## MP Board Class 11th Maths Important Questions Chapter 5 Complex Numbers and Quadratic Equations

### Relations and Functions Important Questions

**Relations and Functions Objective Type Questions**

(A) Choose the correct option :

Question 1.

If x + iy = 2 + 3i, then (x, y) will be :

(a) (3, 2)

(b) (2, 3)

(c) (- 2,- 3)

(d) (3, 3)

Answer:

(b) (2, 3)

Question 2.

The simplest form of \(\frac { 1 + i }{ 1 – i }\) :

(a) – i

(b) i

(c) ± i

(d) None of these.

Answer:

(b) i

Question 3.

If z is a complex number, then z + z is :

(a) Real number

(b) Imaginary number

(c) Nothing can be said

(d) None of these.

Answer:

(a) Real number

Question 4.

The value of \(\sqrt {i}\) = ……………..

(a) ± \(\frac { 1 – i }{ \sqrt { 2 } }\)

(b) ± \(\frac { 1 + i }{ \sqrt { 2 } }\)

(c) ± (1 + i)

(d) ± (1 – i)

Answer:

(b) ± \(\frac { 1 + i }{ \sqrt { 2 } }\)

Question 5.

Number of solution of the equation z + z = 0 is :

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(d) 4

Question 6.

Argument of complex number 0 is :

(a) 0

(b) π

(c) 2π

(d) Not defined

Answer:

(d) Not defined

Question 7.

Conjugate of \(\frac { 1 – i }{ 1 + i }\) is :

(a) \(\frac { 1 – i }{ 1 + i }\)

(b) \(\frac { 1 – i }{ 1 – i }\)

(c) i

(d) – i

Answer:

(c) i

Question 8.

Argument (z_{1}.z_{2}) = ………………

(a) arg (z_{1}) – arg (z_{2})

(b) arg (z_{1}) + arg (z_{2})

(c) arg (z_{1}). arg (z_{2})

(d) None of these

Answer:

(b) arg (z_{1}) + arg (z_{2})

Question 9.

The least positive integer n for which \(\frac { 1 – i }{ 1 + i }\)^{n} = 1 is :

(a) 2

(b) 4

(c) 8

(d) 12

Answer:

(b) 4

Question 10.

If |z – \(\frac { 4}{ z}\)| = 2, then |z| has maximum value:

(a) \(\sqrt {3}\) + 1

(b) \(\sqrt {5}\) + 1

(c) 2

(d) 2 + \(\sqrt {2}\)

Answer:

(b) \(\sqrt {5}\) + 1

(B) Match the following :

Answer:

- (d)
- (a)
- (f)
- (b)
- (c)
- (e)

(c) Fill in the blanks :

- The value of (1 + i)
^{4 }(1 + \(\frac {1}{i}\))^{4}is ………………. - The value of \(\frac {1}{i}\) is ……………….
- The value of z – \(\bar { z }\) is ……………….
- Argument of – 1 – i is ……………….
- polar form of i
^{3}is ………………. - Argument of complex number is \(\frac { 1+\sqrt { 3i } }{ \sqrt { 3 } +i }\) is ……………….
- If |\(\frac { z – i }{ z + i }\)| = 1, then locus of z will be ……………….
- If x + \(\frac {1}{x}\) = 2cosθ, then the value of x
^{n}= \(\frac { 1 }{ { x }^{ n } }\) will be ………………. - If z = 2 – 3i, then the value of z. \(\bar { z }\) is ……………….
- If 2 + (x + yi) = 3 – i, the x = ……………… and y = ……………….

Answer:

- 16
- – i
- Imaginary number
- \(\frac {5π}{4}\)
- cos\(\frac {π}{2}\) – i sin\(\frac {π}{2}\)
- \(\frac {π}{6}\)
- X – axis
- 2cos nθ
- 13
- 1, – 1

(D) Write true / false :

- Additive inverse of complex number – 2 + 5i is 2 – 5 i.
- If | z
^{2}– 1 |= z^{2 }+ 1, then z lies on a circle. - If | z
_{1}| = 12 and | z_{2}– 3 – 4i | = 5, then the least value of | z_{1}– z_{2}| is 7. - If arg(z) < 0, then arg(- z) – arg(z) = π.
- Polar form of 1 + \(\sqrt {- 1}\) is \(\sqrt {2}\)(cos\(\frac {π}{4}\) + isin\(\frac {π}{4}\)).
- If z is a complex number such that | z | ≥ 2, then the minimum value of |z + \(\frac { 1 }{ 2 }\) | equal \(\frac { 5 }{ 2 }\)

Answer:

- True
- False
- False
- True
- True
- False.

(E) Write answer in one word / sentence :

- If Z
_{1}= 2 – i, z_{2}= 1 + i, them find the value \(\frac { { { z }_{ 1 }+{ z }_{ 2 }+1 } }{ { z }_{ 1 }-{ z }_{ 2 }+1 }\) - Modulus of \(\frac {1 – i }{ 1 + i }\) – \(\frac { 1 – i }{ 1 + i }\) will be.
- Polar form of complex number \(\sqrt {3}\) + i will be.
- If Z
_{1}= 2 – i and z_{2}= – 2 + i, then Re(\(\frac { { { z }_{ 1 }{ z }_{ 2 } } }{ { z }_{ 1 } }\)) will be - If the number \(\frac { z – 1 }{ z + 1 }\) is purly imaginary, then the value of |z| will be.

Answer:

- \(\sqrt {2}\)
- 2
- 2(cos\(\frac {π}{6}\) + isin\(\frac {π}{6}\))
- \(\frac { – 2 }{ 5 }\)
- 1

**Complex Numbers and Quadratic Equations Very Short Answer Type Questions**

Question 1.

If 1, ω, ω^{2} are the cube root of unity then find the value of ω^{3n}.

Solution:

Since ω^{3} = 1

ω^{3n} = (ω^{3})^{n} = (1)^{n} = 1.

Question 2.

Find the value of i^{4n}.

Solution:

i^{4n} = (i^{4})^{n} = (i^{2}.i^{2})^{n} = [(- 1)(- 1)]^{n} = (1)^{n} = 1.

Question 3.

Find the condition of two complex numbers x + iy and a + ib are compaired.

Solution:

If x + iy = a + ib

Then, x = a, y = b.

Question 4.

Write conjugate of a complex number z = a – ib.

Solution:

z = conjugate of a – ib = a – (- ib)

= a + ib.

Question 5.

Write the complex number – 6_{3} in ordered form.

Solution:

Ordered form of z = a + ib is (a, b).

Given: z = 0 – 6i

Ordered pair (0, – 6).

Question 6.

Write the complex number 3 – \(\sqrt { 7i }\) in ordered form.

Solution:

Ordered pair of z = a + ib is (a, b).

∴Ordered pair of 3 – \(\sqrt { 7i }\) = (3, – \(\sqrt { 7i }\))

Question 7.

Find the value of i + \(\frac { 1 }{ i }\).

Solution:

i + \(\frac { 1 }{ i }\) = \(\frac { { i }^{ 2 }+1 }{ i }\) = \(\frac { – 1 + i }{ i }\) = 0.

Question 8.

Find the value of i^{-19}.

Solution:

i^{-19} = \(\frac { 1 }{ { i }^{ 19 } }\) = \(\frac { 1 }{ { i }^{ 18 }i } \) = \(\frac { 1 }{ (i^{ 2 })^{ 9 }.i }\)

= \(\frac { 1 }{ -1.i }\) = \(\frac { – 1 }{ i }\)

= \(\frac { – 1.i }{ { i }^{ 2 } }\) = i.

Question 9.

Write (1 – i)^{4} in form of a + ib.

Solution:

(1 – i)^{4} = [ (1 – i)^{2} ]^{2} (NCERT)

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

= (- 2i)^{2} = 4i^{2} = 4(- 1) = – 4

= – 4 + 0i.

Question 10.

Find the modulus of complex number z = – 1 – i \(\sqrt {3}\) .

Solution:

Z = – 1 – i\(\sqrt {3}\)

a = – 1, b = \(\sqrt {3}\)

Modulus = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\) = \(\sqrt {1 + 3}\) = \(\sqrt {2}\)

Question 11.

Find the modulus of complex number 1 – i.

Solution:

z = 1 – i

a = 1, b = – 1

Modulus = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\) = \(\sqrt {1 + 1}\) = \(\sqrt {2}\)

Question 12.

Find the sum of 2 – 3i and its conjugate.

Solution:

z = 2 – 3i and \(\bar { z }\) = 2 + 3i

z + \(\bar { z }\) = 2 – 3i + 2 + 3i = 4.

Question 13.

Solve the quadratic equation x^{2} + 2 = 0. (NCERT)

Solution:

x^{2} + 2 = 0

⇒ x^{2} = – 2

⇒ x^{2} = – 1 x 2

= 2i^{2} [∵ i^{2} = – 1]

x = ± \(\sqrt {2i}\) or + \(\sqrt {2}\)i, – \(\sqrt {2}\)i

Question 14.

Solve the quadratic equation x^{2} + 3 = 0. (NCERT)

Solution:

x^{2} + 3 = 0

⇒ x^{2} = – 3 = – 1 x 3

⇒ x^{2} = 3i^{2}, [∵ i^{2} = – 1]

x^{2} = ± \(\sqrt {3i}\) or + \(\sqrt {3}\)i, – \(\sqrt {3}\)i

Question 15.

Write the triangle inequality for the two complex number Z_{1} and z_{2}.

Solution:

| Z_{1}| and | z_{2}| are the two sides of any triangle

Then, | z_{0} + z_{2}| ≤ | z_{1}| +| z_{2}|

Question 16.

If T and 9 are the modulus and argument of the complex number a + ib, then write its polar form.

Solution:

Polar form of a + ib is r(cosθ + i sinθ).

Question 17.

If z = cosθ + i sinθ where |z| = 1, then find \(\frac { 1 }{ z }\).

Solution:

\(\frac { 1 }{ z }\) = cosθ – isinθ.

Question 18.

Write the statement of De – moivre’s theorem.

Solution:

If polar form of z = a + ib is r(cosθ + isinθ), then De – moivre’s theorem is

(cosθ + isinθ)n = cos nθ + i sin nθ.

Question 19.

Write polar form of i.

Solution:

a + ib = 0 + i. 1

r = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\) =1, θ = tan^{-1}\(\frac { b }{ a}\) = tan^{-1}∞ = \(\frac { π }{ 2}\)

Polar form of i = (cos\(\frac { π }{ 2}\) + i sin \(\frac { π }{ 2}\)).

Question 20.

Find the polar form of (- 2 + 2i).

Solution:

**Complex Numbers and Quadratic Equations Long Answer Type Questions**

Question 1.

Convert (\(\frac { 1 }{ 3}\) + 3i)^{3} in form of a + ib. (NCERT)

Solution:

Question 2.

Write ( – 2 – \(\frac { 1 }{ 3}\)i)^{3} in form of a + ib. (NCERT)

Solution:

Question 3.

Write (5 – 3i)^{3} in the form of a + ib.

Solution:

(5 – 3i)^{3} = (5)^{3} – (3i)^{3} – 3(5^{2}.3i) + 3.5(3i)^{2}

= 125 – 27.i^{3} – 75.3i + 15.9i^{2}

= 125 – 27i.i^{2} – 225i + 15(- 9)

= 125 + 27i – 225i – 135

= – 10 – 198i.

Find the multiplicative inverse of following complex number :

Question 4.

4 – 3i

Solution:

Let the multiplicative inverse of 4 – 3i is a + ib.

Then, (4 – 3i) x (a + ib) = 1

Question 5.

\(\sqrt {5}\) + 3i

Solution:

Let z = \(\sqrt {5}\) + 3i

Be multiplicative inverse

Question 6.

Find the polar form of the following :

(i) \(\frac { 1 + 7i }{ { 2 – i }^{ 2 } }\), (ii) \(\frac { 1 + 3i }{ 1 – 2i }\)

Solution:

Now, let – 1 + i = r(cosθ + i sinθ),

Then, r cosθ = – 1 …. (1)

and r sinθ = 1 …. (2)

Squaring and adding equation (1) and (2),

r^{2} = 1 + 1 = 2

⇒ r = \(\sqrt {2}\)

Put value of r in equation (1) and (2),

Comparing the real and imaginary parts,

r cosθ = – 1

r sinθ = 1

Squaring and adding equation (1) and (2),

∵ Real part is negative and imaginary part is positive of z, hence θ will lie in 4^{th} quadrant

Question 7.

Find modulus and argument of complex number z = – 1 – i\(\sqrt {3}\) (NCERT)

Solution:

Let z = – 1 – i\(\sqrt {3}\) = rcosθ + ir sinθ

Comparing the real and imaginary parts,

rcosθ = – 1 …. (1)

rsinθ = – \(\sqrt {3}\) …. (2)

Squaring and adding equation (1) and (2),

Since both the real and imaginary part of z are negative.

∴ θ lies in third quadrant

Question 8.

Find the modulus and argument of z = – \(\sqrt {3}\)+ i.

Solution:

Let z = – \(\sqrt {3}\)+ i = rcosθ + ir sinθ

Comparing the real and imaginary parts,

r cosθ = – \(\sqrt {3}\)

rsinθ = 1

Squaring and adding eqns. (1) and (2),

rcosθ = – \(\sqrt {3}\) …. (1)

rsinθ = 1 …. (2)

Squaring and adding equation (1) and (2),

Since both the real and imaginary part of z are negative and positive respectively.

∴ θ lies in 2^{2nd} quadrant

Question 9.

Find the modulus and argument of 1 – i.

Solution:

Let z = 1 – i = rcosθ + ir smθ

Comparing the real and imaginary parts,

1 = r cosθ

and – 1 = rsinθ

Squaring and adding equation (1) and (2),

(1)^{2} + (- 1)^{2} = r^{2} cos^{2}0 + r^{2} sin^{2}0

⇒ r^{2}(cos^{2}0 + sin^{2}0) = 1 + 1,

⇒ r^{2}= 2 ⇒ r = \(\sqrt {2}\) (modulus)

Dividing equation (2) by equation (1),

Question 10.

Find square root of complex number 3 + 4i.

Solution:

Let \(\sqrt {3 + 4i}\) = x + iy

On squaring,

3 + 4 i = (x + iy)^{2}

⇒ 3 + 4i = x^{2} + 2ixy + i^{2}y^{2}

⇒ 3 + 4i = x^{2} + 2ixy – y^{2}

⇒ 3 + 4i = (x^{2} – y^{2}) + i(2xy)

x^{2} – y^{2} = 3 …. (1)

2xy = 4 …. (2)

(x^{2} + y^{2})^{2} = (x^{2} – y^{2})^{2} + 4x^{2}y^{2}

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

= 25

∴ x^{2} + y^{2} = 5 …. (3)

Now adding eqns. (1) and (3),

2x^{2} = 8

⇒ x^{2} = 4

⇒ x = ± 2

From equation (2), 2(± 2)y = 4

∴ y = ± 1

\(\sqrt { x + iy }\) = ± 2 ± i.

Question 11.

Find square root of complex number 8 – 6i.

Solution:

Let \(\sqrt { 8 – 6i }\) = x + iy

On squaring,

8 – 6i = (x + iy)^{2}

⇒ 8 – 6i = x^{2} + 2ixy + i^{2}y^{2} .

⇒ 8 – 6i = (x^{2} – y^{2}) + 2ixy

∴ 8 = x^{2} – y^{2} …. (1)

and – 6 = 2 xy …. (2)

(x^{2} + y^{2})^{2} = (x^{2} – y^{2})^{2} + 4x^{2}y^{2}

(x^{2}+y^{2})^{2} = (8)^{2} + (- 6)^{2}

= 64 + 36 = 100

∴ x^{2} + y^{2} = 10 …. (3)

Now adding equation (1) and (3),

2x^{2} = 18

⇒ x^{2} = 9

⇒ x = ±3

From equation (3), y^{2} = 10 – 9 = 1

∴ y = ± 1

\(\sqrt { 8 – 6i}\) = ± 3 ± i.

Question 12.

x^{2} + 3x + 9 = 0.

Solution:

Given : x^{2} + 3x + 9 = 0

Comparing the above equation by ax^{2} + bx + c = 0,

a = 1, b = 3, c = 9

Now, D = b^{2} – 4ac

= 3^{2} – 4 x 1 x 9 = 9 – 36 = – 27 < 0

Question 13.

– x^{2} + x – 2 = 0.

Solution:

Given : – x^{2} + x – 2 = 0

Comparing the above equation by ax^{2} + bx + c = 0,

Question 14.

x^{2} + 3x + 5 = 0.

Solution:

Given : x^{2} + 3x + 5 = 0

Comparing the above equation by ax^{2} + bx+c = 0,

a = 1, b = 3, c = 5

Question 15.

x^{2} – x + 2 = 0

Solution:

Given : x^{2} – x + 2 = 0

Comparing the above equation by ax^{2} + bx + c = 0,

a = 1, b = – 1, c = 2

Now, D = b^{2} – 4ac

= (- 1)^{2} – 4 x 1 x 2 = 1 – 8 = – 7 < 0

Question 16.

If (\(\frac { 1 + i }{ 1 – i }\)) = 1, then And the minimum positive integer number of m. (NCERT)

Solution: