OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Practicing OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(f) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(f)

Question 1.
Find the square root of the following complex numbers.
(i) 3 + 4i
(ii) – 8 + 6i
(iii) – 40 – 42i
(iv) i
(v) \(\left(\frac{2+3 i}{5-4 i}+\frac{2-3 i}{5-4 i}\right)\)
Solution:
(i) \(\sqrt{3+4 i}\) = x + iy; where x, y ∈ R
On squaring both sides ; we have
3 + 4i = (x + iy)²
⇒ 3 + 4i = x² – y² + 2ixy
On comparing real and imaginary parts on both sides ; we have
x² – y² = 3 …(1)
and 2xy = 4 …(2)
∴ x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{3^2+4^2}\) = 5 … (3)
On adding (1) and (3); we have
2x² = 8 ⇒ x² = 4 ⇒ x = ± 2
eqn. (3) – eqn. (1) gives ;
2y² = 2 ⇒ y = ± 1
Since xy be +ve
∴ x and y both are of same sign
Hence x = 2, y = 1 or x = – 2, y = – 1
∴ \(\sqrt{3+4 i}\) = 2 + i or – (2 + i)

(ii) \(\sqrt{- 8 + 6 i}\) = x + iy where x, y ∈ R
On squaring both sides ; we have
– 8 + 6i = (x + iy)² = x² – y² + 2ixy
On ωmparing real and imaginary parts on both sides, we get
x² – y² = – 8 … (1)
and 2xy = 6 … (2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{(-8)^2+6^2}\)
= \(\sqrt{64+36}\) = 10 … (3)
On adding (1) and (3) ; we have
2x² = 2 ⇒ x = ± 1
eqn. (3) – eqn. (1) gives ;
2y² = 18 ⇒ y² = 9 ⇒ y = ± 3
Since xy be +ve ∴ both x and y are of same sign.
∴ x = 1; y = 3 or x = – 1, y = – 3
Thus \(\sqrt{-8+6 i}\) = 1 + 3i or – (1 + 3i)

(iii) \(\sqrt{-40-42 i}\) = x – iy where x, y ∈ R
On squaring both sides ; we have
– 40 – 42i = (x – iy)² = x² – y² – 2ixy
On comparing real and imaginary parts on both sides ; we have
– 40 = x² – y² …(1)
and 2xy = 42 …(2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{(-40)^2+42^2}\)
= \(\sqrt{1600+1764}\)
= \(\sqrt{3364}\) = 58 … (3)
On adding (1) and (3) ; we have
2x² =18 ⇒ x = ± 3
eqn. (3) – eqn. (1) gives ;
2y² = 98 ⇒ y² = 49 ⇒ y = ± 7
Since xy be of +ve sign ∴ both x and y are of same sign
i.e. when x = 3, y = 7 or x = – 3, y = – 7
Thus, \(\sqrt{-40-42 i}\)
= 3 – 7i or – 3 + 7i
= ±(3 – 7i)

(iv) Let \(\sqrt{i}\) = x + iy;
On squaring; we have i = (x + iy)² = x² – y² + 2ixy
On comparing real and imaginary parts on both sides, we have
x² – y² = 0 …(1)
and 2xy = 1 …(2)
∴ x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{0^2+1^2}\) = 1 … (3)
On adding eqn. (1) and eqn. (3); we have

Question 2.
If ω is a cube root of unity, then
(i) ω + ω² = ….
(ii) 1 + ω = …
(iii) 1 + ω³ = ….
(iv) ω³ = ….
Solution:
Now ω be the cube root of unity
∴ ω = \(\frac{-1+\sqrt{3} i}{2}\) and ω² = \(\frac{-1-\sqrt{3} i}{2}\)

(i) 1 + ω² = \(\frac{-1+\sqrt{3} i}{2}+\frac{-1-\sqrt{3} i}{2}\)
= \(\frac { -2 }{ 2 }\) = – 1

(ii) 1 + ω = 1 + \(\frac{-1+\sqrt{3} i}{2}=\frac{2-1+\sqrt{3} i}{2}=\frac{1+\sqrt{3} i}{2}\) = – ω²

(iii) 1 + ω² = 1 + \(\frac{-1-\sqrt{3} i}{2}=\frac{2-1-\sqrt{3} i}{2}=\frac{1-\sqrt{3} i}{2}=-\left(\frac{-1+\sqrt{3} i}{2}\right)\) = – ω

(iv) ω³ = ω². ω = \(\left(\frac{-1-\sqrt{3} i}{2}\right)\left(\frac{-1+\sqrt{3} i}{2}\right)=\frac{(-1)^2-(\sqrt{3} i)^2}{4}=\frac{1+3}{4}\) = 1

Question 3.
If 1, ω, ω² are three cube roots of unity, prove that
(i) (1 + ω²)⁴ = ω
(ii) (1 + ω – ω²)³ = (1 – ω + ω²)³ = – 8
(iii) (1 – ω) (1 – ω²) = 3
(iv) \(\frac{1}{1+\omega}+\frac{1}{1+\omega^2}\) = 1
Solution:
(i) (1 + ω²)⁴ = (- ω)⁴= ω³ . ω = ω [∵ 1 + ω + ω² = 0 and ω³ = 1]

(ii) (1 + ω – ω²)² = (- ω² – ω²)³ = (- 2ω²)³ = – 8ω⁶ = – 8 (ω³)² = – 8
and (1 – ω + ω²)³ = (- ω – ω)³ = (- 2ω)³ = – 8ω³ = – 8 x 1 = – 8 [∵ 1 + ω + ω² = 0 and ω³ = 1]

(iii) (1 – ω) (1 – ω²) = 1 – ω – ω² + ω³ = 1 – (ω + ω²) + 1 [∵ ω³ = 1]
= 1 – (- 1) + 1 [∵ 1 + ω + ω² = 0]
= 1 + 1 + 1 = 3

(iv) \(\frac{1}{1+\omega}+\frac{1}{1+\omega^2}=\frac{1}{-\omega^2}+\frac{1}{-\omega}\) [∵ 1 + ω + ω² = 0]
= \(\frac{1+\omega}{-\omega^2}=\frac{-\omega^2}{-\omega^2}\) = 1

Question 4.
(i) (1 – ω – ω²)⁶ = 64
(ii) (1 + ω – ω²) (1 – ω + ω²) = 4
Solution:
(i) (1 – ω – ω²)⁶ = [1 – (ω + ω²)]⁶ = [1 – (- 1)]⁶ = 2⁶ = 64 [∵ 1 + ω + ω² = 0]

(ii) (1 + ω – ω²) (1 – ω + ω²) = (- ω² – ω²) (- ω – ω) [∵ 1 + ω + ω² = 0]
= (- 2ω²) (- 2ω)
= 4ω³
= 4 x 1 = 4

Question 5.
(3 + 5ω + 3ω²)⁶ = (3 + 5ω² + 3ω)⁶ = 64
Solution:
(3 + 5ω + 3ω²)⁶ = (3 + 5ω² + 3ω)⁶
= [3 (- ω) + 5ω]⁶ [∵ 1 + ω + ω² = 0]
= (2ω)⁶ = 64 (ω³)² = 64
and (3 + 5ω² + 3ω)⁶ = [3 (1 + ω) + 5ω²]⁶ = [3 (- ω²) + 5ω²]⁶ [∵ 1 + ω + ω² = 0]
= (2ω²)⁶ = 64ω¹² = 64 (ω³)⁴ = 64 [∵ ω³ = 1]

Question 6.
ω²⁸ + ω²⁹ + 1 = 0
Solution:
ω²⁸ + ω²⁹ + 1 = (ω³)⁹ . ω + (ω³)⁹ . ω² + 1
= 1⁹ . ω + 1⁹ . ω² + 1 [∵ ω³ = 1]
= ω + ω² + 1 = 0

Question 7.
Prove that \(\left(\frac{-1+i \sqrt{3}}{2}\right)^n+\left(\frac{-1-i \sqrt{3}}{2}\right)^n\) is equal to 2 if n be a multiple of 3 and is equal to – 1 if n be any other integer.
Or
If 1, ω, ω² are the cube roots of unity, prove that ωⁿ + ω²ⁿ = 2 or – 1 acωrding as n is a multiple of 3 or any other integer.
Solution:
We know that cube root of unity are 1, ω, ω²
where ω = \(\frac{-1+\sqrt{3} i}{2}\) and ω² = \(\frac{-1-\sqrt{3} i}{2}\)
∴ \(\left(\frac{-1+i \sqrt{3}}{2}\right)^n+\left(\frac{-1-i \sqrt{3}}{2}\right)^n=\omega^n+\left(\omega^2\right)^n=\omega^n+\omega^{2 n}\)

Case-I : When n be a multiple of 3
∴ n = 3k
∴ ωⁿ + ω²ⁿ = ω^3k + ω^6k = (ω³) + (ω³)^2k = 1^k + 1^2k =1 + 1 = 2

Case-II : When n be not a multiple of 3
∴ n = 3k + 1, 3k + 2

Subcase – I : When n = 3k + 1
ωⁿ + ω²ⁿ = ω^3k+1 + ω^2(3k+1) = ω^3k . ω + ω^6k . ω²
= 1 . ω + 1 . ω² = ω + ω² = – 1 [∵ ω + ω² +1 = 0 and ω^3k = (ω³)^k = 1]

Subcase-II : When n = 3k+2
∴ ωⁿ + ω²ⁿ = ω^3k+1 + ω^2(3k+2) = ω^3k. ω² + ω^6k + ω⁴ = (ω³)^k ω² + (ω³)^2k . ω³ . ω
= ω² + ω = – 1 [∵ (ω³)^k = 1^k = 1]
Thus, ωⁿ + ω²ⁿ = 2 or – 1
according as n is a multiple of 3 or any other integer.

Prove the following:

Question 8.
(1 – ω + ω²) (1 + ω – ω²) (1 – ω – ω²) = 8.
Solution:
(1 – ω + ω²) (1 + ω – ω²) (1 – ω – ω²) = (1 + ω² – ω) (1 + ω – ω²) (1 – (ω + ω²))
= (- ω – ω) (- ω² – ω²) (1 – (- 1)) [∵ 1 + ω + ω² = 0]
= (- 2ω) (- 2ω²)² = 8ω³ = 8 [∵ ω³ = 1]

Question 9.
(i) (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸) …. to 2n factors = 1
(ii) (1 – ω + ω²) (1 – ω² + ω⁴) (1 – ω⁴ + ω⁸)…… to 2n factors = 2²ⁿ.
Solution:
(i) L.H.S = (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸) …. to 2n factors
= (1 + ω) (1 + ω²) (1 + ω³ . ω) (1 + ω⁶ . ω²) … 2n factors
= (1 + ω) (1 + ω²) (1 + ω) (1 + ω²) …. 2n factors
= (1 + ω)ⁿ (1 + ω²)ⁿ = [(1 + ω) (1 + ω²)]ⁿ
= [1 + ω + ω² + ω³]ⁿ = [0 + 1]ⁿ = 1ⁿ = 1 [∵ 1 + ω + ω² = 0 and ω³ = 1]

(ii) (1 – ω + ω²) (1 – ω² + ω⁴) (1 – ω⁴ + ω⁸) to 2n factors
= (1 – ω + ω²) (1 – ω² + ω³ . ω) (1 – ω³ . ω + ω⁶ . ω²) to 2n factors
= (1 – ω + ω²) (1 – ω² + ω) (1 – ω + ω²)…. 2n factors [∵ ω³ = 1]
= [(1 – ω + ω²) …. n factors] [(1 – ω² + ω) …. n factors
= (- 2ω)ⁿ (- 2ω²)ⁿ [∵ 1 + ω + ω² = 0]
= 2ⁿ (ω . ω²)ⁿ (- 1)²ⁿ
= 2²ⁿ (ω³)ⁿ
= 2²ⁿ1ⁿ
= 2²ⁿ

Question 10.
\(\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}\) = ω
Solution:

Question 11.
\(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}\) = – 1
Solution:

Question 12.
If ω is a cube root of unity and n is a positive integer which is not a multiple of 3, then show that (1 + ωⁿ + ω²ⁿ) = 0.
Solution:
When n is not a multiple of 3
∴ n be of the form 3k + 1 or 3k + 2
Case – I. When n = 3k + 1
1 + ωⁿ + ω²ⁿ = 1 + ω^3k+1 + ω^6k+2 = 1 + (ω³)^k ω + (ω³)^2k . ω² = 1 + 1^k . ω + 1^2k ω²
= 1 + ω + ω² = 0

Case-II. When n = 3k+2
1 + ωⁿ + ω²ⁿ = 1 + ω^3k+2 + ω^2(3k+2)= 1 + ω^3k . ω² + ω^6k . ω⁴
= 1 + (ω³)^k . ω² + (ω³)^2k . ω³ω = 1 + ω² + ω = 0 [∵ ω³ = 1]

Question 13.
Show that (x + ωy + ω²z) (x + ω²y + ωz) = x² + y² + z² – yz – zx – xy.
Solution:
L.H.S = (x + ωy + ω²z) (x + ω²y + ωz)
= x² + (ω² + ω) xy + xz (ω + ω²) + ω³y² + yz (ω² + ω⁴) + ω³z²
= x² – xy + xz(- 1) + y² + yz (- 1) + z² [∵ 1 + ω + ω² = 1]
= x² + y² + z² – xy – yz – zx = R.H.S

Question 14.
Show that x³ + y³ = (x + y) (ωx + ω²y) (ω²x + ωy).
Solution:
R.H.S = (x + y) (ωx + ω²y) (ω²x + ω²y) = (x + y) (ω³x² + ω²xy + ω⁴xy + ω³y²)
= (x + y) [1 . x² + xy (ω² + ω³ . ω) + 1 . y²] [∵ ω³ = 1]
= (x + y) (x² + xy (ω² + ω) + y²) = (x + y) (x² – xy + y²) [∵ 1 + ω + ω² = 1]
= x³ + y³ = L.H.S

Question 15.
If 1, ω, ω² are cube roots of unity, prove that 1, ω² are vertices of an equilateral triangle.
Solution:
Since 1, ω and ω² are the cube root of unity.

Thus, 1, ω and ω² are the vertices of an equilateral triangle.