Complex Numbers And Quadratic Equations

Given Complex Number:



Multiplying it,

⇒

⇒ - 3 i²

We know that i² = -1

So,

⇒ - 3 × -1

= 3

⇒ 1×i + 1×(-i)

⇒ i + (-i)

⇒ 0

i ^-39

Now, - 39 = - 36 - 3

So we have,

Rationalizing it,

Multiplying like-like terms,

⇒ 21 + 21i + 7i +

⇒ 21 + 28i + 7× (-1)

⇒ 14 + 28i

(1-i) – (-1 + i6)

⇒ 1-i + 1-i6

⇒ 2-7i

Given complex Number:



Now to Express the complex number in the form a + i b, we need to separate the real parts and imaginary parts for both the complex numbers.

Separating like–like terms,

⇒ -4

Given Complex number:

Applying the formula: (a + b)³ = a³ + b³ + 3a²b + 3ab²

Therefore,

Taking -1 outside the bracket,

Let z = 4 - 3 i

We know that multiplicative inverse of z is 1/z.

So the multiplicative inverse of given complex number will be,


Now rationalizing the number we get,



As i ² = -1

To Find: Multiplicative Inverse of √5 + 3 i

Concept: Multiplicative inverse of x is 1/x

Multiplicative inverse of a number x is given by 1/x.

To Find: Multiplicative inverse of - i

Therefore,


Multiplying and dividing the expression by i we get,


we know that, i² = -1

Putting this value we get,



Hence, multiplicative inverse of - i is i.

Use formula: (a + b)(a - b) = a² - b² in the numerator we get,

Rationalizing it,

Or,

Given: z = –1 – i√ 3

Formulas:

Modulus of a complex number z = x + i y is given by,

|z| = √(x² + y²)

So modulus of given z is,

|z| = √[(-1)² + (√3)²]

|z| = √4 = 2

Now for argument let us find out the quadrant in which the complex number is present.
As real and imaginary parts are negative, z lies in third quadrant.

Thus Argument of z in third quadrant is,

arg(z) = α - π

And where,





α = π/3

arg(z) = π/3 - π


= - 2π/3

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, now, let –√3 = r cos θ and 1 = r sin θ ……….(i)

Squaring both sides, we get

3 = r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

3 + 1 = r² cos² θ + r² sin² θ

⇒ 4 = r² (cos² θ + sin² θ)

⇒ 4 = r² or r² = 4 [∵ sin² θ + cos² θ = 1]

⇒ r = √4

⇒ r = 2 (conventionally, r>0) ……….(ii)

Substituting r = 2 in (i), we get

–√3 = 2 cos θ and 1 = 2 sin θ

∵ We know that the complex number –√3 + i lies in the second quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

……….(iii)

From (ii) and (iii), we have

r = 2 and

Let 1 = r cos θ and –1 = r sin θ ……….(i)

Squaring both sides, we get

1 = r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

1 + 1 = r² cos² θ + r² sin² θ

⇒ 2 = r² (cos² θ + sin² θ)

⇒ 2 = r² or r² = 2 [∵ sin² θ + cos² θ = 1]

⇒ r = √2

⇒ r = √2 (conventionally, r>0)

Substituting r = √2 in (i), we get

1 = √2 cos θ and – 1 = √2 sin θ

∵ We know that the complex number 1 – i lies in the third quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is .

Let – 1 = r cos θ and 1 = r sin θ ……….(i)

Squaring both sides, we get

1= r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

1 + 1 = r² cos² θ + r² sin² θ

⇒ 2 = r² (cos² θ + sin² θ)

⇒ 2 = r² or r² = 2 [∵ sin² θ + cos² θ = 1]

⇒ r = √2

⇒ r = √2 (conventionally, r>0)

Substituting r = √2 in (i), we get

– 1 = √2 cos θ and 1 = √2 sin θ

∵ We know that the complex number – 1 + i lies in the second quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is .

Let – 1 = r cos θ and – 1 = r sin θ ……….(i)

Squaring both sides, we get

1= r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

1 + 1 = r² cos² θ + r² sin² θ

⇒ 2 = r² (cos² θ + sin² θ)

⇒ 2 = r² or r² = 2 [∵ sin² θ + cos² θ = 1]

⇒ r = √2

⇒ r = √2 (conventionally, r>0)

Substituting r = √2 in (i), we get

– 1 = √2 cos θ and – 1 = √2 sin θ

∵ We know that the complex number –1 – i lies in the third quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is .

Let – 3 = r cos θ and 0 = r sin θ ……….(i)

Squaring both sides, we get

9= r² cos² θ and 0 = r² sin² θ

Adding both the equations, we get

9 + 0 = r² cos² θ + r² sin² θ

⇒ 9 = r² (cos² θ + sin² θ)

⇒ 9 = r² or r² = 9 [∵ sin² θ + cos² θ = 1]

⇒ r = √9

⇒ r = 3 (conventionally, r>0)

Substituting r = √2 in (i), we get

– 3 = 3 cos θ and 0 = 3 sin θ

⇒ cos θ = – 1 and sin θ = 0

⇒ θ = cos^-1 (-1) and θ = sin^-1 0

∵ We know that the complex number – 3 lies on the real axis (x-axis) and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

∴ θ = cos^-1 cos π and θ = sin^-1 sin π

⇒ θ = π

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is z = 3 (cos π + i sin π)

Let √3 = r cos θ and 1 = r sin θ ……….(i)

Squaring both sides, we get

3= r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

3 + 1 = r² cos² θ + r² sin² θ

⇒ 4 = r² (cos² θ + sin² θ)

⇒ 4 = r² or r² = 4 [∵ sin² θ + cos² θ = 1]

⇒ r = √4

⇒ r = 2 (conventionally, r>0)

Substituting r = 2 in (i), we get

√3 = 2 cos θ and 1 = 2 sin θ

∵ We know that the complex number √3 + i lies in the first quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is .

Let 0 = r cos θ and 1 = r sin θ……….(i)

Squaring both sides, we get

0= r² cos² θ and 1 = r² sin² θ

Adding both the equations, we get

0 + 1 = r² cos² θ + r² sin² θ

⇒ 1 = r² (cos² θ + sin² θ)

⇒ 1 = r² or r² = 1 [∵ sin² θ + cos² θ = 1]

⇒ r = √1

⇒ r = 1 (conventionally, r>0)

Substituting r = √2 in (i), we get

0 = cos θ and 1 = sin θ

⇒ cos θ = 0 and sin θ = 1

⇒ θ = cos^-1 0 and θ = sin^-1 1

∵ We know that the complex number i lies on the imaginary axis (y-axis) and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.

As we know that the polar representation of a complex number z = x + iy is

z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.

So, the required polar form is .

It is given in the question that,

=

=

=

=

=

= [- 1 – i]³

= (- 1)³ [1 + i]³

= - [1³ + i³ + 3 × 1 × i (1 + i)]

= - [1 + i³ + 3i + 3i²]

= - [1 – i + 3i – 3]

= - [- 2 + 2i]

= 2 – 2i

It is given in the question that,

z₁ and z₂ are two complex numbers and we have to prove that:

Re (z₁z₂) = Rez₁ Rez₂ – Imz₁ Imz₂

For this, firstly let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

Thus, z₁z₂ = (x₁ + iy₁) (x₂ + iy₂)

= x₁ (x₂ + iy₂) + iy₁ (x₂ + iy₂)

= x₁x₂ + ix₂y₂ + iy₁x₂ + i²y₁y₂

= x₁x₂ + ix₂y₂ + iy₁x₂ – y₁y₂ (i² = - 1)

= (x₁x₂ – y₁y₂) + i (x₁y₂ + y₁x₂)

Re (z₁z₂) = x₁x₂ – y₁y₂

∴ Re (z₁z₂) = Rez₁ Rez₂ – Imz₁ Imz₂

Hence, proved

We have,

=

=

=

=

=

=

Now by multiplying numerator and denominator by (14 + 5i), we get:

=

=

=

=

=

Hence, this is the required standard form

It is given in the question that,

_{We have to prove that,}

Proof:

_{Now by multiplying the numerator and denominator by (c + id), we get:}

Thus,

Now, by comparing the real and the imaginary parts we get:

=

=

=

=

=

=

Hence, proved

We have,

=

Now, by multiplying numerator and denominator by 3 + 4i we get:

=

=

=

=

= - 1 + i

Let us assume and

Now by squaring and adding the given values, we get:

We know that,

∴ r² = 2

As conventionally r > 0, thus

Also,

As lies in the 2^nd quadrant

∴

Hence,

=

=

This is the required polar form of the given equation

_{We have,}

Now, by multiplying numerator and denominator by 1 + 2i we get:

=

=

=

= - 1 + i

Let us assume and

Now by squaring and adding the given values, we get:

We know that,

∴ r² = 2

As conventionally r > 0, thus

Also,

As lies in the 2^nd quadrant

∴

Hence,

=

=

This is the required polar form of the given equation

We have,

This equation can be rewritten as follows:

9x² – 12x + 20 = 0

Now, in order to solve the equation we have to compare this with ax² + bx + c = 0 where a = 9, b = - 12 and c = 20

Thus, the discriminant of the above given equation is:

D = b² – 4ac

= (- 12)² – 4 × 9 × 20

= 144 – 720

= - 576

Hence, the required solutions are:

=

=

=

=

=

We have,

This equation can be rewritten as follows:

2x² – 4x + 3 = 0

Now, in order to solve the equation we have to compare this with ax² + bx + c = 0 where a = 2, b = - 4 and c = 3

Thus, the discriminant of the above given equation is:

D = b² – 4ac

= (- 4)² – 4 × 2 × 3

= 16 – 24

= - 8

Hence, the required solutions are:

=

=

=

We have,

Now, in order to solve the equation we have to compare this with ax² + bx + c = 0 where a = 27, b = - 10 and c = 1

Thus, the discriminant of the above given equation is:

D = b² – 4ac

= (- 10)² – 4 × 27 × 1

= 100 – 108

= - 8

Hence, the required solutions are:

=

=

=

We have,

Now, in order to solve the equation we have to compare this with ax² + bx + c = 0 where a = 21, b = - 28 and c = 10

Thus, the discriminant of the above given equation is:

D = b² – 4ac

= (- 28)² – 4 × 21 × 10

= 784 – 840

= - 56

Hence, the required solutions are:

=

=

=

=

It is given in the question that,

_z1 = 2 – i

_z2 = 1 + i

∴

=

=

=

=

=

=

=

=

=

Hence, the value of is .

It is given in the question that,

=

=

=

_{Now, by comparing the real and the imaginary parts we obtain:}

∴

=

=

=

∴

Hence, proved

(i) It is given in the question that,

z₁ = 2 – 1 and z₂ = - 2 + i

Now, we have:

z₁z₂ = (2 – i) (- 2 + i)

= - 4 + 2i + 2i – i^2

= - 4 + 4i – (- 1)

= - 3 + 4i

∴

Now, by multiplying numerator and denominator by (2 – i) we get:

=

=

=

=

Now, on comparing the real parts we get:

(ii) We have,

=

=

Now, on comparing the imaginary parts we get:

Let us assume,

Now multiplying the numerator and denominator by 1 + 3i, we get:

=

=

=

=

Let us now assume,

∴

Now, by squaring the both sides we get:

As conventionally r > 0, thus

∴

Hence,

Thus, modulus of the given complex number =

And, argument of given complex number =

Let us assume, z = (x – iy) (3 + 5i)

z = 3x + 5xi – 3yi – 5yi²

= 2x + 5xi – 3yi + 5y

= (3x + 5y) + i (5x – 3y)

∴ = (3x + 5y) – i (5x – 3y)

It is given in the question that, = - 6 – 24i

Now by equating the real and imaginary parts of the equation, we get:

3x + 5y = - 6 (i)

5x – 3y = 24 (ii)

Now, by multiply equation (i) by 3 and equation (ii) by 5 and then by adding them both we get:

34x = 102

∴

= 3

Now putting the value of x in (i), we get:

3 × 3 + 5y = - 6

5y = - 15

∴

Hence, x = 3 and y = - 3

Given in the question,

Thus, in order to find the modulus we will follow following steps:

It is given in the question that,

(x + iy)³ = u + iv

x³ + i³y³ + 3 × x × iy (x + iy) = u + iv

x³ + i³y³ + 3x²yi + 3xi²y²= u + iv

x³ – iy³ + 3x²yi – 3xy² = u + iv

(x³ – 3xy²) + i(-y³ + 3x²y)= u + iv

Now, equating the imaginary and real part we will get,

u = (x³ – 3xy²), v = (-y³ + 3x²y)

According to question,

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

Thus,

Hence, proved

It is given that_and β are different complex numbers,

And, |β|=1

⇒ = 1

Now,

=

= 1

We have equation as:

|1-i|^x = 2^x

2 ^x/2 = 2 ^x

On equating powers, we get:

x = 2x

2x - x = 0

x = 0

Hence, for the given solution 0 is the only integral solution.

It is given in the question that,

(a + ib) (c + id) (e + if) (g + ih) = A + iB

Using, [|z₁z₂|=|z₁||z₂|]

|(a + ib)| × | (c + id) | × |(e + if) | × |(g + ih)| = |A + iB|

Now, we will square on both sides,

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Hence, proved

It is given in the question that,

i^m = 1

We can also write,

i^m = i^4k

On equating the powers,

Thus, m = 4k, Where k is some integer.

∴ 1 is the least positive integer.

Least positive integral value of m is 4 × 1 = 4

It is given in the question that,

x² + 3 = 0

∴

Hence,

It is given in the question that,

2x² + x + 1 = 0

Now, comparing the given value with ax² + bx + c = 0 where a = 2, b = 1 and c = 1

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

x² + 3x + 9 = 0

Now, comparing the given value with ax² + bx + c = 0 where a = 1, b = 3 and c = 9

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

- x² + x - 2 = 0

Now, comparing the given value with ax² + bx + c = 0 where a = - 1, b = 1 and c = - 2

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

x² + 3x + 5 = 0

Now, comparing the given value with ax² + bx + c = 0 where a = 1, b = 3 and c = 5

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

x² - x + 2 = 0

Now, comparing the given value with ax² + bx + c = 0 where a = 1, b = - 1 and c = 2

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

Now, comparing the given value with ax² + bx + c = 0 where , b = 1 and

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

Now, comparing the given value with ax² + bx + c = 0 where , and

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

Now, comparing the given value with ax² + bx + c = 0 where , and

Thus,

=

=

=

Hence, the value of x should be and

It is given in the question that,

Now, comparing the given value with ax² + bx + c = 0 where , and

Thus,

=

=

=

=

Hence, the value of x should be and