Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1

Question 1.
Simplify:
(i) √-16 + 3√-25 + √-36 – √-625
Solution: = 4i + 3(5i) + 6i – 25i
= 25i – 25i
= 0

(ii) 4√-4 + 5√-9 – 3√-16
Solution: Question 2.
Write the conjugates of the following complex numbers
(i) 3 + i
Solution:
Conjugate of (3 + i) is (3 – i).

(ii) 3 – i
Solution:
Conjugate of (3 – i) is (3 + i).

(iii) √-5 – √7 i
Solution:
Conjugate of (√-5 – √7 i) is (√-5 + √7 i).

(iv) -√-5
Solution:
-√-5 = -√5 × √-1 = -√5 i
Conjugate of (-√-5) is √5 i (v) 5i
Solution:
Conjugate of (5i) is (-5i).

(vi) √5 – i
Solution:
Conjugate of (√5 – i) is (√5 + i).

(vii) √2 + √3 i
Solution:
Conjugate of (√2 + √3 i) is (√2 – √3 i)

(viii) cos θ + i sin θ
Solution:
Conjugate of (cos θ + i sin θ) is (cos θ – i sin θ)

Question 3.
Find a and b if
(i) a + 2b + 2ai = 4 + 6i
Solution:
a + 2b + 2ai = 4 + 6i
Equating real and imaginary parts, we get
a + 2b = 4 …..(i)
2a = 6 ……(ii)
∴ a = 3
Substituting, a = 3 in (i), we get
3 + 2b = 4
∴ b = $$\frac{1}{2}$$
∴ a = 3 and b = $$\frac{1}{2}$$

Check:
For a = 3 and b = $$\frac{1}{2}$$
Consider, L.H.S. = a + 2b + 2ai
= 3 + 2($$\frac{1}{2}$$) + 2(3)i
= 4 + 6i
= R.H.S.

(ii) (a – b) + (a + b)i = a + 5i
Solution:
(a – b) + (a + b)i = a + 5i
Equating real and imaginary parts, we get
a – b = a ……(i)
a + b = 5 ……(ii)
From (i), b = 0
Substituting b = 0 in (ii), we get
a + 0 = 5
∴ a = 5
∴ a = 5 and b = 0 (iii) (a + b) (2 + i) = b + 1 + (10 + 2a)i
Solution:
(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
∴ 2a + b = 1 ……(i)
and a + b = 10 + 2a
-a + b = 10 …….(ii)
Subtracting equation (ii) from (i), we get
3a = -9
∴ a = -3
Substituting a = – 3 in (ii), we get
-(-3) + b = 10
∴ b = 7
∴ a = -3 and b = 7

(iv) abi = 3a – b + 12i
Solution:
abi = 3a – b + 12i
∴ 0 + abi = (3a – b) + 12i
Equating real and imaginary parts, we get
3a – b = 0
∴ 3a = b …..(i)
and ab = 12
∴ b = $$\frac{12}{a}$$ ……..(ii)
Substituting b = $$\frac{12}{a}$$ in (i), we get
3a = $$\frac{12}{a}$$
3a2 = 12
a2 = 4
a = ±2
When a = 2, b = $$\frac{12}{a}$$ = $$\frac{12}{2}$$ = 6
When a = -2, b = $$\frac{12}{a}$$ = $$\frac{12}{-2}$$ = -6
∴ a = 2 and b = 6 or a = -2 and b = -6

(v) $$\frac{1}{a+i b}$$ = 3 – 2i
Solution: (vi) (a + ib) (1 + i) = 2 + i
Solution:
(a + ib)(1 + i) = 2 + i
a + ai + bi + bi2 = 2 + i
a + (a + b)i + b(-1) = 2 + i ……(∵ i2 = -1)
(a – b) + (a + b)i = 2 + i
Equating real and imaginary parts, we get
a – b = 2 ……(i)
a + b = 1 …….(ii)
Adding equations (i) and (ii), we get
2a = 3
∴ a = $$\frac{3}{2}$$
Substituting a = $$\frac{3}{2}$$ in (ii), we get
$$\frac{3}{2}$$ + b = 1
∴ b = 1 – $$\frac{3}{2}$$ = $$\frac{-1}{2}$$
∴ a = $$\frac{3}{2}$$ and b = $$\frac{-1}{2}$$

Question 4.
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:
(i) (1 + 2i)(-2 + i)
Solution:
(1 + 2i)(-2 + i) = -2 + i – 4i + 2i2
= -2 – 3i + 2(-1) ……[∵ i2 = -1]
∴ (1 + 2i)(-2 + i) = -4 – 3i
∴ a = -4 and b = -3 (ii) (1 + i)(1 – i)-1
Solution: (iii) $$\frac{i(4+3 i)}{1-i}$$
Solution: (iv) $$\frac{(2+i)}{(3-i)(1+2 i)}$$
Solution: (v) $$\left(\frac{1+i}{1-1}\right)^{2}$$
Solution: (vi) $$\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}$$
Solution: (vii) (1 + i)-3
Solution: (viii) $$\frac{2+\sqrt{-3}}{4+\sqrt{-3}}$$
Solution: (ix) (-√5 + 2√-4 ) + (1 – √-9 ) + (2 + 3i)(2 – 3i)
Solution:
(-√5 + 2√-4) + (1 – √-9) + (2 + 3i)(2 – 3i)
= (-√5 + 2√4.√-1) + (1 – √9.√-1) + 4 – 9i2
= [-√5 + 2(2)i] + (1 – 3i) + 4 – 9i2
= -√5 + 4i + 1 – 3i + 4 – 9(-1) ……[∵ i2 = -1]
= (14 – √5) + i
∴ a = 14 – √5 and b = 1 (x) (2 + 3i)(2 – 3i)
Solution:
(2 + 3i)(2 – 3i)
= 4 – 9i2
= 4 – 9(-1) …[∵ i2 = -1]
= 4 + 9
= 13
∴ (2 + 3i)(2 – 3i) = 13 + 0i
∴ a = 13 and b = 0

(xi) $$\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}$$
Solution: Question 5.
Show that (-1 + √3i)3 is a real number.
Solution: Question 6.
Find the value of (3 + $$\frac{2}{\mathrm{i}}$$) (i6 – i7) (1 + i11).
Solution: Question 7.
Evaluate the following:
(i) i35
(ii) i888
(iii) i93
(iv) i116
(v) i403
(vi) $$\frac{1}{i^{58}}$$
(vii) i-888
(viii) i30 + i40 + i50 + i60
Solution: Question 8.
Show that 1 + i10 + i20 + i30 is a real number.
Solution: = 1 – 1 + 1 – 1
= 0, which is a real number.

Question 9.
Find the value of
(i) i49 + i68 + i89 + i110
(ii) i + i2 + i3 + i4
Solution:  Question 10.
Simplify: $$\frac{\mathbf{i}^{592}+\mathbf{i}^{590}+\mathbf{i}^{588}+\mathbf{i}^{586}+\mathbf{i}^{584}}{\mathbf{i}^{582}+\mathbf{i}^{580}+\mathbf{i}^{578}+\mathbf{i}^{576}+\mathbf{i}^{574}}$$
Solution: Question 11.
Find the value of 1 + i2 + i4 + i6 + i8 + …… + i20.
Solution: Question 12.
Show that 1 + i10 + i100 – i1000 = 0.
Solution: Question 13.
Is (1 + i14 + i18 + i22) a real number? Justify your answer.
Solution: Question 14.
Evaluate: $$\left(\mathbf{i}^{37}+\frac{1}{\mathbf{i}^{67}}\right)$$
Solution: Question 15.
Prove that: (1 + i)4 × $$\left(1+\frac{1}{\mathrm{i}}\right)^{4}$$ = 16
Solution: Question 16.
Find the value of $$\frac{\mathbf{i}^{6}+\mathbf{i}^{7}+\mathbf{i}^{8}+\mathbf{i}^{9}}{\mathbf{i}^{2}+\mathbf{i}^{3}}$$
Solution:  Question 17.
If a = $$\frac{-1+\sqrt{3} i}{2}$$, b = $$\frac{-1-\sqrt{3} i}{2}$$, then show that a2 = b and b2 = a.
Solution: Question 18.
If x + iy = (a + ib)3, show that $$\frac{x}{a}+\frac{y}{b}$$ = 4(a2 – b2)
Solution:
x + iy = (a + ib)3
x + iy = a3 + 3a2bi + 3ab2i2 + b3i3
x + iy = a3 + 3a2bi – 3ab2 – b3i ……[∵ i2 = -1, i3 = -i]
x + iy = (a3 – 3ab2) + (3a2b – b3)i
Equating real and imaginary parts, we get
x = a3 – 3ab2 and y = 3a2b – b3
$$\frac{x}{a}$$ = a2 – 3b2 and $$\frac{y}{b}$$ = 3a2 – b2
$$\frac{x}{a}+\frac{y}{b}$$ = a2 – 3b + 3a2 – b2
$$\frac{x}{a}+\frac{y}{b}$$ = 4a2 – 4b2
$$\frac{x}{a}+\frac{y}{b}$$ = 4(a2 – b2)

Alternate Method:
x + iy = (a + ib)3
x + iy = a3 + 3a2bi + 3ab2i2 + b3i3
x + iy = a3 + 3a2bi – 3ab2 – b3i …..[∵ i2 = -1, i3 = -i]
x + iy = (a3 – 3ab2) + (3a2b – b3)i
Equating real and imaginary parts, we get
x = a3 – 3ab2 and y = 3a2b – b3
Consider Question 19.
If $$\frac{a+3 i}{2+i b}$$ = 1 – i, show that (5a – 7b) = 0.
Solution:
$$\frac{a+3 i}{2+i b}$$ = 1 – i
a + 3i = (1 – i)(2 + ib)
= 2 + bi – 2i – bi2
= 2 + (b – 2)i – b(-1) ……[∵ i2 = -1]
a + 3i = (2 + b) + (b – 2)i
Equating real and imaginary parts, we get
a = 2 + b and 3 = b – 2
a = 2 + b and b = 5
a = 2 + 5 = 7
5a – 7b = 5(7) – 7(5) = 35 – 35 = 0 Question 20.
If x + iy = $$\sqrt{\frac{a+i b}{c+i d}}$$, prove that $$\left(x^{2}+y^{2}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$$
Solution:  Question 21.
If (a + ib) = $$\frac{1+i}{1-i}$$, then prove that a2 + b2 = 1.
Solution: ∴ a + bi = 0 + i
Equating real and imaginary parts, we get
a = 0 and b = 1
a2 + b2 = 02 + 12 = 1

Question 22.
Show that $$\left(\frac{\sqrt{7}+i \sqrt{3}}{\sqrt{7}-i \sqrt{3}}+\frac{\sqrt{7}-i \sqrt{3}}{\sqrt{7}+i \sqrt{3}}\right)$$ is real.
Solution: Question 23.
If (x + iy)3 = y + vi, then show that $$\frac{y}{x}+\frac{v}{y}$$ = 4(x2 – y2).
Solution:  Question 24.
Find the values of x and y which satisfy the following equations (x, y ∈ R)
(i) (x + 2y) + (2x – 3y)i + 4i = 5
Solution:
(x + 2y) + (2x – 3y)i + 4i = 5
(x + 2y) + (2x – 3y)i = 5 – 4i
Equating real and imaginary parts, we get
x + 2y = 5 ……(i)
and 2x – 3y = -4 …..(ii)
Equation (i) x 2 – equation (ii) gives
7y = 14
∴ y = 2
Substituting y = 2 in (i), we get
x + 2(2) = 5
x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2

Check:
For x = 1 and y = 2
Consider, L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.

(ii) $$\frac{x+1}{1+i}+\frac{y-1}{1-i}=i$$
Solution: (x + y) + (y – x – 2)i = 2i
(x + y) + (y – x – 2)i = 0 + 2i
Equating real and imaginary parts, we get
x + y = 0 and y – x – 2 = 2
∴ x + y = 0 …….(i)
and -x + y = 4 …….(ii)
Adding (i) and (ii), we get
2y = 4
∴ y = 2
Substituting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

(iii) $$\frac{x+i y}{2+3 i}+\frac{2+i}{2-3 i}=\frac{9}{13}(1+i)$$
Solution: (2x + 3y + 1) + (8 – 3x + 2y)i = 9 + 9i
Equating real and imaginary parts, we get
2x + 3y + 1 = 9 and 8 – 3x + 2y = 9
2x + 3y = 8 ……(i)
and 3x – 2y = – 1 ……(ii)
Equation (i) × 2 + equation (ii) × 3 gives
13x = 13
∴ x = 1
Substituting x = 1 in (i), we get
2(1) + 3y = 8
3y = 6
∴ y = 2
∴ x = 1 and y = 2 (iv) If x(1 + 3i) + y(2 – i) – 5 + i3 = 0, find x + y
Solution:
x(1 + 3i) + y(2 – i) – 5 + i3 = 0
x + 3xi + 2y – yi – 5 – i = 0 ……[∵ i3 = -i]
(x + 2y – 5) + (3x – y – 1)i = 0 + 0i
Equating real and imaginary parts, we get
x + 2y – 5 = 0 …..(i)
and 3x – y – 1 = 0 ……(ii)
Equation (i) + equation (ii) × 2 gives
7x – 7 = 0
7x = 1
∴ x = 1
Substituting x = 1 in (i), we get
1 + 2y – 5 = 0
2y = 4
y = 2
∴ x = 1 and y = 2
∴ x + y = 1 + 2 = 3

(v) If x + 2i + 15i6y = 7x + i3(y + 4), find x + y
Solution:
x + 2i + 15i6y = 7x + i3(y + 4)
x + 2i + 15(i2)3 y = 7x + i3(y + 4)
x + 2i + 15(-1)3 y = 7x – i(y + 4) ……[∵ i2 = -1, i3 = -i]
x + 2i – 15y – 7x + iy + 4i = 0
(-6x – 15y) + i(y + 6) = 0 + 0i
Equating real and imaginary parts, we get
-6x – 15y = 0 and y + 6 = 0
-6x – 15y = 0 and y = -6
-6x – 15(-6) = 0
-6x + 90 = 0
∴ x = 15
∴ x + y = 15 – 6 = 9