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

## Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1

(I) Select the correct answer from the given alternatives.

Question 1.

If n is an odd positive integer, then the value of 1 + (i)^{2n} + (i)^{4n} + (i)^{6n} is:

(A) -4i

(B) 0

(C) 4i

(D) 4

Answer:

(B) 0

Hint:

1 + (i^{2})^{n} + (i^{4})^{n} + (i^{2})^{3n}

= 1 – 1 + 1 – 1 …..(n odd positive integer)

= 0

Question 2.

The value of \(\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}\) is equal to:

(A) -2

(B) 1

(C) 0

(D) -1

Answer:

(D) -1

Hint:

Question 3.

√-3 √-6 is equal to

(A) -3√2

(B) 3√2

(C) 3√2 i

(D) -3√2 i

Answer:

(A) -3√2

Hint:

√-3 √-6

= (√3 i) (√6 i)

= 3√2 (-1)

= -3√2

Question 4.

If ω is a complex cube root of unity, then the value of ω^{99} + ω^{100} + ω^{101} is:

(A) -1

(B) 1

(C) 0

(D) 3

Answer:

(C) 0

Hint:

ω^{99} + ω^{100} + ω^{101}

= ω^{99} (1 + ω + ω^{2})

= ω^{99} (0)

= 0

Question 5.

If z = r(cos θ + i sin θ), then the value of \(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\) is

(A) cos 2θ

(B) 2cos 2θ

(C) 2cos θ

(D) 2sin θ

Answer:

(B) 2cos 2θ

Hint:

Question 6.

If ω(≠1) is a cube root of unity and (1 + ω)^{7} = A + Bω, then A and B are respectively the numbers

(A) 0, 1

(B) 1, 1

(C) 1, 0

(D) -1, 1

Answer:

(B) 1, 1

Hint:

(1 + ω)^{7}

= (-ω^{2})^{7}

= -ω^{14}

= -ω^{2}(ω^{3})^{4}

= -ω^{2}

= 1 + ω

A = 1, B = 1

Question 7.

The modulus and argument of (1 + i√3)^{8} are respectively

(A) 2 and \(\frac{2 \pi}{3}\)

(B) 256 and \(\frac{8 \pi}{3}\)

(C) 256 and \(\frac{2 \pi}{3}\)

(D) 64 and \(\frac{4 \pi}{3}\)

Answer:

(C) 256 and \(\frac{2 \pi}{3}\)

Hint:

Question 8.

If arg (z) = θ, then arg \(\overline{(\mathrm{z})}\) =

(A) -θ

(B) θ

(C) π – θ

(D) π + θ

Answer:

(A) -θ

Hint:

Let z = \(\mathrm{re}^{\mathrm{i} \theta}\), then \(\overline{\mathrm{z}}=\mathrm{r} \mathrm{e}^{-\mathrm{i} \theta}\)

∴ arg \(\overline{\mathbf{z}}\) = -θ.

Question 9.

If -1 + √3 i = \(\mathrm{re}^{\mathrm{i} \theta}\), then θ =

(A) –\(\frac{2 \pi}{3}\)

(B) \(\frac{\pi}{3}\)

(C) –\(\frac{\pi}{3}\)

(D) \(\frac{2 \pi}{3}\)

Answer:

(D) \(\frac{2 \pi}{3}\)

Hint:

Question 10.

If z = x + iy and |z – zi| = 1, then

(A) z lies on X-axis

(B) z lies on Y-axis

(C) z lies on a rectangle

(D) z lies on a circle

Answer:

(D) z lies on a circle

Hint:

|z – zi | = |z| |1 – i| = 1

∴ |z| = \(\frac{1}{\sqrt{2}}\)

∴ x^{2} + y^{2} = \(\frac{1}{2}\)

(II) Answer the following:

Question 1.

Simplify the following and express in the form a + ib.

(i) 3 + √-64

Solution:

3 + √-64

= 3 + √64 √-1

= 3 + 8i

(ii) (2i^{3})^{2}

Solution:

(2i^{3})^{2}

= 4i^{6}

= 4(i^{2})^{3}

= 4(-1)^{3}

= -4 …..[∵ i^{2} = -1]

= -4 + 0i

(iii) (2 + 3i) (1 – 4i)

Solution:

(2 + 3i)(1 – 4i)

= 2 – 8i + 3i – 12i^{2}

= 2 – 5i – 12(-1) …..[∵ i^{2} = -1]

= 14 – 5i

(iv) \(\frac{5}{2}\)i(-4 – 3i)

Solution:

(v) (1 + 3i)^{2} (3 + i)

Solution:

(1 + 3i)^{2} (3 + i)

= (1 + 6i + 9i^{2})(3 + i)

= (1 + 6i – 9)(3 + i) ……[∵ i^{2} = -1]

= (-8 + 6i)(3 + i)

= -24 – 8i + 18i + 6i^{2}

= -24 + 10i + 6(-1)

= -24 + 10i – 6

= -30 + 10i

(vi) \(\frac{4+3 i}{1-i}\)

Solution:

(vii) \(\left(1+\frac{2}{i}\right)\left(3+\frac{4}{i}\right)(5+i)^{-1}\)

Solution:

(viii) \(\frac{\sqrt{5}+\sqrt{3 i}}{\sqrt{5}-\sqrt{3} i}\)

Solution:

(ix) \(\frac{3 i^{5}+2 i^{7}+i^{9}}{i^{6}+2 i^{8}+3 i^{18}}\)

Solution:

(x) \(\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}\)

Solution:

Question 2.

Solve the following equations for x, y ∈ R

(i) (4 – 5i)x + (2 + 3i)y = 10 – 7i

Solution:

(4 – 5i)x + (2 + 3i)y = 10 – 7i

(4x + 2y) + (3y – 5x) i = 10 – 7i

Equating real and imaginary parts, we get

4x + 2y= 10 i.e., 2x + y = 5 ……(i)

and 3y – 5x = -7 ……(ii)

Equation (i) × 3 – equation (ii) gives

11x = 22

∴ x = 2

Putting x = 2 in (i), we get

2(2) + y = 5

∴ y = 1

∴ x = 2 and y = 1

(ii) \(\frac{x+i y}{2+3 i}\) = 7 – i

Solution:

\(\frac{x+i y}{2+3 i}\) = 7 – i

x + iy = (7 – i)(2 + 3i)

x + iy = 14 + 21i – 2i – 3i^{2}

x + iy = 14 + 19i – 3(-1)

x + iy = 17 + 19i

Equating real and imaginary parts, we get

∴ x = 17 and y = 19

(iii) (x + iy) (5 + 6i) = 2 + 3i

Solution:

(iv) 2x + i^{9} y(2 + i) = x i^{7} + 10 i^{16}

Solution:

2x + i^{9} y(2 + i) = x i^{7} + 10 i^{16}

2x + (i^{4})^{2} . i . y(2 + i) = x(i^{2})^{3} . i + 10 . (i^{4})^{4}

2x + (1)^{2} . iy(2 + i) = x(-1)^{3} . i + 10(1)^{4} ……..[∵ i^{2} = -1, i^{4} = 1]

2x + 2yi + y i^{2} = -xi + 10

2x + 2yi – y + xi = 10

(2x – y) + (x + 2y)i = 10 + 0 . i

Equating real and imaginary parts, we get

2x – y = 10 ……(i)

and x + 2y = 0 ……..(ii)

Equation (i) × 2 + equation (ii) gives, we get

5x = 20

∴ x = 4

Putting x = 4 in (i), we get

2(4) – y = 10

y = 8 – 10

∴ y = -2

∴ x = 4 and y = -2

Question 3.

Evaluate

(i) (1 – i + i^{2})^{-15}

Solution:

(ii) i^{131} + i^{49}

Solution:

i^{131} + i^{49}

= (i^{4})^{32} . i^{3} + (i^{4})^{12} . i

= (1)^{32} (-i) + (1)^{12} . i

= -i + i

= 0

Question 4.

Find the value of

(i) x^{3} + 2x^{2} – 3x + 21, if x = 1 + 2i

Solution:

(ii) x^{4} + 9x^{3} + 35x^{2} – x + 164, if x = -5 + 4i

Solution:

Question 5.

Find the square roots of

(i) -16 + 30i

Solution:

Let \(\sqrt{-16+30 \mathrm{i}}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

-16 + 30i = a^{2} + b^{2} i^{2} + 2abi

-16 + 30i = (a^{2} – b^{2}) + 2abi …..[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = -16 and 2ab = 30

a^{2} – b^{2} = -16 and b = \(\frac{15}{a}\)

(ii) 15 – 8i

Solution:

Let \(\sqrt{15-8 i}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

15 – 8i = a^{2} + b^{2} i^{2} + 2abi

15 – 8i = (a^{2} – b^{2}) + 2abi …..[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 15 and 2ab = -8

a^{2} – b^{2} = 15 and b = \(\frac{-4}{a}\)

When a = 4, b = \(\frac{-4}{4}\) = -1

When a = -4, b = \(\frac{-4}{-4}\) = 1

∴ \(\sqrt{15-8 i}\) = ±(4 – i)

(iii) 2 + 2√3 i

Solution:

Let \(\sqrt{2+2 \sqrt{3}}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

2 + 2√3 i = a^{2} + b^{2} i^{2} + 2abi

2 + 2√3 i = a^{2} – b^{2} + 2abi …..[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 2 and 2ab = 2√3

a^{2} – b^{2} = 2 and b = \(\frac{\sqrt{3}}{a}\)

(iv) 18i

Solution:

Let √18i = a + bi, where a, b ∈ R.

Squaring on both sides, we get

18i = a^{2} + b^{2} i^{2} + 2abi

0 + 18i = a^{2} – b^{2} + 2abi …..[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 0 and 2ab = 18

a^{2} – b^{2} = 0 and b = \(\frac{9}{a}\)

\(a^{2}-\left(\frac{9}{a}\right)^{2}=0\)

\(a^{2}-\frac{81}{a^{2}}=0\)

a^{4} – 81 = 0

(a^{2} – 9) (a^{2} + 9) = 0

a^{2} = 9 or a^{2} = -9

But a ∈ R

∴ a^{2} ≠ -9

∴ a^{2} = 9

∴ a = ± 3

When a = 3, b = \(\frac{9}{3}\) = 3

When a = -3, b = \(\frac{9}{-3}\) = -3

∴ √18i = ±(3 + 3i) = ±3(1 + i)

(v) 3 – 4i

Solution:

Let \(\sqrt{3-4 i}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

3 – 4i = a^{2} + b^{2} i^{2} + 2abi

3 – 4i = a^{2} – b^{2} + 2abi ……[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 3 and 2ab = -4

a^{2} – b^{2} = 3 and b = \(\frac{-2}{a}\)

(vi) 6 + 8i

Solution:

Let \(\sqrt{6+8 i}\) = a + bi, where a, b ∈ R.

Squaring on both sides, we get

6 + 8i = a^{2} + b^{2} i^{2} + 2abi

6 + 8i = a^{2} – b^{2} + 2abi ……[∵ i^{2} = -1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 6 and 2ab = 8

Question 6.

Find the modulus and argument of each complex number and express it in the polar form.

(i) 8 + 15i

Solution:

(ii) 6 – i

Solution:

(iii) \(\frac{1+\sqrt{3} \mathbf{i}}{2}\)

Solution:

(iv) \(\frac{-1-\mathbf{i}}{\sqrt{2}}\)

Solution:

(v) 2i

Solution:

(vi) -3i

Solution:

(vii) \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \mathbf{i}\)

Solution:

Question 7.

Represent 1 + 21, 2 – i, -3 – 2i, -2 + 3i by points in Argand’s diagram.

Solution:

The complex numbers 1 + 2i, 2 – i, -3 – 2i, -2 + 3i will be represented by the points A(1, 2), B(2, -1), C(-3, -2), D(-2, 3) respectively as shown below:

Question 8.

Show that z = \(\frac{5}{(1-i)(2-i)(3-i)}\) is purely imaginary number.

Solution:

Question 9.

Find the real numbers x and y such that \(\frac{x}{1+2 i}+\frac{y}{3+2 i}=\frac{5+6 i}{-1+8 i}\)

Solution:

\(\frac{x}{1+2 i}+\frac{y}{3+2 i}=\frac{5+6 i}{-1+8 i}\)

(3x + y) + 2(x + y)i = 5 + 6i

Equating real and imaginary parts, we get

3x + y = 5 ……(i)

and 2(x + y) = 6

i.e., x + y = 3 …….(ii)

Subtracting (ii) from (i), we get

2x = 2

∴ x = 1

Putting x = 1 in (ii), we get

1 + y = 3

∴ y = 2

∴ x = 1, y = 2

Question 10.

Show that \(\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{10}+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{10}=0\)

Solution:

Question 11.

Show that \(\left(\frac{1+i}{\sqrt{2}}\right)^{8}+\left(\frac{1-i}{\sqrt{2}}\right)^{8}=2\)

Solution:

Question 12.

Convert the complex numbers in polar form and also in exponential form.

(i) z = \(\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}\)

Solution:

(ii) z = -6 + √2 i

Solution:

z = -6 + √2 i

∴ a = -6, b = √2

i.e. a < 0, b > 0

(iii) \(\frac{-3}{2}+\frac{3 \sqrt{3} i}{2}\)

Solution:

Question 13.

If x + iy = \(\frac{a+i b}{a-i b}\), prove that x^{2} + y^{2} = 1.

Solution:

Question 14.

Show that z = \(\left(\frac{-1+\sqrt{-3}}{2}\right)^{3}\) is a rational number.

Solution:

Question 15.

Show that \(\frac{1-2 i}{3-4 i}+\frac{1+2 i}{3+4 i}\) is real.

Solution:

Question 16.

Simplify

(i) \(\frac{\mathrm{i}^{29}+\mathrm{i}^{39}+\mathrm{i}^{49}}{\mathrm{i}^{30}+\mathrm{i}^{40}+\mathrm{i}^{50}}\)

Solution:

(ii) \(\left(\mathrm{i}^{65}+\frac{1}{\mathrm{i}^{145}}\right)\)

Solution:

(iii) \(\frac{\mathrm{i}^{238}+\mathrm{i}^{236}+\mathrm{i}^{234}+\mathrm{i}^{232}+\mathrm{i}^{230}}{\mathrm{i}^{228}+\mathrm{i}^{226}+\mathrm{i}^{224}+\mathrm{i}^{222}+\mathrm{i}^{220}}\)

Solution:

Question 17.

Simplify \(\left[\frac{1}{1-2 i}+\frac{3}{1+i}\right]\left[\frac{3+4 i}{2-4 i}\right]\)

Solution:

Question 18.

If α and β are complex cube roots of unity, prove that (1 – α) (1 – β) (1 – α^{2}) (1 – β^{2}) = 9.

Solution:

α and β are the complex cube roots of unity.

Question 19.

If ω is a complex cube root of unity, prove that (1 – ω + ω^{2})^{6} + (1 + ω – ω^{2})^{6} = 128.

Solution:

ω is the complex cube root of unity.

∴ ω^{3} = 1 and 1 + ω + ω^{2} = 0

Also, 1 + ω^{2} = -ω, 1 + ω = -ω^{2}

∴ L.H.S. = (1 – ω + ω^{2})^{6} + (1 + ω – ω^{2})^{6}

= [(1 + ω^{2}) – ω]^{6} + [(1 + ω) – ω^{2}]^{6}

= (-ω – ω))^{6} + (-ω^{2} – ω^{2})6

= (-2ω)^{6} + (-2ω^{2})^{6}

= 64ω^{6} + 64ω^{12}

= 64(ω^{3})^{2} + 64(ω^{3})^{4}

= 64(1)^{2} + 64(1)^{4}

= 128

= R.H.S.

Question 20.

If ω is the cube root of unity, then find the value of \(\left(\frac{-1+\mathbf{i} \sqrt{3}}{2}\right)^{18}+\left(\frac{-1-\mathbf{i} \sqrt{3}}{2}\right)^{18}\)

Solution:

If ω is the complex cube root of unity, then

Given Expression = ω^{18} + (ω^{2})^{18}

= ω^{18} + ω^{36}

= (ω^{3})^{6} + (ω^{3})^{12}

= (1)^{6} + (1)^{12}

= 2