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Class 11 (FYJC / HSC)Mathematics & Statistics2026-27 Syllabus

Chapter 1 Complex Numbers Ex 1.4 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 1 Complex Numbers Ex 1.4. Step-by-step solved exercises, numerical problems, and digest answers.

8 Solved Questions• 24 Diagrams• 1640 words

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4

Question 1 Maharashtra Board Solution
Find the value of (i) ω18 (ii) ω21 (iii) ω-30 (iv) ω-105
Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution
If ω is the complex cube root of unity, show that (i) (2 – ω)(2 – ω2) = 7
Solution & Step-by-Step Answer:
ω is the complex cube root of unity. ω3 = 1 and 1 + ω + ω2 = 0 Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1 L.H.S. = (2 – ω)(2 – ω2) = 4 – 2ω2 – 2ω + ω3 = 4 – 2(ω2 + ω) + 1 = 4 – 2(-1) + 1 = 4 + 2 + 1 = 7 = R.H.S.

(ii) (1 + ω – ω2)6= 64
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(iii) (1 + ω)3– (1 + ω2)3= 0
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(iv) (2 + ω + ω2)3– (1 – 3ω + ω2)3= 65
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(v) (3 + 3ω + 5ω2)6– (2 + 6ω + 2ω2)3= 0
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(vi) = ω2
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(vii) (a + b) + (aω + bω2) + (aω2+ bω) = 0
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(viii) (a – b)(a – bω)(a – bω2) = a3– b3
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

(ix) (a + b)2+ (aω + bω2)2+ (aω2+ bω)2= 6ab
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1

Question 3 Maharashtra Board Solution
If ω is the complex cube root of unity, find the value of (i) ω +
Solution & Step-by-Step Answer:
ω is the complex cube root of unity. ω3 = 1 and 1 + ω + ω2 = 0 Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1

(ii) ω2+ ω3+ ω4
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1
ω2+ ω3+ ω4
= ω2(1 + ω + ω2)
= ω2(0)
= 0

(iii) (1 + ω2)3
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1
(1 + ω2)3
= (-ω)3
= -ω3
= -1

(iv) (1 – ω – ω2)3+ (1 – ω + ω2)3
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1
(1 – ω – ω2)3+ (1 – ω + ω2)3
= [1 – (ω + ω2)]3+ [(1 + ω2) – ω]3
= [1 – (-1)]2+ (-ω – ω)3
= 23+ (-2ω)3
= 8 – 8ω3
= 8 – 8(1)
= 0

(v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)
Solution:
ω is the complex cube root of unity.
ω3= 1 and 1 + ω + ω2= 0
Also, 1 + ω2= -ω, 1 + ω = -ω2and ω + ω2= -1
(1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)
= (1 + ω)(1 + ω2)(1 + ω)(1 + ω2) …..[∵ ω3= 1, ω4= ω]
= (-ω2)(-ω)(-ω2)(-ω)
= ω6
= (ω3)2
= (1)2
= 1

Question 4 Maharashtra Board Solution
If α and β are the complex cube roots of unity, show that (i) α2 + β2 + αβ = 0 (ii) α4 + β4 + α-1β-1 = 0
Solution & Step-by-Step Answer:
α and β are the complex cube roots of unity.

Question 5 Maharashtra Board Solution
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are complex cube roots of unity, show that xyz = a3 + b3.
Solution & Step-by-Step Answer:
x = a + b, y = αa + βb, z = aβ + bα α and β are the complex cube roots of unity. ∴ α = and β =

Question 6 Maharashtra Board Solution
Find the equation in cartesian coordinates of the locus of z if (i) |z| = 10
Solution & Step-by-Step Answer:
Let z = x + iy |z| = 10 |x + iy| = 10 = 10 ∴ x2 + y2 = 100

(ii) |z – 3| = 2
Solution:
Let z = x + iy
|z – 3| = 2
|x + iy – 3| = 2
|(x – 3) + iy| = 2
= 2
∴ (x – 3)2+ y2= 4

(iii) |z – 5 + 6i| = 5
Solution:
Let z = x + iy
|z – 5 + 6i| = 5
|x + iy – 5 + 6i| = 5
|(x – 5) + i(y + 6)| = 5
= 5
∴ (x – 5)2+ (y + 6)2= 25

(iv) |z + 8| = |z – 4|
Solution:
Let z = x + iy
|z + 8| = |z – 4|
|x + iy + 8| = |x + iy – 4|
|(x + 8) + iy | = |(x – 4) + iy|

(x + 8)2+ y2= (x – 4)2+ y2
x2+ 16x + 64 + y2= x2– 8x + 16 + y2
16x + 64 = -8x + 16
24x + 48 = 0
∴ x + 2 = 0

(v) |z – 2 – 2i | = |z + 2 + 2i|
Solution:
Let z = x + iy
|z – 2 – 2i| = |z + 2 + 2i|
|x + iy – 2 – 2i | = |x + iy + 2 + 2i |
|(x – 2) + i(y – 2)| = |(x + 2) + i(y + 2)|

(x – 2)2+ (y – 2)2= (x + 2)2+ (y + 2)2
x2– 4x + 4 + y2– 4y + 4 = x2+ 4x + 4 + y2+ 4y + 4
-4x – 4y = 4x + 4y
8x + 8y = 0
x + y = 0
y = -x

(vi)
Solution:
Let z = x + iy

x2+ (y + 3)2= x2+ (y – 6)2
y2+ 6y + 9 = y2– 12y + 36
18y – 27 = 0
2y – 3 = 0

Question 7 Maharashtra Board Solution
Use De Moivre’s theorem and simplify the following: (i)
Solution & Step-by-Step Answer:

(ii)
Solution:

(iii)
Solution:

Question 8 Maharashtra Board Solution
Express the following in the form a + ib, a, b ∈ R, using De Moivre’s theorem. (i) (1 – i)5
Solution & Step-by-Step Answer:

(ii) (1 + i)6
Solution:

(iii) (1 – √3 i)4
Solution:

(iv) (-2√3 – 2i)5
Solution:

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Chapter 1 Complex Numbers Ex 1.3
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Chapter 1 Complex Numbers Miscellaneous Exercise 1