Complex Numbers And Quadratic Equations

Given

= (1 – i)ⁿ (1 + i)ⁿ

= (1 – i²)ⁿ

= 2ⁿ

Given

= (1 + i) (1 + i² + i³ + i⁴ + i⁵ + i⁶ + i⁷ + i⁸ + i⁹ + i¹⁰ + i¹¹ + i¹² + i¹³)

= (1 + i) i

= i + i²

= i – 1

Given

= i³ – (-i³)

= 2i³

= 0 – 2i

Thus, (x, y) = (0, -2)

Given

Rationalizing the denominator,

Thus,

Hence,

Given

= (i⁴)²⁵

= 1

∴ (a, b) = (1, 0)

Given a = cos θ + i sin θ

∴ If a = cos θ + i sin θ, the value of

Given (1 + i) z = (1 – i) z̅

Rationalizing the denominator,

= -iz̅

Hence proved.

Given z = x + iy

⇒ z̅ = x – iy

Now, z z̅ + 2 (z + z̅) + b = 0

⇒ (x + iy) (x – iy) + 2 (x + iy + x – iy) + b = 0

⇒ x² + y² + 4x + b = 0

This is the equation of a circle.

Let z = x + iy

Now,

Given that real part is 4.

⇒ x² + x – 2 + y² = 4 (x² – 2x + 1 + y²)

⇒ 3x² + 3y² – 9x + 6 = 0

Which represents a circle.

Hence, locus of z is a circle.

Let z = x + iy

Given

⇒ arg (z – 1) – arg (z + 1) = π/4

⇒ arg (x + iy – 1) – arg (x + iy + 1) = π/4

⇒ arg (x – 1 + iy) – arg (x + 1 + iy) = π/4

⇒ x² + y² – 1 = 2y

⇒ x² + y² – 2y – 1 = 0

Which represents a circle.

Given |z| = z + 1 + 2i

Putting z = x + iy, we get

⇒ |x + iy| = x + iy + 1 + 2i

We know that

Comparing real and imaginary parts,

And 0 = y + 2

⇒ y = -2

Putting this value of y in ,

⇒ x² + (-2)² = (x + 1)²

⇒ x² + 4 = x² + 2x + 1

∴ x = 3/2

∴ z = x + iy

= 3/2 – 2i

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

Putting z = x + iy, we get

⇒ |x + iy + 1| = x + iy + 2 (1 + i)

We know that

Comparing real and imaginary parts,

And 0 = y + 2

⇒ y = -2

Putting this value of y in ,

⇒ (x + 1)² + (-2)² = (x + 2)²

⇒ x² + 2x + 1 + 4 = x² + 4x + 4

⇒ 2x = 1

∴ x = 1/2

∴ z = x + iy

= 1/2 – 2i

Let z = x + iy

Given arg (z – 1) = arg (z + 3i)

⇒ arg (x + iy – 1) = arg (x + iy + 3i)

⇒ arg (x – 1 + iy) = arg (x + I (y) = π/4

⇒ xy = xy – y + 3x – 3

⇒ 3x – 3 = y

∴ (x – 1): y = 1: 3

Given

Substituting z = x + iy, we get

⇒ |x – 2 + iy| = 2 |x – 3 + iy|

⇒ x² – 4x + 4 + y² = 4 (x² – 6x + 9 + y²)

⇒ 3x² + 3y² – 20x + 32 = 0

Hence, centre of circle is (10/3, 0) and radius is 2/3.

Let z = x + iy

Now,

Given that is purely imaginary.

⇒ x² – 1 + y² = 0

⇒ x² + y² = 1

∴ |z| = 1

Let z₁ = |z₁| (cos θ₁ + I sin θ₁) and z₂ = |z₂| (cos θ₂ + I sin θ₂)

Given that |z₁| = |z₂|

And arg (z₁) + arg (z₂) = π

⇒ θ₁ + θ₂ = π

⇒ θ₁ = π – θ₂

Now, z₁ = |z₂| (cos (π - θ₂) + I sin (π - θ₂))

⇒ z₁ = |z₂| (-cos θ₂ + I sin θ₂)

⇒ z₁ = -|z₂| (cos θ₂ – I sin θ₂)

⇒ z₁ = - [|z₂| (cos θ₂ – I sin θ₂)]

∴ z₁ = -z̅ ₂

Hence proved.

Let z₁ = x + iy

Now,

Since x² + y² = 1

∴ Hence, the real part of z₂ is zero.

Given z₁ and z₂ are conjugate complex numbers.

⇒ z₂ = z̅ ₁ = |z|/2pi is,

he negative side of the real axis

= r₁ (cos θ₁ - i sin θ₁)

= r₁ [cos (-θ₁) + I sin (-θ₁)]

Similarly, z₃ = r₂ (cos θ₂ - i sin θ₂)

⇒ z₄ = r₂ [cos (-θ₂) + I sin (-θ₂)]

= θ₁ – (-θ₂) + (-θ₁) – θ₂

= θ₁ + θ₂ – θ₁ – θ₂

= 0

Given |z₁| = |z₂| = … = |z_n| = 1

⇒ |z₁|² = |z₂|² = … = |z_n|² = 1

⇒ z₁ z̅ ₁= z₂ z̅ ₂= z₃ z̅ ₃= … = z_n z̅ _n= 1

Now,

Hence proved.

Let z₁ = |z₁| (cos θ₁ + I sin θ₁) and z₂ = |z₂| (cos θ₂ + I sin θ₂)

Given arg (z₁) - arg (z₂) = 0

⇒ θ₁ - θ₂ = 0

⇒ θ₁ = θ₂

Now, z₂ = |z₂| (cos θ₁ + I sin θ₁)

⇒ z₁ – z₂ = ((|z₁|cos θ₁ - |z₂| cos θ₁) + i (|z₁| sin θ₁ - |z₂| sin θ₁))

We know that cos² θ + sin² θ = 1

∴ |z₁ – z₂| = |z₁| - |z₂|

Hence proved.

Given Re (z²) = 0, |z| = 2

Let z = x + iy.

Then

Given

⇒ x² + y² = 4 … (1)

Also, z² = x² + 2ixy – y²

= (x² - y²) + 2ixy

Now, Re (z²) = 0

⇒ x² – y² = 0 … (2)

Solving (1) and (2), we get

⇒ x² = y² = 2

⇒ x = y = ±√2

∴ z = x + iy

= ±√2 ± i√2

Hence, we have four complex numbers.

Given equation z + √2 |(z + 1)| + i = 0 … (1)

Putting z = x + iy, we get

⇒ x + iy + √2 |x + iy + 1| + i = 0

Comparing real and imaginary parts to zero, we get

… (2)

And y + 1 = 0

⇒ y = -1

Putting y = -1 into (2), we get

⇒ 2x² + 4x + 4 = x²

⇒ x² + 4x + 4 = 0

⇒ (x + 2)² = 0

⇒ x = -2

∴ z = x + iy

= -2 – i

Given

Let z = |z| (cos θ₁ + I sin θ₁) and w = |w| (cos θ₂ + I sin θ₂)

Given |zw| = |z| |w| = 1

Also arg (z) – arg (w) = π/2

⇒ θ₁ - θ₂ = π/2

Now, z̅ w = |z| (cos θ₁ - I sin θ₁) |w| (cos θ₂ + I sin θ₂)g | = 1

e get

ts to zero, we get

= |z| |w| (cos (-θ₁) + I sin (-θ₁)) (cos θ₂ + I sin θ₂)

= 1 [cos (θ₂ – θ₁) + I sin (θ₂ – θ₁)]

= [cos (-π/2) + I sin (-π/2)]

= 1 [0 – i]

= -i

Hence proved.

Given |az₁ – bz₂|² + |bz₁ + az₂|²

= |az₁|² + |bz₂|² – 2 Re (az₁ × bz̅ ₂) + |bz₁|² + |az₂|² + 2 Re (bz₁ × az̅ ₂)

= (a² + b²) |z₁|² + (a² + b²) |z₂|²

= (a² + b²) (|z₁|² + |z₂|²)bz = |

Given √-25 × √-9

= i √25 × i √9

= i² (5 × 3)

= -15

Given

= -2

Given i + i² + i³ + …. Upto 1000 terms

= (i + i² + i³ + i⁴) + (i⁵ + i⁶ + i⁷ + i⁸) + … 250 brackets

Since iⁿ + iⁿ⁺¹ + i^{n + 2} + i^{n + 3} = 0 where n ∈ N

= 0 + 0 + 0 … + 0

= 0

Explanation:

Given 1 + i

= 1/2 (1 – i)

∴ Multiplication inverse of 1 + i is 1/2 (1 – i).

Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

⇒ z₁ + z₂ = (x₁ + x₂) + i (y₁ + y₂) which is real

⇒ y₁ + y₂ = 0

⇒ y₁ = - y₂

Assuming x₁ = x₂

Since z₂ = x₁ – iy₁

∴ z₂ = z̅ ₁

⇒ arg (z) + arg z̅

= θ + (-θ)

= 0

Given |z + 4| ≤ 3

For greatest value of |z + 1|,

⇒ |z + 1| = |z + 4 – 3|

⇒ |z + 1| ≤ |z + 4 – 3|

⇒ |z + 1| ≤ 3 + 3

⇒ |z + 1| ≤ 6

So, greatest value of |z + 1| is 6.

We know that least value of the modulus of a complex number is zero.

So, the least value of |z + 1| is zero.

Explanation:

Given

⇒ 6 |x – 2 + iy| = π |x + 2 + iy|

Squaring on both sides,

⇒ 36 |x – 2 + iy|² = π² |x + 2 + iy|²

⇒ 36 [x² – 4x + 4 + y²] = π² [x² – 4x + 4 + y²]

⇒ (36 – π²) x² + (36 – π²) y² – (144 + 4π²) x + 144 – 4π² = 0

Which is a circle.

Given |z| = 4 and arg (z) = 5π/6

Let z = |z| (cos θ + i sin θ) where θ = arg (z)

= -2√3 + 2i

∴ z = -2√3 + 2i

False

Explanation:

We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.

False

Explanation:

Let z = x + iy, where x, y > 0

i.e. z or point A (x, y) lies in first quadrant.

Now, -iz = -I (x + iy)

= -ix – i²y

= y – ix

Now, point B (y, -x) lies in fourth quadrant.

Also ∠AOB = 90°

Thus, B is obtained by rotating A in clockwise direction about origin.

True

Explanation:

Given |z| + |z – 1|

We know that |z₁| + |z₂| ≥ |z₁ – z₂|

⇒ |z| + |z – 1| ≥ |z – (z – 1) |

⇒ |z| + |z – 1| ≥ 1

So, minimum value of |z| + |z – 1| is 1.

True

Explanation:

Given |z – 1| = |z – i|

Putting z = x + iy,

⇒ |x – 1 + iy| = |x – i (1 – y) |

⇒ (x – 1)² + y² = x² + (1 – y)²

⇒ x² - 2x + 1 + y² = x² + 1 + y² – 2y

⇒ -2x + 1 = 1 – 2y

⇒ -2x + 2y = 0

⇒ x – y = 0

Now, equation of a line through the points (1, 0) and (0, 1) is

⇒ x + y = 1

This line is perpendicular to the line x – y = 0.

False

Explanation:

Given z ≠ 0 and Re (z) = 0

Let z = x + iy

Then x = 0

∴ z = iy

⇒ Im (z²) = i²y² = -y² ≠ 0

True

Explanation:

Given |z – 4| < |z – 2|

Putting z = x + iy,

⇒ |x – 4 + iy| < |x – 2 + iy|

⇒ (x – 4)² + y² < (x – 2)² + y²

⇒ x² – 8x + 16 + y² < x² – 4x + 4 + y²

⇒ -8x + 16 < -4x + 4

⇒ 4x > 12

∴ x > 3

False

Explanation:

Given |z₁ + z₂| = |z₁| + |z₂|

⇒ |z₁ + z₂|² = |z₁|² + |z₂|² + 2 |z₁| |z₂|

⇒ |z₁|² + |z₂|² + 2 Re (z₁z̅ ₂) = |z₁|² + |z₂|² + 2 |z₁| |z₂|

⇒ 2 Re (z₁ z̅ ₂) = 2 |z₁| |z₂|

⇒ cos (θ₁ – θ₂) = 1

⇒ θ₁ – θ₂ = 0

∴ arg (z₁) – arg (z₂) = 0

False

Explanation:

We know that any real number is also a complex number.

∴ 2 is a complex number.

(a) Given z = i + √3

So, |z| = |i + √3|

= 2

Also, z lies in the first quadrant.

= π/6

∴ The polar form of z is

(b) Given z = -1 + √-3

= -1 + i √3

Here z lies in the second quadrant.

⇒ arg (z) = amp (z)

= π – tan^-1 √3

= π - π/3

= 2π/3

(c) Given |z + 2| = |z – 2|

⇒ |x + 2 + iy| = |x – 2 + iy|

⇒ (x + 2)² + y² = (x – 2)² + y²

⇒ x² + 4x + 4 = x² – 4x + 4

⇒ 8x = 0

∴ x = 0

It is a straight line which is a perpendicular bisector of segment joining the points (-2, 0) and (2, 0).

(d) Given |z + 2i| = |z – 2i|

⇒ |x + i(y + 2)| = |x + i(y - 2)|

⇒ (x)² + (y + 2)² = (x)² + (y – 2)²

⇒ 4y = 0

∴ y = 0

It is a straight line which is a perpendicular bisector of segment joining the points (0, -2) and (0, 2).

(e) Given |z + 4i| ≥ 3

⇒ |x + iy + 4i| ≤ 3

⇒ |x + i(y + 4)|≤ 3

⇒ (x)² + (y + 4)² ≤ 9

⇒ x² + y² + 8y + 16 ≤ 9

⇒ x² + y² + 8y + 7 ≤ 0

This represents the region on or outside circle having centre (0, -4) and radius 3.

(f) Given |z + 4| ≤ 3

⇒ |x + iy + 4| ≤ 3

⇒ |x + 4 + iy|≤ 3

⇒ (x + 4)² + y² ≤ 9

⇒ x² + 8x + 16 + y² ≤ 9

⇒ x² + 8x + y² + 7 ≤ 0

This represents the region on or inside circle having centre (-4, 0) and radius 3.

(g) Given

Hence z̅ lies in the third quadrant.

(h) Given z̅ = 1 - i

lies in the third quadrant.

,outr inside circle having centre (-4, 0) and radius 3.nts (-2, x numbers is not possible or has no

= 1/2 (1 + i)

∴ Reciprocal of z lies in first quadrant.

Explanation:

Given

Rationalizing the denominator,

Explanation:

Given |z₁| = |z₂|

If |z₁| = |z₂| then z₁ and z₂ are at the same distance from origin.

But if arg (z₁) ≠ arg (z₂) then z₁ and z₂ are different.

So, if |z₁| = |z₂|, then it is not necessary that z₁ = z₂.

For example: z₁ = 3 + 4i and z₂ = 4 + 3i

Here |z₁| = |z₂| = 5 but z₁ ≠ z₂.

Explanation:

Given

Explanation:

Given |z| = 4 and arg (z) = 5π/6

Let z = |z| (cos θ + i sin θ) where θ = arg (z)

= -2√3 + 2i

∴ z = -2√3 + 2i

Explanation:

Given

= 1

Explanation:

Given z = (1 + i√3)²

= 1 – 3 + 2i√3

= -2 + i (2√3)

So, z lies in second quadrant.

= π – tan^-1 √3

= π - π/3

= 2π/3

Explanation:

Given

⇒ |z – 5i| = |z + 5i|

⇒ |x + iy – 5i| = |x + iy + 5i|

⇒ |x + i (y – 5) |² = |x + i (y + 5) |²

⇒ x² + (y – 5)² = x² + (y + 5)²

⇒ 20y = 0

⇒ y = 0

∴ z lies on the x – axis.

Given that sin x + i cos 2x and cos x – i sin 2x are conjugate to each other.

⇒ sin x - i cos 2x = cos x – i sin 2x

On comparing real and imaginary parts of both sides, we get

⇒ sin x = cos x and cos 2x = sin 2x

⇒ tan x = 1 and tan 2x = 1

Consider tan 2x = 1

We know that

This is not satisfied by tan x = 1.

Hence, no value of x is possible.

Given

Rationalizing the denominator,

It is given that z is purely real.

⇒ -3 sin α = 0

⇒ sin α = 0

∴ α = nπ, n ∈ I

Since z = x + iy lies in the third quadrant, we get

X < 0 and y < 0

Now

Since also lies in third quadrant, we get

⇒ x² – y² < 0 and -2xy < 0

⇒ x² < y² and xy > 0

But x, y < 0

∴ y < x < 0

Given (z + 3) (z̅ + 3) = (x + iy + 3) (x –iy + 3)

= (x + 3)² – (iy)²

= (x + 3)² + y²

= |x + 3 + iy|²

= |z + 3|²

Given

Rationalizing the denominator,

⇒ i^x = 1

∴ x = 4n, n ∈ N

Given

Rationalizing the denominator,

… (1)

… (2)

Multiplying equation (1) and (2),

So, α² + β² = 1

Let z₁ = |z₁| (cos θ₁ + i sin θ₁) and z₂ = |z₂| (cos θ₂ + i sin θ₂)

Now, z₁z₂ = |z₁| |z₂| (cos θ₁ + i sin θ₁) (cos θ₂ + i sin θ₂)

= |z₁| |z₂| [cos θ₁ cos θ₂ + i sin θ₁ cos θ₂ + i cos θ₁ sin θ₂ + i² sin θ₁ sin θ₂]

= |z₁| |z₂| [cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

⇒ |z₁ z₂| = |z₁| |z₂|

And arg (z₁ z₂) = θ₁ + θ₂ = arg (z₁) + arg (z₂)

⇒ |z₁ + z₂| = |z₁| + |z₂| is true only when z₁, z₂ and O are collinear.

Also, |z₁ + z₂| ≥ ||z₁| - |z₂||

Given z = 2 – i

If z rotated through an angle of π/2 about the origin in clockwise direction.

Then the new position = z. e^-(π/2)

= (2 – i) e^-(π/2)

= (2 – i) [cos (-π/2) + i sin (-π/2)]

= (2 – i) (0 – i)

= -1 – 2i

X + yi is a non – real complex number if y ≠ 0 if x, y ? R.

Given a + ib = c + id

⇒ |a + ib| = |c + id|

Squaring on both sides,

∴ a² + b² = c² + d²

Given

Let z = x + yi

⇒ |x + (y + 1) i| = |-x – (y – 1) i|

⇒ x² + (y + 1)² = x² + (y – 1)²

⇒ (y + 1)² = (y – 1)²

⇒ y² + 2y + 1 = y² – 2y + 1

⇒ 2y = -2y

⇒ 4y = 0

⇒ y = 0 (x – axis)

Given z is a complex number.

Let z = x + yi

⇒ |z| = |x + yi| and |z|² = |x + yi|²

⇒ |z|² = x² + y² … (1)

Now z² = x² + y²i² + 2xyi

⇒ z² = x² – y² + 2xyi

So, |z|² = x² + y² = |z|²

∴ |z|² = |z²|

Let z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂)

Since |z₁ + z₂| = |z₁| + |z₂|

⇒ z₁ + z₂ = r₁ cos θ₁ + ir₁ sin θ₁+ r₂ cos θ₂ + ir₂ sin θ₂

But |z₁ + z₂| = |z₁| + |z₂|

Squaring both sides,

⇒ 2r₁r₂ – 2r₁r₂ cos (θ₁ – θ₂) = 0

⇒ 1 – cos (θ₁ – θ₂) = 0

⇒ cos (θ₁ – θ₂) = 1

⇒ (θ₁ – θ₂) = 0

⇒ θ₁ = θ₂

∴ arg (z₁) = arg (z₂)

Let

Rationalizing the denominator,

If z is a real number,

⇒ 3 cos θ = 0

⇒ cos θ = 0

∴ θ = (2n + 1) π/2, n ∈ N

Let z = x + 0i and x < 0

Since the point (-x, 0) lies on the negative side of the real axis,

∴ Principal argument (z) = π

Given where z = 1 + 2i

= |z|/2