RBSE Solutions for Class 11 Maths Chapter 5 Complex Numbers

RBSE Solutions for Class 11 Maths Chapter 5 Complex Numbers

Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Ex 5.1

Question 1.
Write the following in simplest form:
(i) i⁵²
(ii) √-2 √-3
(iii) (1 + i)⁵ (1 – i)⁵
Solution:

Question 2.
Find the additive and multiplicative inverse of following numbers:
(i) 1 + 2i
(ii) 1/(3 + 4i)
(iii) (3 + i)²
Solution:
(i) Let z = 1 + 2i
Then additive inverse = -z = -(1 + 2i) = -1 – 2i
And multiplicative inverse = 1/z

Question 3.
Find the conjugate number of complex number (2+i)3/3+i
Solution:

Question 4.
Find the modulus of the following:
(i) 4 + i
(ii) -2 – 3i
(iii) 1/(3 – 2i).
Solution:

Question 5.
If a² + b² = 1 then find the value of 1+b+ia/1+b−ia
Solution:

Question 6.
If a = cos θ + i sin θ, then find the value of (1+a)/(1−a)
Solution:

Question 7.
Find the value of x and y which satisfy the equation

Solution:


Comparing the real and imaginary part on both sides
4x + 9y – 3 = 0 and 4x + 9y = 3 …..(i)
2x – 7y – 3 = 10 and 2x – 7y = 13 ……(ii)
Solving equation (i) and (ii), we get
x = 3 and y = -1

Question 8.
If z₁ and z₂ are any two complex numbers, then prove that
|z₁ + z₂|² + |z₁ – z₂|² = 2|z₁|² + 2|z₂|².
Solution:

Question 9.

Solution:

Question 10.
If (x + iy)^1/3 = a + ib, then prove that x/a + y/b = 4(a² – b²).
Solution:

Question 11.

Solution:

Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Ex 5.2

Question 1.
Find the arguments of the following numbers:

Solution:

Question 2.
Express the following complex numbers into the polar form:

Solution:

Question 3.
If z₁ and z₂ are two non-zero complex numbers, then prove that arg z₁z2¯ = arg z₁ – arg z₂
Solution:

Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Ex 5.3

Question 1.
Find the square root of the following complex numbers:
(i) -5 + 12i
(ii) 8 – 6i
(iii) -i
Solution:
Formula

Question 2.
Find the value of √4+3√−20 + √4−3√−20
Solution:
From formula √a+ib=

Question 3.
Find the cube root of the following:
(i) -216
(ii) -512
Solution:
(i) -216 = (-6) (1)^1/3 = -6, -6ω, -6ω²
Hence, cube root of -216 = -6, -6ω, -6ω²
(ii) -512 = (-8) (1)^1/3 = -8, -8ω, -8ω²
Hence, cube root of -512 = -8, -8ω, – 8ω²

Question 4.
Prove that:
(i) 1 + ωⁿ + ω²ⁿ = 0, whereas n = 2, 4
(ii) 1 + ωⁿ + ω²ⁿ = 3, whereas n is multiple of 3.
Solution:

Question 5.
Prove that:

(ii) (1 + 5ω² + ω) (1 + 5ω + ω²) (5 + ω + ω²) = 64
Solution:
(i) ω is cube root of unity.

Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Ex 5.4

Question 1.
Find the solution of following equations by vedic method
(i) x² + 4x + 13 = 0
(ii) 2x² + 5x + 4 = 0
(iii) ix² + 4x – 15/2 = 0
Solution:
(i) Given equation
x² + 4x + 13 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = 4, c = 13
First derivative = Discriminant

Question 2.
Find quadratic equation which has the following roots
(i) 5 and -2
(ii) 1 + 2i
Solution:
(i) root α = 5 and β = – 2
Then, sum of roots = α + β = 5 – 2
⇒ α + β = 3
and product of roots = αβ = 5 × (-2)
⇒ αβ = -10
Hence, required equation whose roots are 5 and -2.
x² – (sum of roots) x + product of roots = 0
⇒ x² – 3x + (-10) = 0
⇒ x² – 3x – 10 = 0
Hence, required equation is x² – 3x – 10 = 0 whose roots are 5 and -2.

(ii) Roots α = 1 + 2i and β = 1 – 2i
Then, sum of roots α + β = 1 + 2i + 1 – 2i = 2
and product of roots αβ = (1 + 2i)(1 – 2i) = 1 – 4i² = 1 + 4 = 5
Hence, required equation whose roots are 1 + 2i and 1 – 2i,
x² – (sum of roots) x + Product of roots = 0
⇒ x² – 2x + 5 = 0.
Remark: If one root is 1 + 2i, then second not will be 1 – 2i.

Question 3.
If one root of equation x² – px + q = 0 is twice the other then prove that 2p² = 9q.
Solution:
Given equation
x² – px + q = 0
Let its roots be α and β, then
According to question α = 2β then

Question 4.
Find that condition for which equation ax² + bx + c = 0 has roots in the ratio m : n.
Solution:
Given equation,
ax² + bx + c = 0
Let its roots be α and β, then
According to question,
α : β = m : n

Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Miscellaneous Exercise

Question 1.
Real and imaginary parts of complex number 1+i1−i respectively are:
(A) 1, 1
(B) 0, 0
(C) 0, 1
(D) 1, 0
Solution:

Real part = 0
Imaginary part = 1
Hence, option (C) is correct.

Question 2.
If 2 + (2a + 5ib) = 8 + 10i, then:
(A) a = 2, b = 3
(B) a = 2, b = -3
(C) a = 3, b = 2
(D) a = 3, b = -2
Solution:
2 + (2a + 5ib) = 8 + 10i
⇒ (2 + 2a) + 5bi = 8 + 10i
On comparing
2 + 2a = 8
⇒ 2a =6
⇒ a = 3
and 5b = 10
⇒ b = 2
So, a = 3 and b = 2
Hence, option (C) is correct.

Question 3.
The multiplicative inverse of 3 – i is:

Solution:
Let z = 3 – i
Let multiplicative inverse of z

Question 4.

Solution:

Question 5.
If z₁, z₂ ∈ C then which statement is true:
(A) |z₁ – z₂| ≥ |z₁| + |z₂|
(B) |z₁ – z₂| ≤ |z₁| + |z₂|
(C) |z₁ + z₂| ≥ |z₁ – z₂|
(D) |z₁ + z₂| ≤ |z₁ – z₂|
Solution:
statement |z₁ – z₂| ≤ |z₁| + |z₂| is true.
Hence, option (B) is correct.

Question 6.
If |z – 3| = |z + 3|, then z lies in
(A) at x-axis
(B) at y-axis
(C) on line x = y
(D) on line x = -y
Solution:
Let z = x + iy

It is situated on the axis.
Hence, option (B) is correct.

Question 7.
Write the principal argument of -2.
Solution:
-2 = -2 + 0i
Let -2 + 0i = r(cos θ + i sin θ)
⇒ r cos θ = -2 ….(i)
r sin θ = 0 ………(ii)
⇒ tan θ = 0/−2 = 0
Hence, principal arument of -z is π.

Question 8.
Write polar form of 5√3/2+5/2 i
Solution:

Question 9.
What is the value of 4 + 5ω⁴ + 3ω⁵?
Solution:

Question 10.
Write 1/1−cosθ+isinθ in the form of a + ib.
Solution:

Question 11.
What is the number of non-zero integer roots of |1 – i|^x = 2^x?
Solution:

Comparing power of both sides
x/2 = x, which is not possible (because x is not zero)
Thus, the root of equation |1 – i|^x = 2^x is not possible.
Hence, the number of roots is zero.

Question 12.
If z₁, z₂ ∈ C prove that
(i) |z₁ – z₂| ≤ |z₁| + |z₂|
(ii) |z₁ + z₂| ≥ |z₁| – |z₂|
(iii) |z₁ + z₂ + ….. + z_n| ≤ |z₁| + |z₂| + ……. + |z_n|
Solution:

Question 13.
If |z₁| = 1 = |z₂| then prove that

Solution:
According to question,

Question 14.

Solution:

Question 15.
If |z₁| = |z₂| and arg z₁ + arg z₂ = 0 then prove that z₁ = z2¯.
Solution:

Question 16.
If θ₁ and θ₂ are arguments of complex numbers z₁ and z₂ respectively then prove that:
z₁z2¯ + z1¯z₂ = 2|z₁||z₂| cos(θ₁ – θ₂)
Solution:

Question 17.
Prove that:
(i) (a + bω + cω²) (a + bω² + cω) = a² + b² + c² – ab – bc – ca
(ii) (a + b + c) (a + bω + cω²) (a + bω² + cω) = a³ + b³ + c³ – 3abc
Solution:

Question 18.
If α, β are two complex numbers and |β| = 1 then prove that ∣∣β−α/1−α¯β∣∣=1
Solution:

Question 19.
If α and β roots of equation px² – qx – r = 0, then find the equation which has roots 1/α and 1/β
Solution:
α, β are roots of equation px² – qx – r = 0

⇒ rx² + qx – p = 0
Hence, required equation is rx² + qx – p = 0

Question 20.
Find conditions for which equation lx² – 2mx + n = 0 has roots such that one root is P times the other roots.
Solution:
Given equation lx² – 2mx + n = 0
Let it’s one root is a and another root is Pα.
Then, the sum of roots

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