Complex Numbers & Quadratic Equations Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. Express the following expression in the form of a + bi

(1 – i) – (-1 + i6)

Ans. (1 – i) – (-1 + i6) = 1 – i + 1 – i6

= 2 – 7i

Ques. Find the conjugate of √-3 + 4i².

Ans. Simplify the expression -

√-3 + 4i² = √3i - 4

Thus, the conjugate will be \(\bar{Z}\) = -√3i - 4.

Ques. Express (5 - 3i)³ in the form of a + bi.

Ans. (5 - 3i)³ = 5³ - 3 × 5² × (3i) + 3 × 5(3i)² - (3i)³

= 125 - 225i - 135 + 27i

= 10 - 198i

Ques. Solve: -i

Ans. -i = 1/-i

= 1/-i × i/i = i/-i²

= i/1 = i

Ques. Find the value of √(-16).

Ans. √(-16) = √(16) × √(-1)

= 4i [ i = √(-1) ]

Short Answer Questions [2 Marks Questions]

Ques. Solve: (1+i⁴).

Ans. (1+i⁴) = [(1+ i)²]²

= (1+ i² + 2i)²

= (1 - 1 + 2i)²

= (2i)²

= 4i²

= 4(-1) = -4

Ques. Solve : i^-39.

Ans.  i^-39 = 1/i³⁹

= 1/ (i⁴)⁹ . i³

= 1 / 1 × (-i) [ i⁴ = 1, i³ = -i]

= 1/-i × i/i

= i/-i² = i / -(-1) = i [i² = -1]

Ques. Express the following expression in the form of a + bi

i⁹ + i¹⁹

Ans. i⁹ + i¹⁹ = i^4×2+1  + i^4×4+3

= (i⁴)² . i + (i⁴)⁴ . i³

= 1 × i + 1 × (-i) [ i⁴ = 1, i³ = -i]

= i + (-i)

= 0

Ques. Find the value of √(-25) + 3√(-4) + 2√(-9).

Ans. √(-25) + 3√(-4) + 2√(-9)

= √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9}

= √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9}

= 5i + 3×2i + 2×3i [√(-1) = i]

= 5i + 6i + 6i

= 17i.

Ques. Express the following in the form of a + bi

1. (-5i) (1/8 i)
2. (-i) (2i) (-1/8 i)³

Ans. (i) (-5i) (1/8i) = \(-\frac{5}{8}\) i²

= \(-\frac{5}{8}\) (-1) 

= \(\frac{5}{8}\)

= \(\frac{5}{8}\) + i0

(ii) (-i) (2i) (-1/8 i)³ = 2 × 1/ 8 × 8 × 8 × i³

= 1/ 256 (i²) . i

= -1/ 256 i

Long Answer Questions [3 Marks Questions]

Ques. Find the least value for n, in the expression {(1 + i) / (1 - i)}n, where n is a real number.

Ans. {(1 + i)/(1 – i)}n

= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]n

= [{(1 + i)²}/{(1 – i²)}]n

= [(1 + i² + 2i)/{1 – (-1)}]n

= [(1 – 1 + 2i)/{1 + 1}]n

= [2i/2]n

= in

Now, in is real when n = 2 [ i² = -1]

So, the least value of n is 2.

Ques. Find the value of i^-999.

Ans. i^-999 = 1/i⁹⁹⁹

= 1/(i⁹⁹⁶ × i³)

= 1/{(i⁴)²⁴⁹ × i³}

= 1/{1²⁴⁹ × i³} [ i⁴ = 1]

= 1/i³

= i⁴/i³ [ i⁴ = 1]

= i

Thus, i^-999 = i

Ques. Find the value of x and y in the expression -

(3y – 2) + i(7 – 2x) = 0

Ans. (3y – 2) + i(7 – 2x) = 0

First, compare the real and imaginary parts, we find

3y – 2 = 0

⇒ y = 2/3

and 7 – 2x = 0

⇒ x = 7/2

So, the value of x = 7/2 and y = 2/3

Ques. If {(1 + i) / (1 -i)}n = 1, then find the least value of n.

Ans. {(1 + i) / (1 -i)}n = 1

⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]n = 1

⇒ [{(1 + i)²}/{(1 – i²)}]n = 1

⇒ [(1 + i² + 2i)/{1 – (-1)}]n = 1

⇒ [(1 – 1 + 2i)/{1 + 1}]n = 1

⇒ [2i/2]n = 1

⇒ in = 1

Now, in is 1 when n = 4

Thus, the least value of n is 4.

Ques. Find the modulus of 5 + 4i.

Ans. let z = 5 + 4i

Now, the modulus of z will be calculated as

|Z| = √(5² + 4²)

|Z| = √(25 + 16)

|Z| = √41

Thus, the modulus of 5 + 4i = √41.

Ques. Find the multiplicative inverse of the complex number √5 + 3i.

Ans. Let z = √5 + 3i

Then, √5 + \(\bar{Z}\) = √5 + 3i

And |Z|² = (√5 )² + (3)² 

= 5 + 9 = 14

So, the multiplicative inverse of √5 + 3i is given by-

Z^-1 = \(\bar{Z}\) / |Z|² = √5 + 3i / 14

= √5 / 14 – 3i / 14

Ques. Express the following expression in the form of a + bi

(\(\frac{1}{3}\) + 3i)³

Ans. (\(\frac{1}{3}\) + 3i)³ = (1/3)³ + (3i)³ + 3(1/3) (3i) (1/3 +3i)

= (1/27) + 27i³ + 3i(1/3 +3i)

= 1/27 + 27(-i) + i + 9i² [i³ = -i]

= 1/27 – 27i + i – 9 [ i² = -1]

= (1/27 – 9) + i(-27 + 1)

= -242/27 – 26i

Very Long Answer Questions [5 Marks Questions]

Ques. Find the sum of the complex numbers -√3 + √2i and 2√3 - i

Ans. z₁ + z₂ = -√3 + √2i + 2√3 – i

= √3 + (√2 -1)i

z₁z₂ = (-√3 + √2i) (2√3 - i)

= -6 + √3i + 2√6i - √2 i² 

= -6 + √3i + 2√6i + √2

= (-6 + √2) + (√3 + 2√6)i

Ques. Find the value of x²⁰⁰⁰ + 1/x²⁰⁰⁰, if x + 1/x = 1.

Ans. Given x + 1/x = 1

⇒ (x² + 1) = x

⇒ x² – x + 1 = 0

⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1)

⇒ x = {1 ± √(1 – 4)}/2

⇒ x = {1 ± √(-3)}/2

⇒ x = {1 ± √(-1)×√3}/2

⇒ x = {1 ± i√3}/2 [ i = √(-1)]

⇒ x = -w, -w²

Now, put x = -w, we get

x²⁰⁰⁰ + 1/x²⁰⁰⁰ = (-w)²⁰⁰⁰ + 1/(-w)²⁰⁰⁰ 

= w²⁰⁰⁰ + 1/w²⁰⁰⁰ 

= w²⁰⁰⁰ + 1/w²⁰⁰⁰

= {(w³)⁶⁶⁶ × w²} + 1/{(w³)⁶⁶⁶ × w²}

= w² + 1/w² [w³ = 1]

= w² + w³ /w²

= w² + w

= -1 [1 + w + w² = 0]

Thus, x²⁰⁰⁰ + 1/x²⁰⁰⁰ = -1

Ques. Solve the quadratic equation - 2x² + x + 1 =0

Ans. 2x² + x + 1 = 0 (given)

Let’s compare this quadratic equation with the general form ax² + bx + c = 0

On comparing, we get

a = 2, b = 1 and c = 1

So, the discriminant of the equation is -

D = b² – 4ac

Now, substitute the values in the above formula

D = (1)² – 4(2)(1)

D = 1 – 8

D = -7

Therefore, the required solution for the given quadratic equation will be

x =[-b ± √D]/2a

x = [-1 ± √-7]/2(2)

We know that, √-1 = i

x = [-1 ± √7i] / 4

Thus, the solution for the given quadratic equation is (-1 ± √7i) / 4.

Ques. Express the following term in the form of a + ib: 

(3i – 7) + (7 – 4i) – (6 + 3i) + i²³

Ans. (3i – 7) + (7 – 4i) – (6 + 3i) + i²³

= 3i – 7 + 7 – 4i – 6 – 3i + i²³

= -4i – 6 + i²²⁺¹

= -4i – 6 + (i²)¹¹ . i

= -4i – 6 + (-1)¹¹ . i

= -4i – 6 – i

Thus, (3i – 7) + (7 – 4i) – (6 + 3i) + i²³ = -6 – 5i

Ques. The complex numbers sin x + i cos 2x are conjugate to each other. Find the value of x. 

Ans. Complex number = sin x + i cos 2x (given)

Conjugate number = sin x – i cos 2x

Now, sin x + i cos 2x = sin x – i cos 2x

⇒ sin x = cos x and sin 2x = cos 2x [comparing real and imaginary part]

⇒ tan x = 1

and tan 2x = 1

Now, both of them are not possible for the same value of x.

Therefore, x has no value.

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