Mobius Transformations MCQ Quiz - Objective Question with Answer for Mobius Transformations - Download Free PDF

Concept:

1. Loxodromic Transformation:

A Möbius transformation with two distinct, non-real fixed points

It exhibits rotation and scaling simultaneously

It maps circles and lines into spirals or other geometric shapes with a constant angle of intersection

2. Hyperbolic Transformation:

A Möbius transformation with two distinct real fixed points

It maps lines and circles to hyperbolas

The transformation involves scaling along the real axis and rotation in the complex plane

Explanation:

1. 

This is a Möbius transformation of the form    , where a = 3i , b = 4 , c = 1 , and d = -i

To determine the nature of the transformation, we examine the fixed points and their properties

The fixed points of a Möbius transformation    can be found by setting w = z :

  

Multiplying both sides by z - i , we get:

z(z - i) = 3iz + 4

⇒  

⇒   

This is a quadratic equation, and solving for z , we can determine that    has two distinct fixed points

Given that    has two distinct fixed points and the transformation involves rotation and scaling,    is a loxodromic transformation

2.    

This is another Möbius transformation

Finding Fixed Points:

We set   :

Multiplying both sides by z - 7 , we get:

z(z - 7) = z

⇒  

⇒   

⇒ z(z - 8) = 0

Thus, the fixed points are z = 0 and z = 8 , both real and distinct

Since    has two distinct real fixed points,    is a hyperbolic transformation

Conclusion:

  is loxodromic, as it has two distinct fixed points and exhibits both scaling and rotation

  is hyperbolic, as it has two distinct real fixed points and maps lines to hyperbolas

Explanation -

Given . This is a specific Mobius transformation.

For Option (1) - Determine if T Maps the Imaginary Axis to a Circle

For z = iy (pure imaginary axis), 

Now,

This results in a value , which is not necessarily on a circle passing through the origin for all y.

So, Option (1) is incorrect.

For Option (3) - Check the Determinant of the Coefficients

The coefficients matrix for T is .   

The determinant is 3 × 1 - (-4) × 2 = 3 + 8 = 11, not zero.

Hence, Option (3) is incorrect.

For Option (2) and (4) - Fixed Points Calculation

Fixed points satisfy T(z) = z :

⇒ 

So, both are distinct but not real. 

So Option (2) is correct and Option (4) is incorrect.

Concept -

Mobius Transformation - A function of a complex variable that maps the extended complex plane to itself, preserving the cross-ratio and transforming circles and lines into circles and lines.

Explanation -

Given . This is a specific Mobius transformation.

For Option (1) - Check Mapping of Real Axis  

For z on the real axis z = x, calculate

This is not real unless x = 0. Therefore, T does not map the real axis onto itself.

So, Option (1) is incorrect.

For Option (2) - Identify Fixed Points  

Fixed points satisfy T(z) = z, so .  
⇒ z + i = z² - iz

⇒ z² - (1+i)z - i = 0

This quadratic equation in z can have complex solutions but not necessarily on the unit circle.

Hence Option (2) is incorrect.

For Option (3) - Cross-Ratio Preservation 

Mobius transformations are known to preserve the cross-ratio of any four points. This is a fundamental property.

Hence, Option (3) is correct.

For Option (4) - Check if T is its Own Inverse 

For T to be its own inverse, T(T(z)) should equal z.  

This does not simplify to z,

So Option (4) is incorrect.

Concept Used:

A one-to-one function, also known as an injective function, is a type of function in mathematics where each distinct element in the domain maps to a distinct element in the codomain. For a function f: A → B is said to be one-to-one if, 

An onto function, also known as a surjective function, is a type of function in mathematics where every element in the codomain is mapped to by at least one element in the domain. For a function f: A → B is said to be onto if, 

Explanation:

Given function  where ω is a fixed complex number in the open unit disk  D = {z ∈ ℂ | |z|

For  , we need to show that  z₁ = z₂.

Let's assume    and then simplify:

Hence,  F_ω is one-to-one.

Therefore, the statement "F is one-to-one" is true.

For  F_ω(z)  to be onto, every point in the target space ( D ) must be covered. Let's consider w  in D, and try to find z  such that F_ω(z) = w. 

As z  is well-defined for any w in D. 

Therefore,  F_ω  is onto.

Therefore, the statement "F is onto" is true.

Concept Used:

Points in the upper half of plane have a positive y coordinate.

Explanation:

Option 1: For 

For z = x + iy with y > 0, consider the transformation  

The imaginary part of w is , which is negative for y > 0. Therefore,  does not map z in H to another point in H.

Option 2: 

Using z = x + iy, we have 

The imaginary part is, which could be positive or negative depending on the sign of x.

So,  can be in H or not, depending on x.

Option 3: 

For z = x + it,

Since y > 0, the imaginary part of w is negative. Thus, does not map z in H to another point in H.

Option 4: 

Let 

Since y > 0, the imaginary part of w is positive, implying that  maps z in H to another point in H.

Thus, only option 4 is correct.

Concept:

(i) H+ = {z ∈ ℂ : y > 0} represents upper half-plane

H- = {z ∈ ℂ : y

L+ = {z ∈ ℂ : x > 0} represents right half-plane

L- = {z ∈ ℂ : x

(ii) We know that under the bilinear transformation , ad – bc > 0

upper half-plane maps to upper half-plane and lower half-plane plan maps to lower half-plane.

Explanation:

Given f(z) =  

Comparing given f(z) with  we get

a = 2, b = 1, c = 5, d = 3

Here ad - bc = 2 × 3 - 5 × 1 = 6 - 5 = 1 > 0

Hence f(z) maps upper half-plane to upper half-plane and lower half-plane plan maps to lower half-plane.

i.e., maps H+ onto H+ & H- onto H- 

(1) is correct

Concept:

(i) H+ = {z ∈ ℂ : y > 0} represents upper half-plane

H- = {z ∈ ℂ : y

L+ = {z ∈ ℂ : x > 0} represents right half-plane

L- = {z ∈ ℂ : x

(ii) We know that under the bilinear transformation , ad – bc > 0

upper half-plane maps to upper half-plane and lower half-plane plan maps to lower half-plane.

Explanation:

f(z) =  = 

Here ad - bc = 1 × 1 - 0 × 3 = 1 - 0 = 1 > 0

Hence f(z) maps upper half-plane to upper half-plane and lower half-plane plan maps to lower half-plane.

i.e., maps H+ onto H+ & H- onto H- 

(1) is correct

Concept -

Mobius Transformation - A function of a complex variable that maps the extended complex plane to itself, preserving the cross-ratio and transforming circles and lines into circles and lines.

Explanation -

Given . This is a specific Mobius transformation.

For Option (1) - Check Mapping of Real Axis  

For z on the real axis z = x, calculate

This is not real unless x = 0. Therefore, T does not map the real axis onto itself.

So, Option (1) is incorrect.

For Option (2) - Identify Fixed Points  

Fixed points satisfy T(z) = z, so .  
⇒ z + i = z² - iz

⇒ z² - (1+i)z - i = 0

This quadratic equation in z can have complex solutions but not necessarily on the unit circle.

Hence Option (2) is incorrect.

For Option (3) - Cross-Ratio Preservation 

Mobius transformations are known to preserve the cross-ratio of any four points. This is a fundamental property.

Hence, Option (3) is correct.

For Option (4) - Check if T is its Own Inverse 

For T to be its own inverse, T(T(z)) should equal z.  

This does not simplify to z,

So Option (4) is incorrect.

Concept:

(i) the linear fractional transformation w = f(z) maps three distinct points z₁, z₂, z₃ uniquely into three distinct point w₁, w₂, w₃ .

The map is determined by the equation    ..........  (i) 

(ii) under linear fractional transformation w = f(z) cross ratio is invariant 

Explanation:

T(z) = , ad - bc ≠ 0 be mobius transformation (linear fractional transformation) such that 

z₁ = 0 & w₁ = 10, so T(0) = 10

z₂ = -i & w₂ = 5-5i, so T(-i) = 5-5i

z₃ = ∞ & w₃ = 5+5i, so T(∞) = 5+5i

Substituting the values of w₁, w₂ & w₃ in equation (i) we get

⇒  ⇒ 

⇒  ⇒ 

Now, taking the set {w ∈ C : |w-5|

It is a circle with center (5,0) with radius 5 

At (0,0), z= -i

At (10,0), z=0

At (5,5), z=∞ 

The image of the set S = {z ∈ C; Re(z)

Hence, Option (3) is true

Explanation:

Let us consider a circle

|z - a| = |a| of centre a and radius |a|, 

Squareing both sides

|z - a|² = |a|²

⇒ (z - a) = |a|2

⇒ (z - a) = |a|2

⇒ (z - a) = |a|2

⇒ z - a - z - |a|2 = |a|2

⇒ ...(i)

Now, given w = f(z) =  ⇒ z =  ⇒ 

Then from (i) we get

⇒  which is a straight line not passing through origin

Hnece (2) is correct

Explanation -

Given . This is a specific Mobius transformation.

For Option (1) - Determine if T Maps the Imaginary Axis to a Circle

For z = iy (pure imaginary axis), 

Now,

This results in a value , which is not necessarily on a circle passing through the origin for all y.

So, Option (1) is incorrect.

For Option (3) - Check the Determinant of the Coefficients

The coefficients matrix for T is .   

The determinant is 3 × 1 - (-4) × 2 = 3 + 8 = 11, not zero.

Hence, Option (3) is incorrect.

For Option (2) and (4) - Fixed Points Calculation

Fixed points satisfy T(z) = z :

⇒ 

So, both are distinct but not real. 

So Option (2) is correct and Option (4) is incorrect.

Concept Used:

A one-to-one function, also known as an injective function, is a type of function in mathematics where each distinct element in the domain maps to a distinct element in the codomain. For a function f: A → B is said to be one-to-one if, 

An onto function, also known as a surjective function, is a type of function in mathematics where every element in the codomain is mapped to by at least one element in the domain. For a function f: A → B is said to be onto if, 

Explanation:

Given function  where ω is a fixed complex number in the open unit disk  D = {z ∈ ℂ | |z|

For  , we need to show that  z₁ = z₂.

Let's assume    and then simplify:

Hence,  F_ω is one-to-one.

Therefore, the statement "F is one-to-one" is true.

For  F_ω(z)  to be onto, every point in the target space ( D ) must be covered. Let's consider w  in D, and try to find z  such that F_ω(z) = w. 

As z  is well-defined for any w in D. 

Therefore,  F_ω  is onto.

Therefore, the statement "F is onto" is true.

Concept Used:

Points in the upper half of plane have a positive y coordinate.

Explanation:

Option 1: For 

For z = x + iy with y > 0, consider the transformation  

The imaginary part of w is , which is negative for y > 0. Therefore,  does not map z in H to another point in H.

Option 2: 

Using z = x + iy, we have 

The imaginary part is, which could be positive or negative depending on the sign of x.

So,  can be in H or not, depending on x.

Option 3: 

For z = x + it,

Since y > 0, the imaginary part of w is negative. Thus, does not map z in H to another point in H.

Option 4: 

Let 

Since y > 0, the imaginary part of w is positive, implying that  maps z in H to another point in H.

Thus, only option 4 is correct.

Concept:

A Mobius transformation maps a circle or straight line to a circle or straight line.

Explanation:

T is a M¨obius transformation such that T(0) = α, T(α) = 0 and T(∞) = −α, where α = (−1 + i)/ √2.

Since T(α) = 0 let us assume that

Let T(z) = 

T(0) = α ⇒ -  = α ⇒ d = - 1

T(∞) = −α ⇒  = −α ⇒ c = 

Hence T(z) =  

L denotes the straight line passing through the origin with slope −1, and C denotes the circle of unit radius centred at the origin.

i.e., L: y = - x

and C: x² + y² = 1 i.e., |z| = 1

Now, α = (−1 + i)/ √2 i.e., (-1/√2, 1/ √2) lies on L.

any point on L is z = x + iy = x - ix = x(1 - i) 

then, T(z) = T(x(1 - i) ) = 

So for the point x = 1 T(z) = ∞ so it can't lie inside a circle.

Hence T maps L to a straight line

Option (1) is true and (2) is false

Let w =  

⇒ 

⇒

⇒ z =

⇒ T^-1(z) =

For any point on circle C it maps to straight line.

T−1 maps C to a straight line 

Option (3) is true and (4) is false

Concept:

(i) A bilinear transformation is of the form

w = T(z) =  such that ad - bc ≠ 0 

(ii) Bilinear transformation that transform z₁, z₂, z₃ to w₁, w₂, w₃ is 

 = 

Explanation:

Given z1 = -1, z2 = 0, z3 = 1 and w1 = 0, w2 = i, w3 = 3i

Let the bilinear transformation be

  = 

⇒  = 

⇒  = 

⇒  = 

⇒ -2wz + 2w = - wz - w + 3iz + 3i

⇒ -wz + 3w = 3iz + 3i

⇒ w(-z + 3) = 3iz + 3i

⇒ w = 

Now, ad - bc = 3i × 3 - (-1)3i = 9i + 3i = 12i ≠ 0

Hence w =  is the required bilinear transformation 

(2) is correct