Complex Numbers And Quadratic Equations

We all know that i = √(-1) .

and = 1

= i (where n is any positive integer )

=

= -1

So,

(i) L.H.S =

=

=

Since it is of the form

Hence the value of .

(ii)

so its solution would be -1

(iii)

⇒i

So, it is of the form of so the solution would be i.

We all know that i = √(-1) .

and = 1

= i (where n is any positive integer )

=

= -1

So,

(i) L.H.S =

=

=

Since it is of the form

Hence the value of .

(ii)

so its solution would be -1

(iii)

⇒i

So, it is of the form of so the solution would be i.

Since i =

(i) L.H.S. =

Since it is of the form = 1 so the solution would be 1

(ii) L.H.S.=

Since it is of the form of = i so the solution would be simply i.

(iii) L.H.S =

Since it is of the form so the solution would be -1

Since i =

(i) L.H.S. =

Since it is of the form = 1 so the solution would be 1

(ii) L.H.S.=

Since it is of the form of = i so the solution would be simply i.

(iii) L.H.S =

Since it is of the form so the solution would be -1

(i)

Since it is of the form so the solution would be -1

(ii)

Since it is of the form of so the solution would be simply -i.

Since it is of the form . so the solution would be i

(i)

Since it is of the form so the solution would be -1

(ii)

Since it is of the form of so the solution would be simply -i.

Since it is of the form . so the solution would be i

(i) =+

(Since = i)

2i

Hence, = 2i

(ii)

+

(since = i

= -1)

0

Hence, = 0

(i) =+

(Since = i)

2i

Hence, = 2i

(ii)

+

(since = i

= -1)

0

Hence, = 0

L.H.S.= 1 + i² + i⁴ + i⁶

ToProve: 1 + i² + i⁴ + i⁶ = 0

1 + (-1) +1 +

Since, = 1

(where n is any positive integer )

1 + -1 + 1 + -1=0

⇒L.H.S = R.H.S

Hence proved.

L.H.S.= 1 + i² + i⁴ + i⁶

ToProve: 1 + i² + i⁴ + i⁶ = 0

1 + (-1) +1 +

Since, = 1

(where n is any positive integer )

1 + -1 + 1 + -1=0

⇒L.H.S = R.H.S

Hence proved.

Given: 6i⁵⁰ + 5i³³ – 2i¹⁵ + 6i⁴⁸

To prove: 6i⁵⁰ + 5i³³ – 2i¹⁵ + 6i⁴⁸ = 7i

6i^4×12+2 + 5i^4×8+1 – 2i^4×3+3 + 6i^4×12

6i² + 5i¹ – 2i³ + 6i⁰

-6+5i+2i+6

7i

Hence proved.

Given: 6i⁵⁰ + 5i³³ – 2i¹⁵ + 6i⁴⁸

To prove: 6i⁵⁰ + 5i³³ – 2i¹⁵ + 6i⁴⁸ = 7i

6i^4×12+2 + 5i^4×8+1 – 2i^4×3+3 + 6i^4×12

6i² + 5i¹ – 2i³ + 6i⁰

-6+5i+2i+6

7i

Hence proved.

Given:

To prove : = 0.

L.H.S.= i^-1 – i^-2 + i^-3 – i^-4

i^-4×1+3 – i^-4×1+2 + i^-4×1+3 – i^-4×1

since = 1

= i

= -1

So,

i¹ – i² + i³ – 1

i +1 – i -1

0

⇒L.H.S = R.H.S

Given:

To prove : = 0.

L.H.S.= i^-1 – i^-2 + i^-3 – i^-4

i^-4×1+3 – i^-4×1+2 + i^-4×1+3 – i^-4×1

since = 1

= i

= -1

So,

i¹ – i² + i³ – 1

i +1 – i -1

0

⇒L.H.S = R.H.S

L.H.S = (1 + i¹⁰ + i²⁰ + i³⁰)

= (1 + + )

since = 1

= i

= -1

= 1 +

= 1 + -1 +1 + -1

= 0, which is a real no.

Hence, (1 + i¹⁰ + i²⁰ + i³⁰) is a real number.

L.H.S = (1 + i¹⁰ + i²⁰ + i³⁰)

= (1 + + )

since = 1

= i

= -1

= 1 +

= 1 + -1 +1 + -1

= 0, which is a real no.

Hence, (1 + i¹⁰ + i²⁰ + i³⁰) is a real number.

L.H.S.=

=

since = 1

= i

=

== -1

=

=

Now, applying the formula = 2ab

= .

= -1 + 1 + 2i

= 2i

L.H.S = R.H.S

Hence proved.

L.H.S.=

=

since = 1

= i

=

== -1

=

=

Now, applying the formula = 2ab

= .

= -1 + 1 + 2i

= 2i

L.H.S = R.H.S

Hence proved.

L.H.S =

since = 1

= i

=

= -1

= .

= .

Applying the formula

We have,

+

i + 3 - 3i - 1

= 2(1-i)

L.H.S = R.H.S

Hence proved .

L.H.S =

since = 1

= i

=

= -1

= .

= .

Applying the formula

We have,

+

i + 3 - 3i - 1

= 2(1-i)

L.H.S = R.H.S

Hence proved .

L.H.S = (1 – i)ⁿ

=

=

Since, . = -1

= .

Applying

= .

=

=

L.H.S = R.H.S

Hce proved.

L.H.S = (1 – i)ⁿ

=

=

Since, . = -1

= .

Applying

= .

=

=

L.H.S = R.H.S

Hce proved.

L.H.S =

Since we know that i = .

So,

= i + 3 i

=4i + 15i + 6i - 25i

= 0

L.H.S = R.H.S

Hence proved.

L.H.S =

Since we know that i = .

So,

= i + 3 i

=4i + 15i + 6i - 25i

= 0

L.H.S = R.H.S

Hence proved.

L.H.S = (1 + i² + i⁴ + i⁶ + i⁸ + …. + i²⁰)

=

= 1 + -1 +1 + -1 + ……….. + 1

As there are 11 times 1 and 6 times it is with positive sign as =1 as this is the extra term and there are 5 times 1 with negative sign

So, these 5 cancel out the positive one leaving one positive value i.e. 1

=

L.H.S = R.H.S

Hence proved.

L.H.S = (1 + i² + i⁴ + i⁶ + i⁸ + …. + i²⁰)

=

= 1 + -1 +1 + -1 + ……….. + 1

As there are 11 times 1 and 6 times it is with positive sign as =1 as this is the extra term and there are 5 times 1 with negative sign

So, these 5 cancel out the positive one leaving one positive value i.e. 1

=

L.H.S = R.H.S

Hence proved.

L.H.S = i⁵³ + i⁷² + i⁹³ + i¹⁰²

=

Since = 1

= i (where n is any positive integer )

=

= -1

= i + 1 + i +

= i+1+i-1

=2i

L.H.S = R.H.S

Hence proved.

L.H.S = i⁵³ + i⁷² + i⁹³ + i¹⁰²

=

Since = 1

= i (where n is any positive integer )

=

= -1

= i + 1 + i +

= i+1+i-1

=2i

L.H.S = R.H.S

Hence proved.

L.H.S =

=

since = 1

= i

= -1

= -1

= i – 1 – i + 1 + i - 1……+i-1

As, all terms will get cancel out consecutively except the first two terms. so that will get remained will be the answer.

= i - 1

L.H.S = R.H.S

Hence proved.

L.H.S =

=

since = 1

= i

= -1

= -1

= i – 1 – i + 1 + i - 1……+i-1

As, all terms will get cancel out consecutively except the first two terms. so that will get remained will be the answer.

= i - 1

L.H.S = R.H.S

Hence proved.

Given: 2(3 + 4i) + i(5 – 6i)

Firstly, we open the brackets

2 × 3 + 2 × 4i + i × 5 – i × 6i

= 6 + 8i + 5i – 6i²

= 6 + 13i – 6(-1) [∵, i² = -1]

= 6 + 13i + 6

= 12 + 13i

Given:

We re – write the above equation

[∵ i² = -1]

= (3 + 4i) – (4 – 3i)

Now, we open the brackets, we get

3 + 4i – 4 + 3i

= -1 + 7i

Given: (–5 + 6i) – (–2 + i)

Firstly, we open the brackets

-5 + 6i + 2 – i

= -3 + 5i

Given: (8 – 4i) – (- 3 + 5i)

Firstly, we open the brackets

8 – 4i + 3 – 5i

= 11 – 9i

Given: (1 – i)² (1 + i) – (3 – 4i)²

= (1 + i² – 2i)(1 + i) – (9 + 16i² – 24i)

[∵(a – b)² = a² + b² – 2ab]

= (1 – 1 – 2i)(1 + i) – (9 – 16 – 24i) [∵ i² = -1]

= (-2i)(1 + i) – (- 7 – 24i)

Now, we open the brackets

-2i × 1 – 2i × i + 7 + 24i

= -2i – 2i² + 7 + 24i

= -2(-1) + 7 + 22i [∵, i² = -1]

= 2 + 7 + 22i

= 9 + 22i

Given:

We re – write the above equation

[∵, i² = -1]

Now, we know that,

(a + b)(a – b) = (a² – b²)

Here, a = 5 and b = i√3

= (5)² – (i√3)²

= 25 – (3i²)

= 25 – [3 × (-1)]

= 25 + 3

= 28 + 0

= 28 + 0i

Given: (3 + 4i) (2 – 3i)

Firstly, we open the brackets

3 × 2 + 3 × (-3i) + 4i × 2 – 4i × 3i

= 6 – 9i + 8i – 12i²

= 6 – i – 12(-1) [∵, i² = -1]

= 6 – i + 12

= 18 – i

Given:

We re – write the above equation

[∵, i² = -1]

Now, open the brackets,

= -2 × (-3) + (-2) × 2i√3 + i√3 × (-3) + i√3 × 2i√3

= 6 – 4i√3 – 3i√3 + 6i²

= 6 – 7i√3 + [6 × (-1)] [∵, i² = -1]

= 6 – 7i√3 – 6

= 0 – 7i√3

Given: (2 - √-3)²

We know that,

(a – b)² = a² + b² – 2ab …(i)

So, on replacing a by 2 and b by √-3 in eq. (i), we get

(2)² + (√-3)² – 2(2)(√-3)

= 4 + (-3) – 4√-3

= 4 – 3 – 4√-3

= 1 – 4√3i² [∵ i² = -1]

= 1 – 4i√3

Given: (5 – 2i)²

We know that,

(a – b)² = a² + b² – 2ab …(i)

So, on replacing a by 5 and b by 2i in eq. (i), we get

(5)² + (2i)² – 2(5)(2i)

= 25 + 4i² – 20i

= 25 – 4 – 20i [∵ i² = -1]

= 21 – 20i

Given: (-3 + 5i)³

We know that,

(-a + b)³ = - a³ + 3a²b – 3ab² + b³ …(i)

So, on replacing a by 3 and b by 5i in eq. (i), we get

-(3)³ + 3(3)²(5i) – 3(3)(5i)² + (5i)³

= -27 + 3(9)(5i) – 3(3)(25i²) + 125i³

= -27 + 135i – 225i² + 125i³

= -27 + 135i – 225 × (-1) + 125i × i²

= -27 + 135i + 225 – 125i [∵ i² = -1]

= 198 + 10i

Given:

We know that,

(– a – b)³ = - a³ – 3a²b – 3ab² – b³ …(i)

So, on replacing a by 2 and b by 1/3i in eq. (i), we get

[∵ i² = -1]

Given: (4 – 3i)^-1

We can re- write the above equation as

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given: (-2 + √-3)^-1

We can re- write the above equation as

[∵ i² = -1]

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given: (2 + i)^-2

Above equation can be re – written as

Now, rationalizing

[∵ (a – b)² = a² + b² – 2ab]

[∵i² = -1]

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

Given: (1 + 2i)^-3

Above equation can be re – written as

Now, rationalizing

We know that,

(a – b)³ = a³ – 3a²b + 3ab² – b³

(a + b)³ = a³ + 3a²b + 3ab² + b³

[∵i² = -1]

[∵i² = -1]

Given: (1 + i)³ – (1 – i)³ …(i)

We know that,

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a – b)³ = a³ – 3a²b + 3ab² – b³

By applying the formulas in eq. (i), we get

(1)³ + 3(1)²(i) + 3(1)(i)² + (i)³ – [(1)³ – 3(1)²(i) + 3(1)(i)² – (i)³]

= 1 + 3i + 3i² + i³ – [1 – 3i + 3i² – i³]

= 1 + 3i + 3i² + i³ – 1 + 3i – 3i² + i³

= 6i + 2i³

= 6i + 2i(i²)

= 6i + 2i(-1) [∵ i² = -1]

= 6i – 2i

= 4i

= 0 + 4i

Given:

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given:

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given:

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

= 1 + 2i√2

Given:

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given:

Solving the denominator, we get

Now, we rationalize the above by multiplying and divide by the conjugate of 6 + 2i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given:

Firstly, we solve the given equation

Now, we rationalize the above by multiplying and divide by the conjugate of 4 + 3i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

[∵ i² = -1]

Given:

Now, we rationalize the above equation by multiply and divide by the conjugate of (2 – i)

[∵(a + b)(a – b) = (a² – b²)]

[∵i² = -1]

Given:

The above equation can be re-written as

[∵(a – b)³ = a³ – b³ – 3a²b + 3ab²]

[∵i² = -1]

= -2 + 0i

Given:

We solve the above equation by using the formula

(a + b)³ = a³ + b³ + 3a²b + 3ab²

[∵ i² = -1]

Now, we rationalize the above by multiplying and divide by the conjugate of 3 + i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

[∵ i² = -1]

Given:

[Taking the LCM]

[putting i² = -1]

Now, rationalizing by multiply and divide by the conjugate of (13 – 13i)

[∵ (a – b)(a + b) = (a² – b²)]

[∵ i² = -1]

[taking 13 common]

(ii) Given:

[Taking the LCM]

Now, rationalizing by multiply and divide by the conjugate of (15 + 8i)

[∵ (a – b)(a + b) = (a² – b²)]

[∵ i² = -1]

Given:

Taking the L.C.M, we get

[∵(a + b)(a – b) = (a² – b²)]

Putting i² = -1

= 0 + 0i

Hence, the given equation is purely real as there is no imaginary part.

(ii) Given:

Taking the L.C.M, we get

…(i)

[∵(a + b)(a – b) = (a² – b²)]

Now, we know that,

(a + b)² + (a – b)² = 2(a² + b²)

So, by applying the formula in eq. (i), we get

Putting i² = -1

Hence, the given equation is purely real as there is no imaginary part.

Since is purely real

Firstly, we need to solve the given equation and then take the imaginary part as 0

We rationalize the above by multiply and divide by the conjugate of (1 -2i cos θ)

We know that,

(a – b)(a + b) = (a² – b²)

[∵ i² = -1]

Since is purely real [given]

Hence, imaginary part is equal to 0

i.e.

⇒ 3 cos θ = 0 × (1 + 4 cos²θ)

⇒ 3 cos θ = 0

⇒ cos θ = 0

⇒ cos θ = cos 0

Since, cos θ = cos y

Then where n Є Z

Putting y = 0

where n Є Z

Hence, for is purely real.

Let z = x + iy

Consider, |z + i| = |z – i|

⇒ |x + iy + i| = |x + iy – i|

⇒ |x + i(y +1)| = |x + i(y – 1)|

Squaring both the sides, we get

⇒ x² + y² + 1 + 2y = x² + y² + 1 – 2y

⇒ x² + y² + 1 + 2y – x² – y² – 1 + 2y = 0

⇒ 2y + 2y = 0

⇒ 4y = 0

⇒ y = 0

Putting the value of y in eq. (i), we get

z = x + i(0)

⇒ z = x

Hence, z is purely real.

Let z₁ = 3 – 4i and z₂ = 4 – 3i

Here, z₁≠ z₂

Now, calculating the modulus, we get,

|z₁|

|z₂|

Given: z = (-5 – 2i)

Here, we have to find the conjugate of (-5 – 2i)

So, the conjugate of (- 5 – 2i) is (-5 + 2i)

Given:

First, we calculate and then find its conjugate

Now, rationalizing

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Hence,

So, a conjugate of is

Given:

Firstly, we calculate and then find its conjugate

[∵ (a + b)² = a² + b² + 2ab]

[∵ i² = -1]

Now, we rationalize the above by multiplying and divide by the conjugate of 3 – i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

[∵ i² = -1]

Hence,

So, the conjugate of is

Given:

Firstly, we calculate and then find its conjugate

[∵ i² = -1]

Now, we rationalize the above by multiplying and divide by the conjugate of 3 + i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

[∵ i² = -1]

Hence,

So, the conjugate of is

Given: z = √-3

The above can be re – written as

[∵ i² = -1]

z = 0 + i√3

So, the conjugate of z = 0 + i√3 is

or

Given: z = √2

The above can be re – written as

z = √2 + 0i

Here, the imaginary part is zero

So, the conjugate of z = √2 + 0i is

or

Given: z = -√-1

The above can be re – written as

[∵ i² = -1]

z = 0 – i

So, the conjugate of z = (0 – i) is

or

Given: z = (2 – 5i)²

First we calculate (2 – 5i)² and then we find the conjugate

(2 – 5i)² = (2)² + (5i)² – 2(2)(5i)

= 4 + 25i² – 20i

= 4 + 25(-1) – 20i [∵ i² = -1]

= 4 – 25 – 20i

= -21 – 20i

Now, we have to find the conjugate of (-21 – 20i)

So, the conjugate of (- 21 – 20i) is (-21 + 20i)

Given: z = (3 + √-5)

The above can be re – written as

z = 3 + i√5 [∵ i² = -1]

Now, we have to find the modulus of (3 + i√5)

So,

Hence, the modulus of (3 + √-5) is √14

Given: z = (-3 – 4i )

Now, we have to find the modulus of (-3 – 4i)

So,

Hence, the modulus of (-3 – 4i) is 5

Given: z = (7 + 24i)

Now, we have to find the modulus of (7 + 24i)

So,

Hence, the modulus of (7 + 24i) is 25

Given: z = 3i

The above equation can be re – written as

z = 0 + 3i

Now, we have to find the modulus of (0 + 3i)

So,

Hence, the modulus of (3i) is 3

Given:

Firstly, we calculate and then find its modulus

[∵ (a + b)² = a² + b² + 2ab]

[∵ i² = -1]

Now, we rationalize the above by multiplying and divide by the conjugate of 4 + 3i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

[∵ i² = -1]

Now, we have to find the modulus of

So,

Hence, the modulus of is

Given:

Firstly, we calculate and then find its modulus

[∵ i² = -1]

Now, we rationalize the above by multiplying and divide by the conjugate of 1 + i

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

[∵ i² = -1]

= 2 – i

Now, we have to find the modulus of (2 – i)

So,

Given: z = 5

The above equation can be re – written as

z = 5 + 0i

Now, we have to find the modulus of (5 + 0i)

So,

Given: z = (1 + 2i)(i – 1)

Firstly, we calculate the (1 + 2i)(i – 1) and then find the modulus

So, we open the brackets,

1(i – 1) + 2i(i – 1)

= 1(i) + (1)(-1) + 2i(i) + 2i(-1)

= i – 1 + 2i² – 2i

= - i – 1 + 2(-1) [∵ i² = - 1]

= - i – 1 – 2

= - i – 3

Now, we have to find the modulus of (-3 - i)

So,

Given: (1 - i√3)

To find: Multiplicative inverse

We know that,

Multiplicative Inverse of z = z^-1

Putting z = 1 - i√3

Now, rationalizing by multiply and divide by the conjugate of (1 - i√3)

Using (a – b)(a + b) = (a² – b²)

[∵ i² = -1]

Given: 2 + 5i

To find: Multiplicative inverse

We know that,

Multiplicative Inverse of z = z^-1

Putting z = 2 + 5i

Now, rationalizing by multiply and divide by the conjugate of (2+5i)

Using (a – b)(a + b) = (a² – b²)

[∵ i² = -1]

Hence, Multiplicative Inverse of (2+5i)is

Given:

To find: Multiplicative inverse

We know that,

Multiplicative Inverse of z = z^-1

Now, rationalizing by multiply and divide by the conjugate of (2+3i)

Using (a – b)(a + b) = (a² – b²)

[∵ i² = -1]

Given:

To find: Multiplicative inverse

We know that,

Multiplicative Inverse of z = z^-1

We solve the above equation

[∵ i² = -1]

Now, we rationalize the above by multiplying and divide by the conjugate of (-1 + 3i)

…(i)

Now, we know that,

(a + b)(a – b) = (a² – b²)

So, eq. (i) become

[∵ i² = -1]

Given:

Consider the given equation,

Now, we rationalize

[Here, we multiply and divide by the conjugate of 1 + i]

Using (a + b)(a – b) = (a² – b²)

[∵ i² = -1]

= (-i)¹⁰⁰

= [(-i)⁴]²⁵

= (i⁴)²⁵

= (1)²⁵

[∵ i⁴ = i² × i² = -1 × -1 = 1]

(a + ib) = 1 + 0i

On comparing both the sides, we get

a = 1 and b = 0

hence, the value of a is 1 and b is 0

Consider,

Now, rationalizing

In denominator, we use the identity

(a – b)(a + b) = a² – b²

= (i)⁹³ – (-i)³

= (i)⁹²⁺¹ – [-(i)³]

= [(i)⁹²(i)] – [-(i² × i)]

= [(i⁴)²³(i)] – [- (-i)]

= [(1)²³(i)] – i

= i - i

x + iy = 0

∴ x = 0 and y = 0

Consider the given equation,

Now, rationalizing

[(a – b)(a + b) = a² – b²]

[i² = -1]

On comparing both the sides, we get

Now, we have to prove that x² + y² = 1

Taking LHS,

x² + y²

Putting the value of x and y, we get

= 1

= RHS

Hence Proved

Consider the given equation,

Now, rationalizing

[(a – b)(a + b) = a² – b²]

[i² = -1]

On comparing both the sides, we get

Now, we have to prove that a² + b² = 1

Taking LHS,

a² + b²

Putting the value of a and b, we get

= 1

= RHS

Now, we have to prove

Taking LHS,

Putting the value of a and b, we get

Hence Proved

To show:

Taking LHS,

[rationalize]

[∵ i² = -1]

= (1 – i)ⁿ(1 + i)ⁿ

= [(1 – i)(1 + i)]ⁿ

= [(1)² – (i)²]ⁿ [(a + b)(a – b) = a² – b²]

= (1 – i²)ⁿ

= [1 – (-1)]ⁿ[∵ i² = -1]

= (2)ⁿ

= 2ⁿ

= RHS

Hence Proved

Given: (1 + i)²ⁿ = (1 – i)²ⁿ

Consider the given equation,

(1 + i)²ⁿ = (1 – i)²ⁿ

Now, rationalizing by multiply and divide by the conjugate of (1 – i)

[(a + b)² = a² + b² + 2ab & (a – b)(a + b) = (a² – b²)]

[i² = -1]

⇒ (i)²ⁿ = 1

Now, i²ⁿ = 1 is possible if n = 2 because (i)²⁽²⁾ = i⁴ = (-1)⁴ = 1

So, the smallest positive integer n = 2

To Prove:

(x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i) = (x⁴ + 4)

Taking LHS

(x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i)

= [(x + 1) + i][(x + 1) – i][(x – 1) + i][(x – 1) – i]

Using (a – b)(a + b) = a² – b²

= [(x + 1)² – (i)²] [(x – 1)² – (i)²]

= [x² + 1 + 2x – i²](x² + 1 – 2x – i²]

= [x² + 1 + 2x – (-1)](x² + 1 – 2x – (-1)] [∵ i² = -1]

= [x² + 2 + 2x][x² + 2 – 2x]

Again, using (a – b)(a + b) = a² – b²

Now, a = x² + 2 and b = 2x

= [(x² + 2)² – (2x)²]

= [x⁴ + 4 + 2(x²)(2) – 4x²] [∵(a + b)² = a² + b² + 2ab]

= [x⁴ + 4 + 4x² – 4x²]

= x⁴ + 4

= RHS

∴ LHS = RHS

Hence Proved

Given: a = cosθ + isinθ

To prove:

Taking LHS,

Putting the value of a, we get

We know that,

1 + cos2θ = 2cos²θ

or

and

Using the above two formulas

Using,

Rationalizing by multiply and divide by the conjugate of

Putting i² = -1, we get

We know that,

cos² θ + sin² θ = 1

= RHS

Hence Proved

Given: z₁ = (2 – i) and z₂ = (1 + i)

To find:

Consider,

Putting the value of z₁ and z₂, we get

Now, rationalizing by multiply and divide by the conjugate of 1 – i

[(a – b)(a + b) = a² – b²]

[Putting i² = -1]

= |2(1 + i)|

= |2 + 2i|

Now, we have to find the modulus of (2 + 2i)

So,

(1 – i) x + (1 + i) y = 1 – 3i

⇒ x – ix + y + iy = 1 – 3i

⇒ (x + y) – i(x – y) = 1 – 3i

Comparing the real parts, we get

x + y = 1 …(i)

Comparing the imaginary parts, we get

x – y = -3 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Adding eq. (i) and (ii), we get

x + y + x – y = 1 + (-3)

⇒ 2x = 1 – 3

⇒ 2x = -2

⇒ x = -1

Putting the value of x = -1 in eq. (i), we get

(-1) + y = 1

⇒ y = 1 + 1

⇒ y = 2

x(3 – 2i) + iy(3 – 2i) = 12 + 5i

⇒ 3x – 2ix + 3iy – 2i²y = 12 + 5i

⇒ 3x + i(-2x + 3y) – 2(-1)y = 12 + 5i [∵ i² = -1]

⇒ 3x + i(-2x + 3y) + 2y = 12 + 5i

⇒ (3x + 2y) + i(-2x + 3y) = 12 + 5i

Comparing the real parts, we get

3x + 2y = 12 …(i)

Comparing the imaginary parts, we get

–2x + 3y = 5 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Multiply eq. (i) by 2 and eq. (ii) by 3, we get

6x + 4y = 24 …(iii)

–6x + 9y = 15 …(iv)

Adding eq. (iii) and (iv), we get

6x + 4y – 6x + 9y = 24 + 15

⇒ 13y = 39

⇒ y = 3

Putting the value of y = 3 in eq. (i), we get

3x + 2(3) = 12

⇒ 3x + 6 = 12

⇒ 3x = 12 – 6

⇒ 3x = 6

⇒ x = 2

Hence, the value of x = 2 and y = 3

Given: x + 4yi = ix + y + 3

or x + 4yi = ix + (y + 3)

Comparing the real parts, we get

x = y + 3

or x – y = 3 …(i)

Comparing the imaginary parts, we get

4y = x …(ii)

Putting the value of x = 4y in eq. (i), we get

4y – y = 3

⇒ 3y = 3

⇒ y = 1

Putting the value of y = 1 in eq. (ii), we get

x = 4(1) = 4

Hence, the value of x = 4 and y = 1

Given: (1 + i) y² + (6 + i) = (2 + i)x

Consider, (1 + i) y² + (6 + i) = (2 + i)x

⇒ y² + iy² + 6 + i = 2x + ix

⇒ (y² + 6) + i(y² + 1) = 2x + ix

Comparing the real parts, we get

y² + 6 = 2x

⇒ 2x – y² – 6 = 0 …(i)

Comparing the imaginary parts, we get

y² + 1 = x

⇒ x – y² – 1 = 0 …(ii)

Subtracting the eq. (ii) from (i), we get

2x – y² – 6 – (x – y² – 1) = 0

⇒ 2x – y² – 6 – x + y² + 1 = 0

⇒ x – 5 = 0

⇒ x = 5

Putting the value of x = 5 in eq. (i), we get

2(5) – y² – 6 = 0

⇒ 10 – y² – 6 = 0

⇒ -y² + 4 = 0

⇒ - y² = -4

⇒ y² = 4

⇒ y = √4

⇒ y = ± 2

Hence, the value of x = 5 and y = ± 2

Given:

⇒ x + 3i = (1 – i)(2 + iy)

⇒ x + 3i = 1(2 + iy) – i(2 + iy)

⇒ x + 3i = 2 + iy – 2i – i²y

⇒ x + 3i = 2 + i(y – 2) – (-1)y [i² = -1]

⇒ x + 3i = 2 + i(y – 2) + y

⇒ x + 3i = (2 + y) + i(y – 2)

Comparing the real parts, we get

x = 2 + y

⇒ x – y = 2 …(i)

Comparing the imaginary parts, we get

3 = y – 2

⇒ y = 3 + 2

⇒ y = 5

Putting the value of y = 5 in eq. (i), we get

x – 5 = 2

⇒ x = 2 + 5

⇒ x = 7

Hence, the value of x = 7 and y = 5

Consider,

Taking LCM

Putting i² = -1

⇒ 4x + 2xi – 3i – 3 + 9y – 7iy = 10i

⇒ (4x – 3 + 9y) + i(2x – 3 – 7y) = 10i

Comparing the real parts, we get

4x – 3 + 9y = 0

⇒ 4x + 9y = 3 …(i)

Comparing the imaginary parts, we get

2x – 3 – 7y = 10

⇒ 2x – 7y = 10 + 3

⇒ 2x – 7y = 13 …(ii)

Multiply the eq. (ii) by 2, we get

4x – 14y = 26 …(iii)

Subtracting eq. (i) from (iii), we get

4x – 14y – (4x + 9y) = 26 – 3

⇒ 4x – 14y – 4x – 9y = 23

⇒ -23y = 23

⇒ y = -1

Putting the value of y = -1 in eq. (i), we get

4x + 9(-1) = 3

⇒ 4x – 9 = 3

⇒ 4x = 12

⇒ x = 3

Hence, the value of x = 3 and y = -1

Given: (x – iy) (3 + 5i) is the conjugate of (-6 – 24i)

We know that,

Conjugate of – 6 – 24i = - 6 + 24i

∴ According to the given condition,

(x – iy) (3 + 5i) = -6 + 24i

⇒ x(3 + 5i) – iy(3 + 5i) = -6 + 24i

⇒ 3x + 5ix – 3iy – 5i²y = -6 + 24i

⇒ 3x + i(5x – 3y) – 5(-1)y = -6 + 24i [∵ i² = -1]

⇒ 3x + i(5x – 3y) + 5y = -6 + 24i

⇒ (3x + 5y) + i(5x – 3y) = -6 + 24i

Comparing the real parts, we get

3x + 5y = -6 …(i)

Comparing the imaginary parts, we get

5x – 3y = 24 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Multiply eq. (i) by 5 and eq. (ii) by 3, we get

15x + 25y = -30 …(iii)

15x – 9y = 72 …(iv)

Subtracting eq. (iii) from (iv), we get

15x – 9y – 15x – 25y = 72 – (-30)

⇒ -34y = 72 + 30

⇒ -34y = 102

⇒ y = -3

Putting the value of y = -3 in eq. (i), we get

3x + 5(-3) = -6

⇒ 3x – 15 = -6

⇒ 3x = -6 + 15

⇒ 3x = 9

⇒ x = 3

Hence, the value of x = 3 and y = -3

Let z₁ = -3 + iyx²

So, the conjugate of z₁ is

and z₂ = x² + y + 4i

So, the conjugate of z₂ is

Given that:

Firstly, consider

- 3 – iyx² = x² + y + 4i

⇒ x² + y + 4i + iyx² = -3

⇒ x² + y + i(4 + yx²) = -3 + 0i

Comparing the real parts, we get

x² + y = -3 …(i)

Comparing the imaginary parts, we get

4 + yx² = 0

⇒ x²y = -4 …(ii)

Now, consider

-3 + iyx² = x² + y – 4i

⇒ x² + y – 4i – iyx² = - 3

⇒ x² + y + i(-4i – yx²) = - 3 + 0i

Comparing the real parts, we get

x² + y = -3

Comparing the imaginary parts, we get

-4 – yx² = 0

⇒ x²y = -4

Now, we will solve the equations to find the value of x and y

From eq. (i), we get

x² = - 3 – y

Putting the value of x² in eq. (ii), we get

(-3 – y)(y) = -4

⇒ -3y – y² = -4

⇒ y² + 3y = 4

⇒ y² + 3y – 4 = 0

⇒ y² + 4y – y – 4 = 0

⇒ y(y + 4) – 1(y + 4) = 0

⇒ (y – 1)(y + 4) = 0

⇒ y – 1 = 0 or y + 4 = 0

⇒ y = 1 or y = -4

When y = 1, then

x² = - 3 – 1

= - 4 [It is not possible]

When y = -4, then

x² = - 3 –(-4)

= - 3 + 4

⇒ x² = 1

⇒ x = √1

⇒ x = ± 1

Hence, the values of x = ±1 and y = -4

Given: z = 2 – 3i

To Prove: z² – 4z + 13 = 0

Taking LHS, z² – 4z + 13

Putting the value of z = 2 – 3i, we get

(2 – 3i)² – 4(2 – 3i) + 13

= 4 + 9i² – 12i – 8 + 12i + 13

= 9(-1) + 9

= - 9 + 9

= 0

= RHS

Hence, z² – 4z + 13 = 0 …(i)

Now, we have to deduce 4z³ – 3z² + 169

Now, we will expand 4z³ – 3z² + 169 in this way so that we can use the above equation i.e. z² – 4z + 13

= 4z³ – 16z² + 13z² +52z – 52z + 169

Re – arrange the terms,

= 4z³ – 16z² + 52z + 13z² – 52z + 169

= 4z(z² – 4z + 13) + 13 (z² – 4z + 13)

= 4z(0) + 13(0) [from eq. (i)]

= 0

= RHS

Hence Proved

Let z = x + iy

Then,

Now, Given: (1 + i)z = (1 – i)

Therefore,

(1 + i)(x + iy) = (1 – i)(x – iy)

x + iy + xi + i²y = x – iy – xi + i²y

We know that i² = -1, therefore,

x + iy + ix – y = x – iy – ix – y

2xi + 2yi = 0

x = -y

Now, as x = -y

z = -

Hence, Proved.

Given: is purely imaginary number

Let z = x + iy

So,

Now, rationalizing the above by multiply and divide by the conjugate of [(x + 1) + iy]

Using (a – b)(a + b) = (a² – b²)

Putting i² = -1

Since, the number is purely imaginary it means real part is 0

⇒ x² + y² – 1 = 0

⇒ x² + y² = 1

∴ |z| = 1

Given: Re(z²) = 0 and |z| = 2

Let z = x + iy

[given]

Squaring both the sides, we get

x² + y² = 4 …(i)

Since, z = x + iy

⇒ z² = (x + iy)²

⇒ z² = x² + i²y² + 2ixy

⇒ z² = x² + (-1)y² + 2ixy

⇒ z² = x² – y² + 2ixy

It is given that Re(z²) = 0

⇒ x² – y² = 0 …(ii)

Adding eq. (i) and (ii), we get

x² + y² + x² – y² = 4 + 0

⇒ 2x² = 4

⇒ x² = 2

⇒ x = ±√2

Putting the value of x² = 2 in eq. (i), we get

2 + y² = 4

⇒ y² = 2

⇒ y = ±√2

Hence, z = √2 ± i√2, -√2 ± i√2

Given: |z| = z + 1 + 2i

Consider,

|z| = (z + 1) + 2i

Squaring both the sides, we get

|z|² = [(z + 1) + (2i)]²

⇒ |z|² = |z + 1|² + 4i² + 2(2i)(z + 1)

⇒ |z|² = |z|² + 1 + 2z + 4(-1) + 4i(z + 1)

⇒ 0 = 1 + 2z – 4 + 4i(z + 1)

⇒ 2z – 3 + 4i(z + 1) = 0

Let z = x + iy

⇒ 2(x + iy) – 3 + 4i(x + iy + 1) = 0

⇒ 2x + 2iy – 3 + 4ix + 4i²y + 4i = 0

⇒ 2x + 2iy – 3 + 4ix + 4(-1)y + 4i = 0

⇒ 2x – 3 – 4y + i(4x + 2y + 4) = 0

Comparing the real part, we get

2x – 3 – 4y = 0

⇒ 2x – 4y = 3 …(i)

Comparing the imaginary part, we get

4x + 2y + 4 = 0

⇒ 2x + y + 2 = 0

⇒ 2x + y = -2 …(ii)

Subtracting eq. (ii) from (i), we get

2x – 4y – (2x + y) = 3 – (-2)

⇒ 2x – 4y – 2x – y = 3 + 2

⇒ -5y = 5

⇒ y = -1

Putting the value of y = -1 in eq. (i), we get

2x – 4(-1) = 3

⇒ 2x + 4 = 3

⇒ 2x = 3 – 4

⇒ 2x = - 1

Hence, the value of z = x + iy

(i) Let

⇒

(ii) Let z = (2 + 3i)² = (2 + 3i)(2 + 3i)

= 4 + 6i + 6i + 9i²

= 4 + 12i + 9i²

= 4 + 12i - 9

= - 5 + 12i

(iii) Let

=

(iv) Let

(v) Let

(vi) Let

(i) Let

∴ the multiplicative inverse of

(ii) Let

z = - 1 + i

⇒ |z|² = ( - 1)² + (1)² = 1 + 1 = 2

∴ the multiplicative inverse of

(iii) Let

Z = - 1 + i√3

⇒ |z|² = ( - 1)² + (√3)² = 1 + 3 = 4

∴ the multiplicative inverse of

Given that, (x + iy)³ = (u + iv)

⇒ x³ + (iy)³ + 3x²iy + 3xi²y² = u + iv

⇒ x³ - iy³ + 3x²iy - 3xy² = u + iv

⇒ x³ - 3xy² + i(3x²y - y³) = u + iv

On equating real and imaginary parts, we get

U = x³ - 3xy² and v = 3x²y - y³

Now ,

= x² - 3y² + 3x² - y²

= 4x² - 4y²

= 4(x² - y²)

Hence,

Given that, (x + iy)^1/3 = (a + ib)

⇒ (x + iy) = (a + ib)³

⇒ (a + ib)³ = x + iy

⇒ a³ + (ib)³ + 3a²ib + 3ai²b² = x + iy

⇒ a³ - ib³ + 3a²ib - 3ab² = x + iy

⇒ a³ - 3ab² + i(3a²b - b³) = x + iy

On equating real and imaginary parts, we get

x = a³ - 3ab² and y = 3a²b - b³

Now ,

= a² - 3b² + 3a² - b²

= 4a² - 4b²

= 4(a² - b²)

Hence,

We have, (1 – 2i)^–3

We have, (x⁴ + 2xi) – (3x² + iy) = (3 – 5i) + (1 + 2iy).

⇒ x⁴ + 2xi - 3x² + iy = 3 – 5i + 1 + 2iy

⇒ (x⁴ - 3x²) + i(2x - y) = 4 + i(2y - 5)

On equating real and imaginary parts, we get

x⁴ - 3x^{2 =} 4 and 2x - y = 2y - 5

⇒ x⁴ - 3x² - 4 = 0 eq(i) and 2x - y - 2y + 5 = 0 eq(ii)

Now from eq (i), x⁴ - 3x² - 4 = 0

⇒ x⁴ - 4x^{2 +} x² - 4 = 0

⇒ x² (x² - 4) ⁺ 1(x² - 4) = 0

⇒ (x² - 4)(x² + 1) = 0

⇒ x² - 4 = 0 and x² + 1 = 0

⇒ x = ±2 and x = √ - 1

Real value of x = ±2

Putting x = 2 in eq (ii), we get

2x - 3y + 5 = 0

⇒ 2×2 - 3y + 5 = 0

⇒ 4 - 3y + 5 = 0 = 9 - 3y = 0

⇒ y = 3

Putting x = - 2 in eq (ii), we get

2x - 3y + 5 = 0

⇒ 2× - 2 - 3y + 5 = 0

⇒ - 4 - 3y + 5 = 0 = 1 - 3y = 0

⇒

Let z= a + ib

⇒ |z| = √(a² + b²)

Now , z² + |z|² = 0

⇒ (a + ib)² + a² + b² = 0

⇒ a² + 2abi + i²b² + a² + b² = 0

⇒ a² + 2abi - b² + a² + b² = 0

⇒ 2a² + 2abi = 0

⇒ 2a(a + ib) = 0

Either a = 0 or z = 0

Since z≠ 0

a = 0 ⇒ z is purely imaginary.

Let z= a + ib

Now,

Given that is purely imaginary ⇒ real part = 0

⇒ a² + b² - 1 = 0

⇒ a² + b² = 1

⇒ |z| = 1

Hence proved.

Let z₁ = a + ib such that | z₁| = √(a² + b²) = 1

Now,

Thus, the real part of z₂ is 0 and z₂ is purely imaginary.

Let z = a + ib

Hence Proved.

(ii) Let z = a + ib

Hence, Proved.

(iii) Let z = a + ib

Hence Proved.

(iv) Let z = a + ib

Hence , is real.

(v) Case 1. Let z = a + 0i

Case 2. Let z = 0 + bi

Case 2. Let z = a + ib

Thus, is 0 or imaginary.

We have, z₁ = (1 + i) and z₂ = (–2 + 4i)

Now,

= - 4 - 2i

Hence,

We have,

_{Applying componendo and dividendo , we get}

Let Z = 4 = r(cosθ + isinθ)

Now, separating real and complex part, we get

4 = rcosθ……….eq.1

0 = rsinθ…………eq.2

Squaring and adding eq.1 and eq.2, we get

16 = r²

Since r is always a positive no., therefore,

r = 4,

hence its modulus is 4.

now, dividing eq.2 by eq.1, we get,

Tanθ = 0

Since cosθ = 1, sinθ = 0 and tanθ = 0. Therefore the θ lies in first quadrant.

Tanθ = 0, therefore θ = 0°

Representing the complex no. in its polar form will be

Z = 4(cos0° + isin0°)

Let Z = -2 = r(cosθ + isinθ)

Now, separating real and complex part, we get

-2 = rcosθ………. eq.1

0 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

4 = r²

Since r is always a positive no.,therefore,

r = 2,

hence its modulus is 2.

now, dividing eq.2 by eq.1 , we get,

Tanθ = 0

Since cosθ = -1, sinθ = 0 and tanθ = 0. Therefore the lies in second quadrant.

Tanθ = 0, therefore θ = π

Representing the complex no. in its polar form will be

Z = 2(cosπ + isinπ)

Let Z = -i = r(cosθ + isinθ)

Now , separating real and complex part , we get

0 = rcosθ……….eq.1

-1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

1 = r²

Since r is always a positive no., therefore,

r = 1,

hence its modulus is 1.

now, dividing eq.2 by eq.1 , we get,

Tanθ = -∞

Since cosθ = 0 , sinθ = -1 and tanθ = -∞ . therefore the lies in fourth quadrant.

Tanθ = -∞, therefore

Representing the complex no. in its polar form will be

Let Z = 2i = r(cosθ + isinθ)

Now , separating real and complex part , we get

0 = rcosθ ……….eq.1

2 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

4 = r²

Since r is always a positive no., therefore,

r = 2,

hence its modulus is 2.

now, dividing eq.2 by eq.1, we get,

Tanθ = ∞

Since cosθ = 0, sinθ = 1 and tanθ = ∞. Therefore the θ lies in first quadrant.

tanθ = ∞, therefore

Representing the complex no. in its polar form will be

Let Z = 1 - i = r(cosθ + isinθ)

Now , separating real and complex part , we get

1 = rcosθ ……….eq.1

-1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

2 = r²

Since r is always a positive no., therefore,

r = √2,

hence its modulus is √2.

now , dividing eq.2 by eq.1 , we get,

Tanθ = -1

Since , and tanθ = -1 . therefore the θ lies in fourth quadrant.

Tanθ = -1, therefore

Representing the complex no. in its polar form will be

}

Let Z = 1 - i = r(cosθ + isinθ)

Now , separating real and complex part , we get

-1 = rcosθ ……….eq.1

1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

2 = r²

Since r is always a positive no., therefore,

,

hence its modulus is √2.

now, dividing eq.2 by eq.1 , we get,

Tanθ = -1

Since and tanθ = -1. therefore the θ lies in second quadrant.

Tanθ = -1, therefore

Representing the complex no. in its polar form will be

}

Let Z = + i = r(cosθ + isinθ)

Now , separating real and complex part , we get

= rcosθ ……….eq.1

1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

4 = r²

Since r is always a positive no., therefore,

r =2,

hence its modulus is 2.

now, dividing eq.2 by eq.1, we get,

Since , and . therefore the θ lies in first quadrant.

, therefore

Representing the complex no. in its polar form will be

}

Let Z = - 1 = r(cosθ + isinθ)

Now , separating real and complex part , we get

-1 = rcosθ ……….eq.1

= rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

4 = r²

Since r is always a positive no., therefore,

r = 2,

hence its modulus is 2.

now, dividing eq.2 by eq.1 , we get,

Since , and . therefore the θ lies in second quadrant.

Tanθ = , therefore

Representing the complex no. in its polar form will be

}

Let Z = - + 1 = r(cosθ + isinθ)

Now , separating real and complex part , we get

1 = rcosθ ……….eq.1

= rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

4 = r²

Since r is always a positive no., therefore,

r = 2,

hence its modulus is 2.

now, dividing eq.2 by eq.1 , we get,

Since , and . therefore the θ lies in the fourth quadrant.

Tanθ = , therefore θ =

Representing the complex no. in its polar form will be

Z = 2{cos + isin}

Let Z = 2 - 2i = r(cosθ + isinθ)

Now , separating real and complex part , we get

2 = rcosθ ……….eq.1

-2 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

8 = r²

Since r is always a positive no. therefore,

r = 2,

hence its modulus is 2√2.

now, dividing eq.2 by eq.1 , we get,

Tanθ = -1

Since , and tanθ = -1 . therefore the θ lies in the fourth quadrant.

Tanθ = -1, therefore

Representing the complex no. in its polar form will be

}

Let Z = 4√2i - 4 = r(cosθ + isinθ)

Now, separating real and complex part , we get

-4 = rcosθ ……….eq.1

4√3 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

64 = r²

Since r is always a positive no., therefore,

r = 8

hence its modulus is 8.

now, dividing eq.2 by eq.1, we get,

Since , and . therefore the lies in second the quadrant.

Tanθ = -√3, therefore θ= .

Representing the complex no. in its polar form will be

}

Let Z = 3√2i - 3√2 = r(cos + isinθ)

Now, separating real and complex part , we get

-3√2 = rcosθ ……….eq.1

3√2 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

36 = r²

Since r is always a positive no., therefore,

r = 6

hence its modulus is 6.

now, dividing eq.2 by eq.1, we get,

Since , and tanθ = -1 . therefore the θ lies in secothe nd quadrant.

Tanθ = -1 , therefore θ = .

Representing the complex no. in its polar form will be

}

= i

Let Z = i = r(cosθ + isinθ)

Now , separating real and complex part , we get

0 = rcosθ ……….eq.1

1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

1 = r²

Since r is always a positive no., therefore,

r = 1,

hence its modulus is 1.

now, dividing eq.2 by eq.1 , we get,

tanθ = ∞

Since cosθ = 0, sinθ = 1 and tanθ = ∞. Therefore the θ lies in first quadrant.

tanθ = ∞, therefore

Representing the complex no. in its polar form will be

}

= -i

Let Z = -i = r(cosθ + isinθ)

Now, separating real and complex part , we get

0 = rcosθ……….eq.1

-1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

1 = r²

Since r is always a positive no., therefore,

r = 1,

hence its modulus is 1.

now, dividing eq.2 by eq.1 , we get,

Tanθ = -∞

Since cosθ = 0, sinθ = -1 and tanθ = -∞, therefore the θ lies in fourth quadrant.

Tanθ = -∞, therefore

Representing the complex no. in its polar form will be

}

= i - 1

Let Z = 1 - i = r(cosθ + isinθ)

Now , separating real and complex part , we get

-1 = rcosθ ……….eq.1

1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

2 = r²

Since r is always a positive no., therefore,

r =√2,

hence its modulus is √2.

now, dividing eq.2 by eq.1 , we get,

Tanθ = -1

Since , and tanθ = -1 . therefore the θ lies in second quadrant.

Tanθ = -1 , therefore

Representing the complex no. in its polar form will be

}

= -i - 1

Let Z = -1 - i = r(cosθ + isinθ)

Now , separating real and complex part , we get

-1 = rcosθ ……….eq.1

-1 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

2 = r²

Since r is always a positive no., therefore,

r = √2,

hence its modulus is √2.

now , dividing eq.2 by eq.1 , we get,

tanθ = 1

Since , and tanθ = 1 . therefore the θ lies in third quadrant.

Tanθ = 1, therefore

Representing the complex no. in its polar form will be

}