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 = ω3k + ω6k = (ω³) + (ω³)2k = 1k + 12k =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 = ω3k+1 + ω2(3k+1) = ω3k . ω + ω6k . ω²
= 1 . ω + 1 . ω² = ω + ω² = – 1 [∵ ω + ω² +1 = 0 and ω3k = (ω³)k = 1]

Subcase-II : When n = 3k+2
∴ ωn + ω2n = ω3k+1 + ω2(3k+2) = ω3k. ω² + ω6k + ω4 = (ω³)k ω² + (ω³)2k . ω³ . ω
= ω² + ω = – 1 [∵ (ω³)k = 1k = 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 + ω3k+1 + ω6k+2 = 1 + (ω³)k ω + (ω³)2k . ω² = 1 + 1k . ω + 12k ω²
= 1 + ω + ω² = 0

Case-II. When n = 3k+2
1 + ωn + ω2n = 1 + ω3k+2 + ω2(3k+2)= 1 + ω3k . ω² + ω6k . 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.