and denomunator by (14 + 5i)

Question 4: If xiy =  prove that .

[ On Multiplying numerator and denomunator by (c + id)]

Question 5: Convert the following in the polar form:

(i) ,

(ii)

(i) Here,

Let cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2 (cos2 θ + sin2 θ) = 1+ 1

r2 (cos2 θ + sin2 θ) = 2
r2 = 2     [cos2 θ + sin2 θ = 1]

z = r cos θ +  i r sin θ

This is the required polar form.

(ii) Here,

Let cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2 (cos2 θ + sin2 θ) = 1+ 1
r2 (cos2 θ  +sin2 θ) = 2

r2 = 2                        [cos2 θ + sin2 θ = 1]

z = r cos θ  + i r sin θ

This is the required polar form.

Question 6: Solve the equation

This equation can also be written as

On comparing this equation with ax2 +  bx +  c = 0, we obtain

a = 9, b = –12, and c = 20

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576

Therefore, the required solutions are

Question 7: Solve the equation

This equation can also be written as

On comparing this equation with ax2 +  bx +  c = 0, we obtain

a = 2, b = –4, and c = 3

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8

Therefore, the required solutions are

Question 8: Solve the equation 27x2 – 10+ 1 = 0

The given quadratic equation is 27x2 – 10x + 1 = 0

On comparing the given equation with ax2 +  bx +  c = 0, we obtain

a = 27, b = –10, and c = 1

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8

Therefore, the required solutions are

Question 9: Solve the equation 21x2 – 28+ 10 = 0

The given quadratic equation is 21x2 – 28x + 10 = 0

On comparing the given equation with ax2 +  bx +  = 0, we obtain

a = 21, b = –28, and c = 10

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56

Therefore, the required solutions are

Question 9: If   find .

Question 10: If a +  ib = , prove that a2 +  b2 =

On comparing real and imaginary parts, we obtain

Hence, proved.

Question 10: Let  . Find

(i) ,

(ii)

(i)

On multiplying numerator and denominator by (2 – i), we obtain

On comparing real parts, we obtain

(ii)

On comparing imaginary parts, we obtain

Question 11: Find the modulus and argument of the complex number .

Let , then

On squaring and adding, we obtain

Therefore, the modulus and argument of the given complex number are   respectively.

Question 12: Find the real numbers x and y if (xiy) (3+5i) is the conjugate of –6 – 24i.

Let

It is given that,

Equating real and imaginary parts, we obtain

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

Putting the value of x in equation (i), we obtain

Thus, the values of and y are 3 and –3 respectively.

Question 13: Find the modulus of  .

Question 14: If (x  + iy)3 = u  + iv, then show that .

On equating real and imaginary parts, we obtain

Hence, proved.

Question 15: If α and β are different complex numbers with  = 1, then find .

Let α = a  + ib and β = x +  iy

It is given that,

Question 16: Find the number of non-zero integral solutions of the equation .

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.

Question 17: If (a  + ib) (c +  id) (e +  if) (g  + ih) = A + iB, then show that

(a2  + b2) (c2+   d2) (e2 +  f2) (g2 +  h2) = A2  +B2.

On squaring both sides, we obtain

(a2  + b2) (c2 +  d2) (e2 +  f2) (g2  + h2) = A2 + B2

Hence, proved.

Question 18: If , then find the least positive integral value of m.

Therefore, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).

The document NCERT Solutions Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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Mathematics (Maths) for JEE Main & Advanced

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FAQs on NCERT Solutions Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

 1. What are complex numbers?
Ans. Complex numbers are numbers that can be expressed in the form a + bi, where "a" and "b" are real numbers, and "i" is the imaginary unit (√-1).
 2. How do you add complex numbers?
Ans. To add complex numbers, simply add the real parts together and the imaginary parts together. For example, (3 + 2i) + (1 + 4i) = 4 + 6i.
 3. How do you multiply complex numbers?
Ans. To multiply complex numbers, use the distributive property and remember that i^2 = -1. For example, (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i.
 4. How do you find the conjugate of a complex number?
Ans. The conjugate of a complex number a + bi is a - bi. It is found by changing the sign of the imaginary part. For example, the conjugate of 3 + 2i is 3 - 2i.
 5. How do you divide complex numbers?
Ans. To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. Simplify the expression and then separate the real and imaginary parts to get the quotient.

Mathematics (Maths) for JEE Main & Advanced

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