**CBSE Class 11 Maths – MCQ and Online Tests – Unit 5 – Complex Numbers and Quadratic Equations**

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**CBSE Class 11 Maths – MCQ and Online Tests – Unit 5 – Complex Numbers and Quadratic Equations**

Question 1.

If arg (z) < 0, then arg (-z) – arg (z) =

(a) p

(b) -p

(c) -p/2

(d) p/2

## Answer

Answer: (a) p

Hint:

Given, arg (z) < 0

Now, arg (-z) – arg (z) = arg(-z/z)

? arg (-z) – arg (z) = arg(-1)

? arg (-z) – arg (z) = p {since sin p + i cos p = -1, So arg(-1) = p}

Question 2.

if x + 1/x = 1 find the value of x^{2000} + 1/x^{2000} is

(a) 0

(b) 1

(c) -1

(d) None of these

## Answer

Answer: (c) -1

Hint:

Given x + 1/x = 1

? (xВ + 1) = x

? xВ – x + 1 = 0

? x = {-(-1) Б v(1В – 4 з 1 з 1)}/(2 з 1)

? x = {1 Б v(1 – 4)}/2

? x = {1 Б v(-3)}/2

? x = {1 Б v(-1)зv3}/2

? x = {1 Б iv3}/2 {since i = v(-1)}

? x = -w, -wВ

Now, put x = -w, we get

x^{2000} + 1/x^{2000} = (-w)^{2000} + 1/(-w)^{2000}

= w^{2000} + 1/w^{2000}

= w^{2000} + 1/w^{2000}

= {(wГ)^{666} з wВ} + 1/{(wГ)^{666} з wВ}

= wВ + 1/wВ {since wГ = 1}

= wВ + wГ /wВ

= wВ + w

= -1 {since 1 + w + wВ = 0}

So, x^{2000} + 1/x^{2000} = -1

Question 3.

If the cube roots of unity are 1, ?, ?В, then the roots of the equation (x – 1)Г + 8 = 0 are

(a) -1, -1 + 2?, – 1 – 2?В

(b) – 1, -1, – 1

(c) – 1, 1 – 2?, 1 – 2?В

(d) – 1, 1 + 2?, 1 + 2?В

## Answer

Answer: (c) – 1, 1 – 2?, 1 – 2?В

Hint:

Note that since 1, ?, and ?В are the cube roots of unity (the three cube roots of 1), they are the three solutions to xГ = 1 (note: ? and ?В are the two complex solutions to this)

If we let u = x – 1, then the equation becomes

uГ + 8 = (u + 2)(uВ – 2u + 4) = 0.

So, the solutions occur when u = -2 (giving -2 = x – 1 ? x = -1), or when:

uВ – 2u + 4 = 0,

which has roots, by the Quadratic Formula, to be u = 1 Б iv3

So, x – 1 = 1 Б iv3

? x = 2 Б iv3

Now, xГ = 1 when xГ – 1 = (x – 1)(xВ + x + 1) = 0, giving x = 1 and

xВ + x + 1 = 0

? x = (-1 Б iv3)/2

If we let ? = (-1 – iv3)/2 and ?2 = (-1 + iv3)/2

then 1 – 2? and 1 – 2?В yield the two complex solutions to (x – 1)Г + 8 = 0

So, the roots of (x – 1)Г + 8 are -1, 1 – 2?, and 1 – 2?В

Question 4:

The value of v(-25) + 3v(-4) + 2v(-9) is

(a) 13i

(b) -13i

(c) 17i

(d) -17i

## Answer

Answer: (c) 17i

Hint:

Given, v(-25) + 3v(-4) + 2v(-9)

= v{(-1) з (25)} + 3v{(-1) з 4} + 2v{(-1) з 9}

= v(-1) з v(25) + 3{v(-1) з v4} + 2{v(-1) з v9}

= 5i + 3з2i + 2з3i {since v(-1) = i}

= 5i + 6i + 6i

= 17i

So, v(-25) + 3v(-4) + 2v(-9) = 17i

Question 5.

if z lies on |z| = 1, then 2/z lies on

(a) a circle

(b) an ellipse

(c) a straight line

(d) a parabola

## Answer

Answer: (a) a circle

Hint:

Let w = 2/z

Now, |w| = |2/z|

=> |w| = 2/|z|

=> |w| = 2

This shows that w lies on a circle with center at the origin and radius 2 units.

Question 6.

If ? is an imaginary cube root of unity, then (1 + ? – ?В)^{7} equals

(a) 128 ?

(b) -128 ?

(c) 128 ?В

(d) -128 ?В

## Answer

Answer: (d) -128 ?В

Hint:

Given ? is an imaginary cube root of unity.

So 1 + ? + ?В = 0 and ?Г = 1

Now, (1 + ? – ?В)^{7} = (-?В – ?В)^{7}

? (1 + ? – ?^{2})^{7} = (-2?^{2})^{7}

? (1 + ? – ?^{2})^{7} = -128 ?^{14}

? (1 + ? – ?^{2})^{7} = -128 ?^{12} з ?^{2}

? (1 + ? – ?^{2})^{7} = -128 (?^{3})^{4} ?^{2}

? (1 + ? – ?^{2})^{7} = -128 ?^{2}

Question 7.

The least value of n for which {(1 + i)/(1 – i)}^{n} is real, is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

Answer: (b) 2

Hint:

Given, {(1 + i)/(1 – i)}^{n}

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

= [{(1 + i)В}/{(1 – iВ)}]^{n}

= [(1 + iВ + 2i)/{1 – (-1)}]^{n}

= [(1 – 1 + 2i)/{1 + 1}]^{n}

= [2i/2]^{n}

= i^{n}

Now, in is real when n = 2 {since i2 = -1 }

So, the least value of n is 2

Question 8.

Let z be a complex number such that |z| = 4 and arg(z) = 5p/6, then z =

(a) -2v3 + 2i

(b) 2v3 + 2i

(c) 2v3 – 2i

(d) -v3 + i

## Answer

Answer: (a) -2v3 + 2i

Hint:

Let z = r(cos ? + i з sin ?)

Then r = 4 and ? = 5p/6

So, z = 4(cos 5p/6 + i з sin 5p/6)

? z = 4(-v3/2 + i/2)

? z = -2v3 + 2i

Question 9:

The value of i^{-999} is

(a) 1

(b) -1

(c) i

(d) -i

## Answer

Answer: (c) i

Hint:

Given, i^{-999}

= 1/i^{999}

= 1/(i^{996} з iГ)

= 1/{(i^{4})^{249} з i^{3}}

= 1/{1^{249} з i^{3}} {since i^{4} = 1}

= 1/i^{3}

= i^{4}/i^{3} {since i^{4} = 1}

= i

So, i^{-999} = i

Question 10.

Let z_{1} and z_{2} be two roots of the equation zВ + az + b = 0, z being complex. Further assume that the origin, z_{1} and z_{1} form an equilateral triangle. Then

(a) aВ = b

(b) aВ = 2b

(c) aВ = 3b

(d) aВ = 4b

## Answer

Answer: (c) aВ = 3b

Hint:

Given, z_{1} and z_{2} be two roots of the equation zВ + az + b = 0

Now, z_{1} + z_{2} = -a and z_{1} з z_{2} = b

Since z_{1} and z_{2} and z_{3} from an equilateral triangle.

? z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} з z_{2} + z_{2} з z_{3} + z_{1} з z_{3}

? z_{1}^{2}+ z_{2}^{2} = z_{1} з z_{2} {since z_{3} = 0}

? (z_{1} + z_{2})В – 2z_{1} з z_{2} = z_{1} з z_{2}

? (z_{1} + z_{2})В = 2z_{1} з z_{2} + z_{1} з z_{2}

? (z_{1} + z_{2})В = 3z_{1} з z_{2}

? (-a)В = 3b

? aВ = 3b

Question 11.

The value of v(-16) is

(a) -4i

(b) 4i

(c) -2i

(d) 2i

## Answer

Answer: (b) 4i

Hint:

Given, v(-16) = v(16) з v(-1)

= 4i {since i = v(-1) }

Question 12.

The value of v(-144) is

(a) 12i

(b) -12i

(c) Б12i

(d) None of these

## Answer

Answer: (a) 12i

Hint:

Given, v(-144) = v{(-1) з 144}

= v(-1) з v(144)

= i з 12 {Since v(-1) = i}

= 12i

So, v(-144) = 12i

Question 13.

The value of x and y if (3y – 2) + i(7 – 2x) = 0

(a) x = 7/2, y = 2/3

(b) x = 2/7, y = 2/3

(c) x = 7/2, y = 3/2

(d) x = 2/7, y = 3/2

## Answer

Answer: (a) x = 7/2, y = 2/3

Hint:

Given, (3y – 2) + i(7 – 2x) = 0

Compare real and imaginary part, we get

3y – 2 = 0

? y = 2/3

and 7 – 2x = 0

? x = 7/2

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

Question 14.

Find real ? such that (3 + 2i з sin ?)/(1 – 2i з sin ?) is imaginary

(a) ? = np Б p/2 where n is an integer

(b) ? = np Б p/3 where n is an integer

(c) ? = np Б p/4 where n is an integer

(d) None of these

## Answer

Answer: (b) ? = np Б p/3 where n is an integer

Hint:

Given,

(3 + 2i з sin ?)/(1 – 2i з sin ?) = {(3 + 2i з sin ?)з(1 – 2i з sin ?)}/(1 – 4iВ з sinВ ?)

(3 + 2i з sin ?)/(1 – 2i з sin ?) = {(3 – 4sinВ ?) + 8i з sin ?}/(1 + 4sinВ ?) …………. 1

Now, equation 1 is imaginary if

3 – 4sinВ ? = 0

? 4sinВ ? = 3

? sinВ ? = 3/4

? sin ? = Бv3/2

? ? = np Б p/3 where n is an integer

Question 15.

If {(1 + i)/(1 – i)}^{n} = 1 then the least value of n is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

Answer: (d) 4

Hint:

Given, {(1 + i)/(1 – i)}^{n} = 1

? [{(1 + i) з (1 + i)}/{(1 – i) з (1 + i)}]^{n} = 1

? [{(1 + i)В}/{(1 – iВ)}]^{n} = 1

? [(1 + iВ + 2i)/{1 – (-1)}]^{n} = 1

? [(1 – 1 + 2i)/{1 + 1}]^{n} = 1

? [2i/2]^{n} = 1

? i^{n} = 1

Now, i^{n} is 1 when n = 4

So, the least value of n is 4

Question 16.

(1 – w + wВ)з(1 – wВ + w^{4})з(1 – w^{4} + w^{8}) з …………… to 2n factors is equal to

(a) 2^{n}

(b) 2^{2n}

(c) 2^{3n}

(d) 2^{4n}

## Answer

Answer: (b) 2^{2n}

Hint:

Given, (1 – w + wВ)з(1 – wВ + w^{4})з(1 – w^{4} + w^{8}) з …………… to 2n factors

= (1 – w + w^{2})з(1 – w^{2} + w )з(1 – w + w^{2}) з …………… to 2n factors

{Since w^{4} = w, w^{8} = w^{2}}

= (-2w) з (-2wВ) з (-2w) з (-2wВ)з …………… to 2n factors

= (2В wГ)з(2В wГ)з(2В wГ) …………… to 2n factors

= (2В)^{n} {since wГ = 1}

= 2^{2n}

Question 17.

The modulus of 5 + 4i is

(a) 41

(b) -41

(c) v41

(d) -v41

## Answer

Answer: (c) v41

Hint:

Let Z = 5 + 4i

Now modulus of Z is calculated as

|Z| = v(5В + 4В)

? |Z| = v(25 + 16)

? |Z| = v41

So, the modulus of 5 + 4i is v41

Question 18.

The value of v(-144) is

(a) 12i

(b) -12i

(c) Б12i

(d) None of these

## Answer

Answer: (a) 12i

Hint:

Given, v(-144) = v{(-1)з144}

= v(-1) з v(144)

= i з 12 {Since v(-1) = i}

= 12i

So, v(-144) = 12i

Question 19:

The complex numbers sin x + i cos 2x are conjugate to each other for

(a) x = np

(b) x = 0

(c) x = (n + 1/2) p

(d) no value of x

## Answer

Answer: (d) no value of x

Hint:

Given complex number = sin x + i cos 2x

Conjugate of this 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.

So, there exist no value of x

Question 20.

The curve represented by Im(zВ) = k, where k is a non-zero real number, is

(a) a pair of striaght line

(b) an ellipse

(c) a parabola

(d) a hyperbola

## Answer

Answer: (d) a hyperbola

Hint:

Let z = x + iy

Now, zВ = (x + iy)В

? zВ = xВ – yВ + 2xy

Given, Im(zВ) = k

? 2xy = k

? xy = k/2 which is a hyperbola.

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