# Miscellaneous Exercise Complex Numbers and Quadratic Equations – NCERT Class 11 Mathematics Chapter 5

1. Evaluate [i18 + (1/i)25]3

Let’s simplify this equation

$$\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^{3} \\ \text{ } \\ \left[ (i^{4})^{4}i^{2} + \left( \frac{1}{i} \right)^{25} \right]^{3} \\ \text{ } \\ \left[ (i^{4})^{4}i^{2} + \frac{1}{i^{25}} \right]^{3} \\ \text{ } \\ \left[ (i^{4})^{4}i^{2} + \frac{1}{(i^{4})^{6}i} \right]^{3} \\ \text{ } \\ \text{Replacing } i^{4} = 1 \\ \text{ } \\ \left[ (1)^{4}i^{2} + \frac{1}{(1)^{6}i} \right]^{3} \\ \text{ } \\ \left[ i^{2} + \frac{1}{i} \right]^{3} \\ \text{ } \\ \text{Replacing } i^{2} = – 1 \\ \text{ } \\ \left[ – 1 + \frac{1}{i} \right]^{3} \\ \text{ } \\ \text{Replacing } \frac{1}{i} = – i \\ \text{ } \\ \left[ – 1 + (- i) \right]^{3} \\ \text{ } \\ \left[ (-1)(1 + i) \right]^{3} \\ \text{ } \\ (-1)^{3}(1 + i)^{3} \\ \text{ } \\ – (1 + i)^{3} \\ \text{ } \\ \text{Using formula } \\ \text{ } \\ (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b) \\ \text{ } \\ – \left[ 1^{3} + i^{3} + 3 × 1 × i (1 + i) \right] \\ \text{ } \\ – \left[ 1 + i^{3} + 3i(1 + i) \right] \\ \text{ } \\ – \left[ 1 + i^{3} + 3i + 3i^{2} \right] \\ \text{ } \\ \text{Replacing } i^{3} = – i \\ \text{ } \\ – \left[ 1 + (- i) + 3i + 3i^{2} \right] \\ \text{ } \\ \text{Replacing } i^{2} = – 1 \\ \text{ } \\ – \left[ 1 + (- i) + 3i + 3(- 1) \right] \\ \text{ } \\ – \left[ 1 – i + 3i – 3 \right] \\ \text{ } \\ – \left[ – 2 + 2i \right] \\ \text{ } \\ 2 – 2i$$

2. For any two complex numbers z1 and z2, prove that Re(z1 z2) = Re(z1) Re(z2) – Im(z1) Im(z2)

Let’s suppose that
z1 = a + ib
z2 = c + id

Where a, b, c, d are Real Numbers and i is Imaginary Number iota

z1 z2 = (a + ib)(c + id) = a(c + id) + ib(c + id)

z1 z2 = ac + iad + ibc + i2bd

Replacing i2 = – 1

z1 z2 = ac + iad + ibc + (- 1)bd

z1 z2 = (ac – bd) + i(ad + bc)

⇒ Re(z1 z2) = ac – bd [Equation 1]

We have already supposed that z1 = a + ib and z2 = c + id
Re(z1) = a
Re(z2) = c

Im(z1) = b
Im(z2) = d

Replacing these values in [Equation 1]
⇒ Re(z1 z2) = Re(z1) Re(z2) – Im(z1) Im(z2)

Hence if z1 and z2 are two Complex Numbers then Re(z1 z2) = Re(z1) Re(z2) – Im(z1) Im(z2)

4. If x – iy = √a – ib/c – id prove that (x2 + y2)2 = a2 + b2/c2 + d2

5. Convert the following in the polar form:

1. 1 + 7i/(2 – i)2

Converting – 1 + i into Polar Form

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (- 1)2 + (1)2 = 1 + 1 = 2

⇒ r2Cos2θ + r2Sin2θ = 2

r2(Cos2θ + Sin2θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 2

r2 = 2

r = ± √2
Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number z = – 1 + i is r = √2

Now let’s find out Argument of the Complex Number z = – 1 + i
Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = 1

Replacing r = √2 in these equations
√2Cosθ = – 1
√2Sinθ = 1

⇒ Cosθ = – 1/√2
⇒ Sinθ = 1/√2

As Sine is positive but Cosine is negative thus θ must lie in Second Quadrant
So for Cosθ = – 1/√2 and Sinθ = 1/√2 value of θ = 3𝜋/4

By definition argument of a Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 and θ = 3𝜋/4 does lies in this interval hence θ = 3𝜋/4 is argument of given complex number – 1 + i

So
r = √2
θ = 3𝜋/4

Polar Form = rCosθ + i(rSinθ) = √2Cos(3𝜋/4) + i(√2Sin(3𝜋/4))
= √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Thus Polar Form of Complex Number z = – 1 + i is √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Hence Polar Form of equation (1 + 7i)/(2 – i)2 = – 1 + i is √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

2. (1 + 3i)/(1 – 2i)

Converting – 1 + i into Polar Form

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (- 1)2 + (1)2 = 1 + 1 = 2

⇒ r2Cos2θ + r2Sin2θ = 2

r2(Cos2θ + Sin2θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 2

r2 = 2

r = ± √2
Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number z = – 1 + i is r = √2

Now let’s find out Argument of the Complex Number z = – 1 + i
Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = 1

Replacing r = √2 in these equations
√2Cosθ = – 1
√2Sinθ = 1

⇒ Cosθ = – 1/√2
⇒ Sinθ = 1/√2

As Sine is positive but Cosine is negative thus θ must lie in Second Quadrant
So for Cosθ = – 1/√2 and Sinθ = 1/√2 value of θ = 3𝜋/4

By definition argument of a Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 and θ = 3𝜋/4 does lies in this interval hence θ = 3𝜋/4 is argument of given complex number – 1 + i

So
r = √2
θ = 3𝜋/4

Polar Form = rCosθ + i(rSinθ) = √2Cos(3𝜋/4) + i(√2Sin(3𝜋/4))
= √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Thus Polar Form of Complex Number z = – 1 + i is √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Hence Polar Form of equation (1 + 3i)/(1 – 2i) = – 1 + i is √2[Cos(3𝜋/4) + i Sin(3𝜋/4)]

Solve each of the equations in Exercises 6 to 9.

6. 3x2 – 4x + 20/3 = 0

Comparing 9x2 – 12x + 20 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 9
⇒ b = – 12
⇒ c = 20

Finding out Discriminant of quadratic equation 2x2 + x + 1 = 0
D = b2 – 4ac

D = (- 12)2 – 4 × 9 × 20
D = 144 – 720

D = – 576

7. x2 – 2x + 3/2 = 0

This Quadratic Equation can be rewritten as 2x2 – 4x + 3 = 0

Comparing 2x2 – 4x + 3 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 2
⇒ b = – 4
⇒ c = 3

Finding out Discriminant of quadratic equation 2x2 – 4x + 3 = 0
D = b2 – 4ac

D = (- 4)2 – 4 × 2 × 3
D = 16 – 24

D = – 8

8. 27x2 – 10x + 1 = 0

Comparing 27x2 – 10x + 1 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 27
⇒ b = – 10
⇒ c = 1

Finding out Discriminant of quadratic equation 27x2 – 10x + 1 = 0
D = b2 – 4ac

D = (- 10)2 – 4 × 27 × 1
D = 100 – 108

D = – 8

9. 21x2 – 28x + 10 = 0

Comparing 21x2 – 28x + 10 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 21
⇒ b = – 28
⇒ c = 10

Finding out Discriminant of quadratic equation 27x2 – 10x + 1 = 0
D = b2 – 4ac

D = (- 28)2 – 4 × 21 × 10
D = 784 – 840

D = – 56

10. If z1 = 2 – i, z2 = 1 + i then find |z1 + z2 + 1/z1 – z2 + 1|

Hence if z1 = 2 – i and z2 = 1 + i then |z1 + z2 + 1/z1 – z2 + 1| = √2

11. If a + ib = (x + i)2/2x2 + 1 then prove that a2 + b2 = (x2 + 1)2/(2x2 + 1)2

12. Let z1 = 2 – i and z2 = – 2 + i then Find

$$\text { (i) } \operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right) \\ \text{ } \\ \text{Re} \left( \frac { (2 – i)(- 2 + i) } { \overline{(2 – i)} } \right) \\ \text{ } \\ \text{Re} \left( \frac { 2(- 2 + i) – i(- 2 + i) } { \overline{(2 – i)} } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 4 + 2i + 2i – i^{2} } { 2 + i } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 4 + 4i – i^{2} } { 2 + i } \right) \\ \text{ } \\ \text{Replacing } i^{2} = – 1 \\ \text{ } \\ \text{Re} \left( \frac { – 4 + 4i – (- 1) } { 2 + i } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 4 + 4i + 1 } { 2 + i } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 3 + 4i } { 2 + i } \right) \\ \text{ } \\ \text{Multiplying and Dividing by (2 – i)} \\ \text{ } \\ \text{Re} \left( \frac { (- 3 + 4i)(2 – i) } { (2 + i)(2 – i) } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 3(2 – i) + 4i(2 – i) } { (2 + i)(2 – i) } \right) \\ \text{ } \\ \text{Using formula } (a + b)(a – b) = a^{2} – b^{2} \\ \text{ } \\ \text{Re} \left( \frac { – 3(2 – i) + 4i(2 – i) } { 2^{2} – i^{2} } \right) \\ \text{ } \\ \text{Replacing } i^{2} = – 1 \\ \text{ } \\ \text{Re} \left( \frac { – 3(2 – i) + 4i(2 – i) } { 2^{2} – (- 1) } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 6 + 3i + 8i – 4i^{2} } { 4 + 1 } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 6 + 3i + 8i – 4i^{2} } { 5 } \right) \\ \text{ } \\ \text{Replacing } i^{2} = – 1 \\ \text{ } \\ \text{Re} \left( \frac { – 6 + 3i + 8i – 4(- 1) } { 5 } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 6 + 3i + 8i + 4 } { 5 } \right) \\ \text{ } \\ \text{Re} \left( \frac { – 2 + 11i } { 5 } \right) \\ \text{ } \\ \text{Re} \left( \frac{- 2}{5} + \frac{11i}{5} \right) \\ \text{ } \\ \implies \frac{- 2}{5}$$

13. Find the modulus and argument of the complex number 1 + 2i/1 – 3i

Let’s first simplify this complex number and then find out it’s modulus and argument

$$\frac{1 + 2i}{1 – 3i} \\ \text{ } \\ \text{Multiplying and Dividing by } (1 + 3i) \\ \text{ } \\ \frac{(1 + 2i)(1 + 3i)}{(1 – 3i)(1 + 3i)} \\ \text{ } \\ \frac{1(1 + 3i) + 2i(1 + 3i)}{(1 – 3i)(1 + 3i)} \\ \text{ } \\ \frac{1 + 3i + 2i + 6i^{2}}{(1 – 3i)(1 + 3i)} \\ \text{ } \\ \text{Using formula } \\ (a – b)(a + b) = a^{2} – b^{2} \\ \text{ } \\ \frac{1 + 5i + 6i^{2}}{1^{2} – (3i)^{2}} \\ \text{ } \\ \frac{1 + 5i + 6i^{2}}{1 – 3^{2}i^{2}} \\ \text{ } \\ \frac{1 + 5i + 6i^{2}}{1 – 9i^{2}} \\ \text{ } \\ \text{Replacing } i^{2} = – 1 \\ \text{ } \\ \frac{1 + 5i + 6(- 1)}{1 – 9(- 1)} \\ \text{ } \\ \frac{1 + 5i – 6}{1 + 9} \\ \text{ } \\ \frac{- 5 + 5i}{10} \\ \text{ } \\ \frac{- 5}{10} + \frac{5i}{10} \\ \text{ } \\ \frac{- 1}{2} + \frac{i}{2} \\ \text{ } \\$$

So we have simplified Complex Number given in the question now let’s move onto finding out Modulus and Argument of this simplified complex number.

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1/2 + i/2 with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1/2
rSinθ = 1/2

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (-1/2)2 + (1/2)2 = 1/4 + 1/4 = 2/4 = 1/2

⇒ r2Cos2θ + r2Sin2θ = 1/2

r2(Cos2θ + Sin2θ) = 1/2

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 1/2

r2 = 1/2

r = ± 1/√2
Thus r = 1/√2 or – 1/√2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 1/√2

Hence Modulus of Complex Number z = – 1/2 + i/2 is 1/√2

Now let’s find out Argument of the Complex Number z = – 1/2 + i/2

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1/2 + i/2 with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1/2
rSinθ = 1/2

Replacing r = 2 in these equations
√2Cosθ = – 1
√2Sinθ = 1

⇒ Cosθ = – 1/√2
⇒ Sinθ = 1/√2

As Cosine is negative and Sine is positive thus θ must lie in Second quadrant
⇒ θ = 3𝜋/4

By definition Argument of a Complex Number θ lie in range − 𝜋 < 𝜃 ≤ 𝜋 and θ = 3𝜋/4 does lie in this range hence it’s argument of the complex number z = – 1/2 + i/2

Thus Modulus and Argument of Complex Number (1 + 2i)/(1 – 3i) is √2 and 3𝜋/4

14. Find the real numbers x and y if (x – iy)(3 + 5i) is the conjugate of – 6 – 24i

Let’s first simplify (x – iy)(3 + 5i) and rewrite it in usual form of a complex number z = a + ib
(x – iy)(3 + 5i) = x(3 + 5i) – iy(3 + 5i)

= 3x + i(5x) – i(3y) – i2(5y)

Replacing i2 = – 1

= 3x + i(5x) – i(3y) – (- 1)(5y)

= 3x + i(5x) – i(3y) + 5y

= (3x + 5y) + i(5x – 3y)

(x – iy)(3 + 5i) = (3x + 5y) + i(5x – 3y)

As per question it’s given that – 6 – 24i is conjugate of (x – iy)(3 + 5i)

By definition of Conjugate of a complex number z = a + ib we know that it’s same complex number except sign for imaginary part is flipped
⇒ z = a + ib
⇒ Conjugate of z = a – ib

Using this logic but for values given in the question
⇒ z = – 6 – 24i
⇒ Conjugate of z = – 6 + 24i

As per question
⇒ Conjugate of z = (x – iy)(3 + 5i) = (3x + 5y) + i(5x – 3y)

⇒ Conjugate of z = – 6 + 24i = (3x + 5y) + i(5x – 3y)

– 6 + 24i = (3x + 5y) + i(5x – 3y)
Comparing
Real and Imaginary parts on both sides of this equation
We get

⇒ 3x + 5y = – 6
⇒ 5x – 3y = 24

15. Find the modulus of (1 + i)/(1 – i) – (1 – i)/(1 + i)

16. If (x + iy)3 = u + iv then show that u/x + v/y = 4(x2 – y2)

Let’s simplify (x + iy)3 = u + iv

Using formula (a + b)3 = a3 + b3 + 3ab(a + b)

(x + iy)3 = x3 + (iy)3 + 3 × x × iy(x + iy) = u + iv

⇒ x3 + i3y3 + i(3x2y) + i2(3xy2) = u + iv

Replacing
i2 = – 1
i3 = – i

⇒ x3 – iy3 + i(3x2y) – (3xy2) = u + iv

(x3 – 3xy2) + i(3x2y – y3) = u + iv

Comparing Real and Imaginary Parts on both sides of this equation
x3 – 3xy2 = u
3x2y – y3 = v

Simplifying
x(x2 – 3y2) = u
y(3x2 – y2) = v

⇒ x2 – 3y2 = u/x
⇒ 3x2 – y2 = v/y

x2 – 3y2 + 3x2 – y2 = u/x + v/y

4(x2 – y2) = u/x + v/y

Hence if (x + iy)3 = u + iv then u/x + v/y = 4(x2 – y2)

18. Find the number of non-zero integral solutions of the equation |1 – i|x = 2x

Taking log2 of both sides of equation |1 – i|x = 2x

⇒ log2(|1 – i|x) = log22x

x log2(|1 – i|) = x log22

Value of log22 = 1

x log2(|1 – i|) = x(1) = x

x log2(|1 – i|) = x

|1 – i| = Square Root of 12 + 12 = √2

Replacing this value in above equation
x log2(√2) = x

x log2(21/2) = x

x (1/2) log22 = x

log22 = 1

⇒ x/2 = x

x = 2x

0 = 2x – x

0 = x

Thus number of non-zero integral solutions of the equation |1 – i|x = 2x is zero