Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 1.

Write the conjugates of the following complex numbers:

(i) 3 + i

(ii) 3 – i

(iii) -√5 – √7i

(iv) -√-5

(v) 5i

(vi) √5 – i

(vii) √2 + √3i

Solution:

(i) Conjugate of (3 + i) is (3 – i)

(ii) Conjugate of (3 – i) is (3 + i)

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

(iv) -√-5 = -√5 × √-1 = -√5i

Conjugate of -√-5 is √5i

(v) Conjugate of 5i is -5i

(vi) Conjugate of √5 – i is √5 + i

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

Question 2.

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)

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

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

(iv) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)

(v) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)

(vi) (2 + 3i)(2 – 3i)

(vii) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)

Solution:

Question 3.

Show that (-1 + √3i)^{3} is a real number.

Solution:

(-1 + √3i)^{3}

= (-1)^{3} + 3(-1)^{2} (√3i) + 3(-1)(√3i)^{2} +(√3i)^{3} [∵ (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}]

= -1 + 3√3i – 3(3i^{2}) + 3√3 i^{3}

= -1 + 3√3i – 3(-3) – 3√3i [∵ i^{2} = -1, i^{3} = -1]

= -1 + 9

= 8, which is a real number.

Question 4.

Evaluate the following:

(i) i^{35}

(ii) i^{888}

(iii) i^{93}

(iv) i^{116}

(v) i^{403}

(vi) \(\frac{1}{i^{58}}\)

(vii) i^{30} + i^{40} + i^{50} + i^{60}

Solution:

We know that, i^{2} = -1, i^{3} = -i, i^{4} = 1

(i) i^{35} = (i^{4})^{8} (i^{2}) i = (1)^{8} (-1) i = -i

(ii) i^{888} = (i^{4})^{222} = (1)^{222} = 1

(iii) i^{93} = (i^{4})^{23} . i = (1)^{23} . i = i

(iv) i^{116} = (i^{4})^{29} = (1)^{29} = 1

(v) i^{403} = (i^{4})^{100} (i^{2}) i = (1)^{100} (-1) i = -i

(vi) \(\frac{1}{i^{88}}=\frac{1}{\left(i^{4}\right)^{14} \cdot i^{2}}=\frac{1}{(1)^{14}(-1)}=-1\)

(vii) i^{30} + i^{40} + i^{50} + i^{60}

= (i^{4})^{7} i^{2} + (i^{4})^{10} + (i^{4})^{12} i^{2} + (i^{4})^{15}

= (1)^{7} (-1) + (1)^{10} + (1)^{12} (-1) + (1)^{15}

= -1 + 1 – 1 + 1

= 0

Question 5.

Show that 1 + i^{10} + i^{20} + i^{30} is a real number.

Solution:

1 + i^{10} + i^{20} + i^{30}

= 1 + (i^{4})^{2} . i^{2} + (i^{4})^{5} + (i^{4})^{7} . i^{2}

= 1 + (1)^{2} (-1) + (1)^{5} + (1)^{7} (-1) [∵ i^{4} = 1, i^{2} = -1]

= 1 – 1 + 1 – 1

= 0, which is a real number.

Question 6.

Find the value of

(i) i^{49} + i^{68} + i^{89} + i^{110}

(ii) i + i^{2} + i^{3} + i^{4}

Solution:

(i) i^{49} + i^{68} + i^{89} + i^{110}

= (i^{4})^{12} . i + (i^{4})^{17} + (i^{4})^{22} . i + (i^{4})^{27} . i^{2}

= (1)^{12} . i + (1)^{17} + (1)^{22} . i + (1)^{27}(-1) ……[∵ i^{4} = 1, i^{2} = -1]

= i + 1 + i – 1

= 2i

(ii) i + i^{2} + i^{3} + i^{4}

= i + i^{2} + i^{2} . i + i^{4}

= i – 1 – i + 1 [∵ i^{2} = -1, i^{4} = 1]

= 0

Question 7.

Find the value of 1 + i^{2} + i^{4} + i^{6} + i^{8} + …… + i^{20}.

Solution:

1 + i^{2} + i^{4} + i^{6} + i^{8} + ….. + i^{20}

= 1 + (i^{2} + i^{4}) + (i^{6} + i^{8}) + (i^{10} + i^{12}) + (i^{14} + i^{16}) + (i^{18} + i^{20})

= 1 + [i^{2} + (i^{2})^{2}] + [(i^{2})^{3} + (i^{2})^{4}] + [(i^{2})^{5} + (i^{2})^{6}] + [(i^{2})^{7} + (i^{2})^{8}] + [(i^{2})^{9} + (i^{2})^{10}]

= 1 + [-1 + (- 1)^{2}] + [(-1)^{3} + (-1)^{4}] + [(-1)^{5} + (-1)^{6}] + [(-1)^{7} + (-1)^{8}] + [(-1)^{9} + (-1)^{10}] [∵ i^{2} = -1]

= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1)

= 1 + 0 + 0 + 0 + 0 + 0

= 1

Question 8.

Find the values of x and y which satisfy the following equations (x, y ∈ R):

(i) (x + 2y) + (2x – 3y)i + 4i = 5

(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)

Solution:

(i) (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) × 2 – equation (ii) gives

7y = 14

∴ y = 2

Putting y- 2 in (i), we get

x + 2(2) = 5

∴ x + 4 = 5

∴ x = 1

∴ x = 1 and y = 2

Check:

If x = 1 and y = 2 satisfy the given condition, then our answer is correct.

L.H.S. = (x + 2y) + (2x – 3y)i + 4i

= (1 + 4) + (2 – 6)i + 4i

= 5 – 4i + 4i

= 5

= R.H.S.

Thus, our answer is correct.

(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)

(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

Putting y = 2 in (i), we get

x + 2 = 0

∴ x = -2

∴ x = -2 and y = 2

Question 9.

Find the value of:

(i) x^{3} – x^{2} + x + 46, if x = 2 + 3i

(ii) 2x^{3} – 11x^{2} + 44x + 27, if x = \(\frac{25}{3-4 i}\)

Solution:

(i) x = 2 + 3i

∴ x – 2 = 3i

∴ (x – 2)^{2} = 9i^{2}

∴ x^{2} – 4x + 4 = 9(-1) …..[∵ i^{2} = -1]

∴ x^{2} – 4x + 13 = 0 ……(i)

∴ x^{3} – x^{2} + x + 46 = (x^{2} – 4x + 13)(x + 3) + 7

= 0(x + 3) + 7 ……[From (i)]

= 7

(ii) x = \(\frac{25}{3-4 i}\)

∴ x = 3 + 4i

∴ x – 3 = 4i

∴ (x – 3)^{2} = 16i^{2}

∴ x^{2} – 6x + 9 = 16(-1) …….[∵ i^{2} = -1]

∴ x^{2} – 6x + 25 = 0 …….(i)

∴ 2x^{3} – 11x^{2} + 44x + 27

= (x^{2} – 6x + 25) (2x + 1) + 2

= 0 . (2x + 1) + 2 ……[From (i)]

= 0 + 2

= 2