{"id":563218,"date":"2021-12-21T06:59:59","date_gmt":"2021-12-21T06:59:59","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=563218"},"modified":"2021-12-28T06:12:22","modified_gmt":"2021-12-28T06:12:22","slug":"maharashtra-board-solutions-class-11-arts-science-maths-part-2-chapter-1-complex-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-11-arts-science-maths-part-2-chapter-1-complex-numbers\/","title":{"rendered":"Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 1- Complex Numbers"},"content":{"rendered":"\n

Class 11: Maths Chapter 1 solutions. Complete Class 11 Maths Chapter 1 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 1- Complex Numbers<\/h2>\n\n\n\n

Maharashtra Board 11th Maths Chapter 1, Class 11 Maths Chapter 1 solutions<\/p>\n\n\n\n

Ex 1.1<\/h4>\n\n\n\n

Question 1.
Simplify:<\/strong>
(i) \u221a-16 + 3\u221a-25 + \u221a-36 \u2013 \u221a-625
Solution:
<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

= 4i + 3(5i) + 6i \u2013 25i
= 25i \u2013 25i
= 0<\/p>\n\n\n\n

(ii) 4\u221a-4 + 5\u221a-9 \u2013 3\u221a-16
Solution:
<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 2.
Write the conjugates of the following complex numbers
(i) 3 + i
Solution:<\/strong>
Conjugate of (3 + i) is (3 \u2013 i).<\/p>\n\n\n\n

(ii) 3 \u2013 i
Solution:<\/strong>
Conjugate of (3 \u2013 i) is (3 + i).<\/p>\n\n\n\n

(iii) \u221a-5 \u2013 \u221a7 i
Solution:<\/strong>
Conjugate of (\u221a-5 \u2013 \u221a7 i) is (\u221a-5 + \u221a7 i).<\/p>\n\n\n\n

(iv) -\u221a-5
Solution:<\/strong>
-\u221a-5 = -\u221a5 \u00d7 \u221a-1 = -\u221a5 i
Conjugate of (-\u221a-5) is \u221a5 i<\/p>\n\n\n\n

(v) 5i
Solution:<\/strong>
Conjugate of (5i) is (-5i).<\/p>\n\n\n\n

(vi) \u221a5 \u2013 i
Solution:<\/strong>
Conjugate of (\u221a5 \u2013 i) is (\u221a5 + i).<\/p>\n\n\n\n

(vii) \u221a2 + \u221a3 i
Solution:<\/strong>
Conjugate of (\u221a2 + \u221a3 i) is (\u221a2 \u2013 \u221a3 i)<\/p>\n\n\n\n

(viii) cos \u03b8 + i sin \u03b8
Solution:<\/strong>
Conjugate of (cos \u03b8 + i sin \u03b8) is (cos \u03b8 \u2013 i sin \u03b8)<\/p>\n\n\n\n

Question 3.
Find a and b if
(i) a + 2b + 2ai = 4 + 6i
Solution:<\/strong>
a + 2b + 2ai = 4 + 6i
Equating real and imaginary parts, we get
a + 2b = 4 \u2026..(i)
2a = 6 \u2026\u2026(ii)
\u2234 a = 3
Substituting, a = 3 in (i), we get
3 + 2b = 4
\u2234 b = 12
\u2234 a = 3 and b = 12<\/p>\n\n\n\n

Check:
<\/strong>For a = 3 and b = 12
Consider, L.H.S. = a + 2b + 2ai
= 3 + 2(12) + 2(3)i
= 4 + 6i
= R.H.S.<\/p>\n\n\n\n

(ii) (a \u2013 b) + (a + b)i = a + 5i
Solution:<\/strong>
(a \u2013 b) + (a + b)i = a + 5i
Equating real and imaginary parts, we get
a \u2013 b = a \u2026\u2026(i)
a + b = 5 \u2026\u2026(ii)
From (i), b = 0
Substituting b = 0 in (ii), we get
a + 0 = 5
\u2234 a = 5
\u2234 a = 5 and b = 0<\/p>\n\n\n\n

(iii) (a + b) (2 + i) = b + 1 + (10 + 2a)i
Solution:<\/strong>
(a + b) (2 + i) = b + 1 + (10 + 2a)i
2(a + b) + (a + b)i = (b + 1) + (10 + 2a)i
Equating real and imaginary parts, we get
2(a + b) = b + 1
\u2234 2a + b = 1 \u2026\u2026(i)
and a + b = 10 + 2a
-a + b = 10 \u2026\u2026.(ii)
Subtracting equation (ii) from (i), we get
3a = -9
\u2234 a = -3
Substituting a = \u2013 3 in (ii), we get
-(-3) + b = 10
\u2234 b = 7
\u2234 a = -3 and b = 7<\/p>\n\n\n\n

(iv) abi = 3a \u2013 b + 12i
Solution:<\/strong>
abi = 3a \u2013 b + 12i
\u2234 0 + abi = (3a \u2013 b) + 12i
Equating real and imaginary parts, we get
3a \u2013 b = 0
\u2234 3a = b \u2026..(i)
and ab = 12<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Equating real and imaginary parts, we get
-6x \u2013 15y = 0 and y + 6 = 0
-6x \u2013 15y = 0 and y = -6
-6x \u2013 15(-6) = 0
-6x + 90 = 0
\u2234 x = 15
\u2234 x + y = 15 \u2013 6 = 9<\/p>\n\n\n\n

Ex 1.2<\/h4>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Given equation is x2<\/sup> \u2013 4x + 13 = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b2<\/sup> \u2013 4ac
= (-4)2<\/sup> \u2013 4 \u00d7 1 \u00d7 13
= 16 \u2013 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by
\"Maharashtra
\u2234 The roots of the given equation are 2 + 3i and 2 \u2013 3i.<\/p>\n\n\n\n

Question 3.
Solve the following quadratic equations:
(i) x2<\/sup> + 3ix + 10 = 0
Solution:<\/strong>
Given equation is x2<\/sup> + 3ix + 10 = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b2<\/sup> \u2013 4ac
= (3i)2<\/sup> \u2013 4 \u00d7 1 \u00d7 10
= 9i2<\/sup> \u2013 40
= -9 \u2013 40 \u2026\u2026[\u2235 i2<\/sup> = -1]
= -49 < 0
So, the given equation has complex roots.
These roots are given by
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

\u2234 x = 2i or x = -5i
\u2234 The roots of the given equation are 2i and -5i.<\/p>\n\n\n\n

(ii) 2x2<\/sup> + 3ix + 2 = 0
Solution:<\/strong>
Given equation is 2x2<\/sup> + 3ix + 2 = 0
Comparing with ax + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b2<\/sup> \u2013 4ac
= (3i)2<\/sup> \u2013 4 \u00d7 2 \u00d7 2
= 9i2<\/sup> \u2013 16
= -9 \u2013 16 \u2026..[\u2235 i2<\/sup> = -1]
= -25 < 0
So, the given equation has complex roots.
These roots are given by
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

\u2234 The roots of the given equation are 12i and -2i.<\/p>\n\n\n\n

(iii) x2<\/sup> + 4ix \u2013 4 = 0
Solution:<\/strong>
Given equation is x2<\/sup> + 4ix \u2013 4 = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b2<\/sup> \u2013 4ac
= (4i)2<\/sup> \u2013 4 \u00d7 1 \u00d7 (-4)
= 16i2<\/sup> + 16
= -16 + 16 \u2026..[\u2235 i2<\/sup> = -1]
= 0
So, the given equation has equal roots.
These roots are given by
\"Maharashtra
\u2234 x = -2i
\u2234 The root of the given equation is -2i.<\/p>\n\n\n\n

(iv) ix2<\/sup> \u2013 4x \u2013 4i = 0
Solution:<\/strong>
ix2<\/sup> \u2013 4x \u2013 4i = 0
Multiplying throughout by i, we get
i2x2<\/sup> \u2013 4ix \u2013 4i2<\/sup> = 0
-x2<\/sup> \u2013 4ix + 4 = 0 \u2026[\u2235 i2<\/sup> = -1]
x2<\/sup> + 4ix \u2013 4 = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b2<\/sup> \u2013 4ac
= (4i)2<\/sup> \u2013 4 \u00d7 1 \u00d7 (-4)
= 16i2<\/sup> + 16
= -16 + 16
= 0
So, the given equation has equal roots.
These roots are given by
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

\u2234 The root of the given equation is -2i<\/p>\n\n\n\n

Question 4.
Solve the following quadratic equations:
(i) x2<\/sup> \u2013 (2 + i) x \u2013 (1 \u2013 7i) = 0
Solution:<\/strong>
Given equation is x2<\/sup> \u2013 (2 + i)x \u2013 (1 \u2013 7i) = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 \u2013 7i)
Discriminant = b2<\/sup> \u2013 4ac
= [-(2 + i)]2<\/sup> \u2013 4 \u00d7 1 \u00d7 -(1 \u2013 7i)
= 4 + 4i + i2<\/sup> + 4 \u2013 28i
= 4 + 4i \u2013 1 + 4 \u2013 28i \u2026..[\u2235 i2<\/sup> = \u2013 1]
= 7 \u2013 24i
So, the given equation has complex roots.
These roots are given by
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Let 7\u221224i\u2212\u2212\u2212\u2212\u2212\u2212\u221a = a + bi, where a, b \u2208 R
Squaring on both sides, we get
7 \u2013 24i = a2<\/sup> + i2<\/sup>b2<\/sup> + 2abi
7 \u2013 24i = a2<\/sup> \u2013 b2<\/sup> + 2abi
Equating real and imaginary parts, we get
a2<\/sup> \u2013 b2<\/sup> = 7 and 2ab = -24
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(ii) x2<\/sup> \u2013 (3\u221a2 + 2i) x + 6\u221a2 i = 0
Solution:<\/strong>
Given equation is x2<\/sup> \u2013 (3\u221a2 + 2i) x + 6\u221a2 i = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = -(3\u221a2 + 2i), c = 6\u221a2i
Discriminant = b2<\/sup> \u2013 4ac
= [-(3\u221a2 + 2i)]2<\/sup> \u2013 4 \u00d7 1 \u00d7 6\u221a2 i
= 18 + 12\u221a2i + 4i2<\/sup> \u2013 24\u221a2 i
= 18 \u2013 12\u221a2 i \u2013 4 \u2026\u2026[\u2235 i2<\/sup> = -1]
= 14 \u2013 12\u221a2 i
So, the given equation has complex roots.
These roots are given by
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(iii) x2<\/sup> \u2013 (5 \u2013 i) x + (18 + i) = 0
Solution:<\/strong>
Given equation is x2<\/sup> \u2013 (5 \u2013 i)x + (18 + i) = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 1, b = -(5 \u2013 i), c = 18 + i
Discriminant = b2<\/sup> \u2013 4ac
= [-(5 \u2013 i)]2<\/sup> \u2013 4 \u00d7 1 \u00d7 (18 + i)
= 25 \u2013 10i + i2<\/sup> \u2013 72 \u2013 4i
= 25 \u2013 10i \u2013 1 \u2013 72 \u2013 4i \u2026\u2026[\u2235 i2<\/sup> = -1]
= -48 \u2013 14i
So, the given equation has complex roots.
These roots are given by
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(iv) (2 + i) x2<\/sup> \u2013 (5 \u2013 i) x + 2(1 \u2013 i) = 0
Solution:<\/strong>
Given equation is (2 + i) x2<\/sup> \u2013 (5 \u2013 i) x + 2(1 \u2013 i) = 0
Comparing with ax2<\/sup> + bx + c = 0, we get
a = 2 + i, b = -(5 \u2013 i), c = 2(1 \u2013 i)
Discriminant = b2<\/sup> \u2013 4ac
= [-(5 \u2013 i)]2<\/sup> \u2013 4 \u00d7 (2 + i) \u00d7 2(1 \u2013 i)
= 25 \u2013 10i + i2<\/sup> \u2013 8(2 + i) (1 \u2013 i)
= 25 \u2013 10i + i2<\/sup> \u2013 8(2 \u2013 2i + i \u2013 i2<\/sup>)
= 25 \u2013 10i \u2013 1 \u2013 8(2 \u2013 i + 1) \u2026..[\u2235 i2<\/sup> = -1]
= 25 \u2013 10i \u2013 1 \u2013 16 + 8i \u2013 8
= -2i
So, the given equation has complex roots.
These roots are given by<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Let \u22122i\u2212\u2212\u2212\u221a = a + bi, where a, b \u2208 R
Squaring on both sides, we get
-2i = a2<\/sup> + b2<\/sup>i2<\/sup> + 2abi
-2i = a2<\/sup> \u2013 b2<\/sup> + 2abi
Equating real and imaginary parts, we get
a2<\/sup> \u2013 b2<\/sup> = 0 and 2ab = -2
a2<\/sup> \u2013 b2<\/sup> = 0 and b = \u22121a
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 5.
Find the value of
(i) x3<\/sup> \u2013 x2<\/sup> + x + 46, if x = 2 + 3i
Solution:<\/strong>
x = 2 + 3i
x \u2013 2 = 3i
(x \u2013 2)2<\/sup> = 9i2<\/sup>
x2<\/sup> \u2013 4x + 4 = 9(-1) \u2026..[\u2235 i2<\/sup> = -1]
x2<\/sup> \u2013 4x + 13 = 0 \u2026..(i)
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Dividend = Divisor \u00d7 Quotient + Remainder
\u2234 x3<\/sup> \u2013 x2<\/sup> + x + 46 = (x2<\/sup> \u2013 4x + 13) (x + 3) + 7
= 0(x + 3) + 7 \u2026..[from(i)]
= 7<\/p>\n\n\n\n

Alternate Method:
x = 2 + 3i
\u03b1 = 2 + 3i, \u03b1\u00af = 2 \u2013 3i
\u03b1\u03b1\u00af = (2 + 3i)(2 \u2013 3i)
= 4 \u2013 6i + 6i \u2013 9i2<\/sup>
= 4 \u2013 9(-1)
= 4 + 9
= 13
\u03b1 + \u03b1\u00af = 2 + 3i + 2 \u2013 3i = 4
\u2234 Standard form of quadratic equation,
x2<\/sup> \u2013 (Sum of roots) x + Product of roots = 0
x2<\/sup> \u2013 4x + 13 = 0
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Dividend = Divisor \u00d7 Quotient + Remainder
\u2234 x3<\/sup> \u2013 x2<\/sup> + x + 46 = (x2<\/sup> \u2013 4x + 13).(x + 3) + 7
= 0(x + 3) + 7 \u2026..[From (i)]
= 7<\/p>\n\n\n\n

(ii) 2x3<\/sup> \u2013 11x2<\/sup> + 44x + 27, if x = 253\u22124i
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Dividend = Divisor \u00d7 Quotient + Remainder
2x3<\/sup> \u2013 11x2<\/sup> + 44x + 27 = (x2<\/sup> \u2013 6x + 25)(2x + 1) + 2
= 0.(2x + 1) + 2 \u2026..[From (i)]
= 0 + 2
= 2<\/p>\n\n\n\n

(iii) x3<\/sup> + x2<\/sup> \u2013 x + 22, if x = 51\u22122i
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n


Dividend = Divisor \u00d7 Quotient + Remainder
x3<\/sup> + x2<\/sup> \u2013 x + 22 = (x2<\/sup> \u2013 2x + 5)(x + 3) + 7
= 0.(x + 3) + 7 \u2026..[From (i)]
= 0 + 7
= 7<\/p>\n\n\n\n

(iv) x4<\/sup> + 9x3<\/sup> + 35x2<\/sup> \u2013 x + 4, if x = -5 + \u221a-4
Solution:<\/strong>
x = -5 + \u221a-4
x + 5 = \u221a-4
x + 5 = \u221a4 \u221a-1
x + 5 = 2i
(x + 5)2<\/sup> = 4i2<\/sup>
x2<\/sup> + 10x + 25 = 4(-1) \u2026.[\u2235 i2<\/sup> = -1]
x2<\/sup> + 10x + 29 = 0 \u2026..(i)
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Dividend = Divisor \u00d7 Quotient + Remainder
x4<\/sup> + 9x3<\/sup> + 35x2<\/sup> \u2013 x + 4 = (x2<\/sup> + 10x + 29) (x2<\/sup> \u2013 x + 16) \u2013 132x \u2013 460
= 0.(x2<\/sup> \u2013 x + 16) \u2013 132x \u2013 460 \u2026..[From (i)]
= -132 (-5 + 2i) \u2013 460
= 660 \u2013 264i \u2013 460
= 200 \u2013 264i<\/p>\n\n\n\n

(v) 2x4<\/sup> + 5x3<\/sup> + 7x2<\/sup> \u2013 x + 41, if x = -2 \u2013 \u221a3i
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Dividend = Divisor \u00d7 Quotient + Remainder
2x4<\/sup> + 5x3<\/sup> + 7x2<\/sup> \u2013 x + 41 = (x2<\/sup> + 4x + 7) (2x2<\/sup> \u2013 3x + 5) + 6
= 0(2x2<\/sup> \u2013 3x + 5) + 6 \u2026\u2026[From (i)]
= 0 + 6
= 6<\/p>\n\n\n\n

Ex 1.3<\/h4>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(ii) \u221a3 + \u221a2 i
Solution:
Let z = \u221a3 + \u221a2 i
a = \u221a3, b = \u221a2,
i.e. a > 0, b > 0
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iii) -8 + 15i
Solution:<\/strong>
Let z = -8 + 15i
a = -8, b = 15 , i.e. a < 0, b > 0
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iv) -3(1 \u2013 i)
Solution:<\/strong>
Let z = -3(1 \u2013 i) = -3 + 3i
a = -3, b = 3 , i.e. a < 0, b > 0<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(v) -4 \u2013 4i
Solution:<\/strong>
Let z = -4 \u2013 4i
a = -4, b = -4 , i.e. a < 0, b < 0
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(vi) \u221a3 \u2013 i
Solution:<\/strong>
Let z = \u221a3 \u2013 i
a = \u221a3, b = -1, i.e. a > 0, b < 0
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
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\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Ex 1.4<\/h4>\n\n\n\n

Question 1.
Find the value of<\/strong>
(i) \u03c918<\/sup>
(ii) \u03c921<\/sup>
(iii) \u03c9-30<\/sup>
(iv) \u03c9-105<\/sup>
Solution:
<\/strong><\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 2.
If \u03c9 is the complex cube root of unity, show that
(i) (2 \u2013 \u03c9)(2 \u2013 \u03c92<\/sup>) = 7
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
L.H.S. = (2 \u2013 \u03c9)(2 \u2013 \u03c92<\/sup>)
= 4 \u2013 2\u03c92<\/sup> \u2013 2\u03c9 + \u03c93<\/sup>
= 4 \u2013 2(\u03c92<\/sup> + \u03c9) + 1
= 4 \u2013 2(-1) + 1
= 4 + 2 + 1
= 7
= R.H.S.<\/p>\n\n\n\n

(ii) (1 + \u03c9 \u2013 \u03c92<\/sup>)6<\/sup> = 64
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iii) (1 + \u03c9)3<\/sup> \u2013 (1 + \u03c92<\/sup>)3<\/sup> = 0
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iv) (2 + \u03c9 + \u03c92<\/sup>)3<\/sup> \u2013 (1 \u2013 3\u03c9 + \u03c92<\/sup>)3<\/sup> = 65
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(v) (3 + 3\u03c9 + 5\u03c92<\/sup>)6<\/sup> \u2013 (2 + 6\u03c9 + 2\u03c92<\/sup>)3<\/sup> = 0
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(vi) a+b\u03c9+c\u03c92c+a\u03c9+b\u03c92 = \u03c92<\/sup>
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(vii) (a + b) + (a\u03c9 + b\u03c92<\/sup>) + (a\u03c92<\/sup> + b\u03c9) = 0
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(viii) (a \u2013 b)(a \u2013 b\u03c9)(a \u2013 b\u03c92<\/sup>) = a3<\/sup> \u2013 b3<\/sup>
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(ix) (a + b)2<\/sup> + (a\u03c9 + b\u03c92<\/sup>)2<\/sup> + (a\u03c92<\/sup>+ b\u03c9)2<\/sup> = 6ab
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 3.
If \u03c9 is the complex cube root of unity, find the value of
(i) \u03c9 + 1\u03c9
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
\u03c9+1\u03c9=\u03c92+1\u03c9=\u2212\u03c9\u03c9=\u22121<\/p>\n\n\n\n

(ii) \u03c92<\/sup> + \u03c93<\/sup> + \u03c94<\/sup>
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
\u03c92<\/sup> + \u03c93<\/sup> + \u03c94<\/sup>
= \u03c92<\/sup>(1 + \u03c9 + \u03c92<\/sup>)
= \u03c92<\/sup>(0)
= 0<\/p>\n\n\n\n

(iii) (1 + \u03c92<\/sup>)3<\/sup>
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
(1 + \u03c92<\/sup>)3<\/sup>
= (-\u03c9)3<\/sup>
= -\u03c93<\/sup>
= -1<\/p>\n\n\n\n

(iv) (1 \u2013 \u03c9 \u2013 \u03c92<\/sup>)3<\/sup> + (1 \u2013 \u03c9 + \u03c92<\/sup>)3<\/sup>
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
(1 \u2013 \u03c9 \u2013 \u03c92<\/sup>)3<\/sup> + (1 \u2013 \u03c9 + \u03c92<\/sup>)3<\/sup>
= [1 \u2013 (\u03c9 + \u03c92<\/sup>)]3<\/sup> + [(1 + \u03c92<\/sup>) \u2013 \u03c9]3<\/sup>
= [1 \u2013 (-1)]2<\/sup> + (-\u03c9 \u2013 \u03c9)3<\/sup>
= 23<\/sup> + (-2\u03c9)3<\/sup>
= 8 \u2013 8\u03c93<\/sup>
= 8 \u2013 8(1)
= 0<\/p>\n\n\n\n

(v) (1 + \u03c9)(1 + \u03c92<\/sup>)(1 + \u03c94<\/sup>)(1 + \u03c98<\/sup>)
Solution:<\/strong>
\u03c9 is the complex cube root of unity.
\u03c93<\/sup> = 1 and 1 + \u03c9 + \u03c92<\/sup> = 0
Also, 1 + \u03c92<\/sup> = -\u03c9, 1 + \u03c9 = -\u03c92<\/sup> and \u03c9 + \u03c92<\/sup> = -1
(1 + \u03c9)(1 + \u03c92<\/sup>)(1 + \u03c94<\/sup>)(1 + \u03c98<\/sup>)
= (1 + \u03c9)(1 + \u03c92<\/sup>)(1 + \u03c9)(1 + \u03c92<\/sup>) \u2026..[\u2235 \u03c93<\/sup> = 1, \u03c94<\/sup> = \u03c9]
= (-\u03c92<\/sup>)(-\u03c9)(-\u03c92<\/sup>)(-\u03c9)
= \u03c96<\/sup>
= (\u03c93<\/sup>)2<\/sup>
= (1)2<\/sup>
= 1<\/p>\n\n\n\n

Question 4.
If \u03b1 and \u03b2 are the complex cube roots of unity, show that
(i) \u03b12<\/sup> + \u03b22<\/sup> + \u03b1\u03b2 = 0
(ii) \u03b14<\/sup> + \u03b24<\/sup> + \u03b1-1<\/sup>\u03b2-1<\/sup> = 0
Solution:<\/strong>
\u03b1 and \u03b2 are the complex cube roots of unity.
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Question 5.
If x = a + b, y = \u03b1a + \u03b2b and z = a\u03b2 + b\u03b1, where \u03b1 and \u03b2 are complex cube roots of unity, show that xyz = a3<\/sup> + b3<\/sup>.
Solution:<\/strong>
x = a + b, y = \u03b1a + \u03b2b, z = a\u03b2 + b\u03b1
\u03b1 and \u03b2 are the complex cube roots of unity.
\u2234 \u03b1 = \u22121+i3\u221a2 and \u03b2 = \u22121\u2212i3\u221a2
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 6.
Find the equation in cartesian coordinates of the locus of z if
(i) |z| = 10
Solution:<\/strong>
Let z = x + iy
|z| = 10
|x + iy| = 10
x2+y2\u2212\u2212\u2212\u2212\u2212\u2212\u221a = 10
\u2234 x2<\/sup> + y2<\/sup> = 100<\/p>\n\n\n\n

(ii) |z \u2013 3| = 2
Solution:<\/strong>
Let z = x + iy
|z \u2013 3| = 2
|x + iy \u2013 3| = 2
|(x \u2013 3) + iy| = 2
(x\u22123)2+y2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a = 2
\u2234 (x \u2013 3)2<\/sup> + y2<\/sup> = 4<\/p>\n\n\n\n

(iii) |z \u2013 5 + 6i| = 5
Solution:<\/strong>
Let z = x + iy
|z \u2013 5 + 6i| = 5
|x + iy \u2013 5 + 6i| = 5
|(x \u2013 5) + i(y + 6)| = 5
(x\u22125)2+(y+6)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a = 5
\u2234 (x \u2013 5)2<\/sup> + (y + 6)2<\/sup> = 25<\/p>\n\n\n\n

(iv) |z + 8| = |z \u2013 4|
Solution:<\/strong>
Let z = x + iy
|z + 8| = |z \u2013 4|
|x + iy + 8| = |x + iy \u2013 4|
|(x + 8) + iy | = |(x \u2013 4) + iy|
(x+8)2+y2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=(x\u22124)2+y2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a
(x + 8)2<\/sup> + y2<\/sup> = (x \u2013 4)2<\/sup> + y2<\/sup>
x2<\/sup> + 16x + 64 + y2<\/sup> = x2<\/sup> \u2013 8x + 16 + y2<\/sup>
16x + 64 = -8x + 16
24x + 48 = 0
\u2234 x + 2 = 0<\/p>\n\n\n\n

(v) |z \u2013 2 \u2013 2i | = |z + 2 + 2i|
Solution:<\/strong>
Let z = x + iy
|z \u2013 2 \u2013 2i| = |z + 2 + 2i|
|x + iy \u2013 2 \u2013 2i | = |x + iy + 2 + 2i |
|(x \u2013 2) + i(y \u2013 2)| = |(x + 2) + i(y + 2)|
(x\u22122)2+(y\u22122)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a=(x+2)2+(y+2)2\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u221a
(x \u2013 2)2<\/sup> + (y \u2013 2)2<\/sup> = (x + 2)2<\/sup> + (y + 2)2<\/sup>
x2<\/sup>\u2013 4x + 4 + y2<\/sup> \u2013 4y + 4 = x2<\/sup> + 4x + 4 + y2<\/sup> + 4y + 4
-4x \u2013 4y = 4x + 4y
8x + 8y = 0
x + y = 0
y = -x<\/p>\n\n\n\n

(vi) |z+3i||z\u22126i|=1
Solution:<\/strong>
Let z = x + iy
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

x2<\/sup> + (y + 3)2<\/sup> = x2<\/sup> + (y \u2013 6)2<\/sup>
y2<\/sup> + 6y + 9 = y2<\/sup> \u2013 12y + 36
18y \u2013 27 = 0
2y \u2013 3 = 0<\/p>\n\n\n\n

Question 7.
Use De Moivre\u2019s theorem and simplify the following:
(i) (cos2\u03b8+isin2\u03b8)7(cos4\u03b8+isin4\u03b8)3
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(ii) cos5\u03b8+isin5\u03b8(cos3\u03b8\u2212isin3\u03b8)2
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iii) (cos7\u03c013+isin7\u03c013)4(cos4\u03c013\u2212isin4\u03c013)6
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 8.
Express the following in the form a + ib, a, b \u2208 R, using De Moivre\u2019s theorem.
(i) (1 \u2013 i)5<\/sup>
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(ii) (1 + i)6<\/sup>
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iii) (1 \u2013 \u221a3 i)4<\/sup>
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

(iv) (-2\u221a3 \u2013 2i)5<\/sup>
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 1- Complex Numbers<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 1- Complex Numbers PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :<\/strong><\/h2>\n\n\n\n