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

Concept:

Sublevel Set of a Polynomial:

• Let \( P(z) \) be a non-constant polynomial over \( \mathbb{C} \). A sublevel set is defined as \( S_R = \{ z \in \mathbb{C} : |P(z)| < R \} \).
• Open Set: The set \( S_R \) is open because \( |P(z)| \) is continuous and inverse image of open set \( (0, R) \) is open.
• Bounded Set: Since \( |P(z)| \rightarrow \infty \) as \( |z| \rightarrow \infty \) (polynomial growth), the set \( S_R \) is bounded for any finite \( R > 0 \).
• Boundary Points: The boundary of \( S_R \) consists of points \( z \) where \( |P(z)| = R \) due to the definition of open level set, hence the equality holds on boundary.
• Connected Components: Each connected component of \( S_R \) is an open set in which \( |P(z)| < R \). By the argument principle or Rouche’s theorem, each such component must contain at least one zero of \( P(z) \).

Calculation:

Given,

\( S_R = \{z \in \mathbb{C} : |P(z)| < R \} \)

⇒ \( |P(z)| \) is continuous on \( \mathbb{C} \)

⇒ Preimage of open set \( (0, R) \) is open

⇒ \( S_R \) is open

⇒ Option 1 is correct

⇒ Since \( |P(z)| \rightarrow \infty \) as \( |z| \rightarrow \infty \), there exists some radius beyond which \( |P(z)| > R \)

⇒ Hence \( S_R \) lies in a bounded region

⇒ Option 2 is correct

⇒ The boundary of \( S_R \) is defined by \( |P(z)| = R \)

⇒ Option 3 is correct

⇒ If a component of \( S_R \) does not contain a zero, then \( \frac{1}{P(z)} \) is holomorphic and bounded inside it

⇒ violates Liouville’s Theorem unless constant

⇒ contradiction

⇒ So every connected component of \( S_R \) must contain a zero

⇒ Option 4 is correct

∴ All options 1, 2, 3, and 4 are correct.

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. \(w_1 = \frac{3iz + 4}{z - i} \)

This is a Möbius transformation of the form  \(w_1 = \frac{az + b}{cz + d} \)  , 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  \(w = \frac{az + b}{cz + d} \)  can be found by setting w = z :

\(z = \frac{3iz + 4}{z - i} \)  

Multiplying both sides by z - i , we get:

z(z - i) = 3iz + 4

⇒ \(z^2 - iz = 3iz + 4 \) 

⇒ \(z^2 - 4iz - 4 = 0 \)  

This is a quadratic equation, and solving for z , we can determine that  \(w_1 \)  has two distinct fixed points

Given that  \(w_1 \)  has two distinct fixed points and the transformation involves rotation and scaling,  \(w_1 \)  is a loxodromic transformation

2. \(w_2 = \frac{z}{z - 7} \)   

This is another Möbius transformation

Finding Fixed Points:

We set  \(w_2 = z \) :

\(z = \frac{z}{z - 7} \)   

Multiplying both sides by z - 7 , we get:

z(z - 7) = z

⇒ \(z^2 - 7z = z\) 

⇒ \(z^2 - 8z = 0 \)  

⇒ z(z - 8) = 0

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

Since  \(w_2 \)  has two distinct real fixed points,  \(w_2 \)  is a hyperbolic transformation

Conclusion:

\(w_1 \)  is loxodromic, as it has two distinct fixed points and exhibits both scaling and rotation

\(w_2 \)  is hyperbolic, as it has two distinct real fixed points and maps lines to hyperbolas

Explanation -

Given \(T(z) = \frac{3z - 4}{2z + 1}\). 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, \(T(iy) = \frac{3iy - 4}{2iy + 1} \)

This results in a value \(\frac{3iy - 4}{2iy + 1} \), 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 \( \begin{pmatrix} 3 & -4 \\ 2 & 1 \end{pmatrix}\).   

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 :

\(\frac{3z - 4}{2z + 1} = z \)

\( 3z - 4 = 2z^2 + z\)

\(2z^2 - 2z + 4 = 0\)

\(z^2 - z + 2 = 0 \)

⇒ \(z = {1 \pm \sqrt{-7} \over 2} = {1 \pm \sqrt{7}i \over 2} \)

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 \(T(z) = \frac{z + i}{z - i}\). This is a specific Mobius transformation.

For Option (1) - Check Mapping of Real Axis  

For z on the real axis z = x, calculate \(T(x) = \frac{x + i}{x - i} \)

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 \(\frac{z + i}{z - i} = z\) .  
⇒ 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.  

\(T(T(z)) = T\left( \frac{z + i}{z - i} \right) = \frac{\frac{z + i}{z - i} + i}{\frac{z + i}{z - i} - i} \) 

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, 

\(f(x_1) \neq f(x_2) \text{ whenever } x_1 \neq x_2\)

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, 

\(\forall y \in B, \exists x \in A \text{ such that } f(x) = y \)

Explanation:

Given function \(F_ω(z) = \frac{ω - z}{1 - \bar{ω}z}\) where ω is a fixed complex number in the open unit disk  D = {z ∈ ℂ | |z| < 1}.

For \( F_ω(z_1) = F_ω(z_2)\) , we need to show that  z₁ = z₂.

Let's assume  \(F_ω(z_1) = F_ω(z_2)\)  and then simplify:

\(\dfrac{ω - z_1}{1 - \bar{ω}z_1} = \dfrac{ω - z_2}{1 - \bar{ω}z_2} \)

\(\Rightarrow (ω - z_1)(1 - \bar{ω}z_2) = (ω - z_2)(1 - \bar{ω}z_1)\)

\(\Rightarrow ω - z_1 - \bar{ω}z_2 + \bar{ω}z_1z_2 = ω - z_2 - \bar{ω}z_1 + \bar{ω}z_1z_2\)

\(\Rightarrow -z_1 + \bar{ω}z_1z_2 = -z_2 + \bar{ω}z_1z_2 \)

\( \Rightarrow -z_1 = -z_2 \)

\(\Rightarrow z_1 = z_2 \)

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. 

\(\Rightarrow \frac{\omega - z}{1 - \bar{\omega}z} = w \)

\(\Rightarrow \omega - z = w(1 - \bar{\omega}z) \)

\(\Rightarrow \omega - z = w - w\bar{\omega}z \)

\(\Rightarrow -z + w\bar{\omega}z = w - \omega \)

\(\Rightarrow z(1 - w\bar{\omega}) = w - \omega \)

\(\Rightarrow z = \frac{w - \omega}{1 - w\bar{\omega}} \)

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

Therefore,  F_ω  is onto.

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

Concept:

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

H- = {z ∈ ℂ : y < 0} represents lower half-plane

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

L- = {z ∈ ℂ : x < 0} represents left half-plane

(ii) We know that under the bilinear transformation \(f(z)=\frac{a z+b}{c z+d}\), 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) = \(\frac{2z+1}{5z+3}\) 

Comparing given f(z) with \(f(z)=\frac{a z+b}{c z+d}\) 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 < 0} represents lower half-plane

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

L- = {z ∈ ℂ : x < 0} represents left half-plane

(ii) We know that under the bilinear transformation \(f(z)=\frac{a z+b}{c z+d}\), ad – bc > 0

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

Explanation:

f(z) = \(\frac{z}{3z+1}\) = \(\frac{1z+0}{3z+1}\)

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 \(T(z) = \frac{z + i}{z - i}\). This is a specific Mobius transformation.

For Option (1) - Check Mapping of Real Axis  

For z on the real axis z = x, calculate \(T(x) = \frac{x + i}{x - i} \)

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 \(\frac{z + i}{z - i} = z\) .  
⇒ 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.  

\(T(T(z)) = T\left( \frac{z + i}{z - i} \right) = \frac{\frac{z + i}{z - i} + i}{\frac{z + i}{z - i} - i} \) 

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 \(\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \)   ..........  (i) 

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

Explanation:

T(z) = \(\rm \frac{az+b}{cz+d}\), 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

\(\frac{(w-10)(5-5i-5-5i)}{(w-5-5i)(5-5i-10)} = \frac{(z-0)(z_2-z_3)}{(z-z_3)(z_2-z_1)} \)

⇒ \(\frac{(w-10)(10i)}{(w-5-5i)(5i+5)} = \frac{z z_3(\frac{z_2}{z_3}-1)}{z_3(\frac{z}{z_3}-1)(-i-0)} \) ⇒ \(\frac{(w-10)(10i)}{(w-5-5i)(5i+5)} = \frac{z(-1)}{(-1)(-i)}\)

⇒ \(\frac{(w-10)(2i)}{(w-5-5i)(i+1)} = \frac{z}{(-i)} \times \frac{i}{i}\) ⇒ \(\frac{2w-20}{w+wi-10i}=z\)

Now, taking the set {w ∈ C : |w-5|<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) <0} under the map w = f(z) is {w ∈ C : |w-5|<5}.

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)\((\overline{z-a})\) = |a|2

⇒ (z - a)\((\overline{z-a})\) = |a|2

⇒ (z - a)\((\overline{z}-\bar a)\) = |a|2

⇒ z\(\bar z\) - a\(\bar z\) - z\(\bar a\) - |a|2 = |a|2

⇒ \(z\bar z=\bar a z+a\bar z\)...(i)

Now, given w = f(z) = \(\frac1z\) ⇒ z = \(\frac1w\) ⇒ \(\bar z=\frac1{\bar w}\)

Then from (i) we get

\(\frac1 w.\frac1 {\bar w}=\bar a \frac1 w+a\frac1 {\bar w}\)

⇒ \(\bar a\bar w+aw = 1\) which is a straight line not passing through origin

Hnece (2) is correct

Concept:

Sublevel Set of a Polynomial:

• Let \( P(z) \) be a non-constant polynomial over \( \mathbb{C} \). A sublevel set is defined as \( S_R = \{ z \in \mathbb{C} : |P(z)| < R \} \).
• Open Set: The set \( S_R \) is open because \( |P(z)| \) is continuous and inverse image of open set \( (0, R) \) is open.
• Bounded Set: Since \( |P(z)| \rightarrow \infty \) as \( |z| \rightarrow \infty \) (polynomial growth), the set \( S_R \) is bounded for any finite \( R > 0 \).
• Boundary Points: The boundary of \( S_R \) consists of points \( z \) where \( |P(z)| = R \) due to the definition of open level set, hence the equality holds on boundary.
• Connected Components: Each connected component of \( S_R \) is an open set in which \( |P(z)| < R \). By the argument principle or Rouche’s theorem, each such component must contain at least one zero of \( P(z) \).

Calculation:

Given,

\( S_R = \{z \in \mathbb{C} : |P(z)| < R \} \)

⇒ \( |P(z)| \) is continuous on \( \mathbb{C} \)

⇒ Preimage of open set \( (0, R) \) is open

⇒ \( S_R \) is open

⇒ Option 1 is correct

⇒ Since \( |P(z)| \rightarrow \infty \) as \( |z| \rightarrow \infty \), there exists some radius beyond which \( |P(z)| > R \)

⇒ Hence \( S_R \) lies in a bounded region

⇒ Option 2 is correct

⇒ The boundary of \( S_R \) is defined by \( |P(z)| = R \)

⇒ Option 3 is correct

⇒ If a component of \( S_R \) does not contain a zero, then \( \frac{1}{P(z)} \) is holomorphic and bounded inside it

⇒ violates Liouville’s Theorem unless constant

⇒ contradiction

⇒ So every connected component of \( S_R \) must contain a zero

⇒ Option 4 is correct

∴ All options 1, 2, 3, and 4 are correct.

Explanation -

Given \(T(z) = \frac{3z - 4}{2z + 1}\). 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, \(T(iy) = \frac{3iy - 4}{2iy + 1} \)

This results in a value \(\frac{3iy - 4}{2iy + 1} \), 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 \( \begin{pmatrix} 3 & -4 \\ 2 & 1 \end{pmatrix}\).   

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 :

\(\frac{3z - 4}{2z + 1} = z \)

\( 3z - 4 = 2z^2 + z\)

\(2z^2 - 2z + 4 = 0\)

\(z^2 - z + 2 = 0 \)

⇒ \(z = {1 \pm \sqrt{-7} \over 2} = {1 \pm \sqrt{7}i \over 2} \)

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, 

\(f(x_1) \neq f(x_2) \text{ whenever } x_1 \neq x_2\)

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, 

\(\forall y \in B, \exists x \in A \text{ such that } f(x) = y \)

Explanation:

Given function \(F_ω(z) = \frac{ω - z}{1 - \bar{ω}z}\) where ω is a fixed complex number in the open unit disk  D = {z ∈ ℂ | |z| < 1}.

For \( F_ω(z_1) = F_ω(z_2)\) , we need to show that  z₁ = z₂.

Let's assume  \(F_ω(z_1) = F_ω(z_2)\)  and then simplify:

\(\dfrac{ω - z_1}{1 - \bar{ω}z_1} = \dfrac{ω - z_2}{1 - \bar{ω}z_2} \)

\(\Rightarrow (ω - z_1)(1 - \bar{ω}z_2) = (ω - z_2)(1 - \bar{ω}z_1)\)

\(\Rightarrow ω - z_1 - \bar{ω}z_2 + \bar{ω}z_1z_2 = ω - z_2 - \bar{ω}z_1 + \bar{ω}z_1z_2\)

\(\Rightarrow -z_1 + \bar{ω}z_1z_2 = -z_2 + \bar{ω}z_1z_2 \)

\( \Rightarrow -z_1 = -z_2 \)

\(\Rightarrow z_1 = z_2 \)

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. 

\(\Rightarrow \frac{\omega - z}{1 - \bar{\omega}z} = w \)

\(\Rightarrow \omega - z = w(1 - \bar{\omega}z) \)

\(\Rightarrow \omega - z = w - w\bar{\omega}z \)

\(\Rightarrow -z + w\bar{\omega}z = w - \omega \)

\(\Rightarrow z(1 - w\bar{\omega}) = w - \omega \)

\(\Rightarrow z = \frac{w - \omega}{1 - w\bar{\omega}} \)

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 \(\frac{1}{z} \in H\)

For z = x + iy with y > 0, consider the transformation \(w = \frac{1}{z} = \frac{1}{x+iy}\) 

\(\Rightarrow w = \frac{x-iy}{x^2 + y^2}\)

The imaginary part of w is \(-\frac{y}{x^2 + y^2}\), which is negative for y > 0. Therefore, \(\frac{1}{z}\) does not map z in H to another point in H.

Option 2: \(\frac{1}{z^2} \in H \)

Using z = x + iy, we have \(w = \frac{1}{z^2} = \frac{1}{(x+iy)^2}\)

\(\Rightarrow w = \frac{1}{x^2 - y^2 + 2ixy} \)

\(\Rightarrow w = \frac{x^2 - y^2 - 2ixy}{(x^2 - y^2 + 2ixy)(x^2 - y^2 - 2ixy)}\)

\(\Rightarrow w = \frac{x^2 - y^2 - 2ixy}{(x^2 - y^2)^2 + (2xy)^2}\)

The imaginary part is\( \frac{-2xy}{(x^2 - y^2)^2 + (2xy)^2}\), which could be positive or negative depending on the sign of x.

So, \(\frac{1}{z^2}\) can be in H or not, depending on x.

Option 3: \(\frac{-z}{z+1} \in H\)

For z = x + it,

\(w = \frac{-z}{z+1} = \frac{-(x+iy)}{x+iy+1}\)

\(\Rightarrow w = \frac{-(x+iy)(x-iy+1)}{(x+1)^2 + y^2}\)

\(\Rightarrow \text{Im}(w) = \frac{-y((x+1)^2 + y^2)}{(x+1)^2 + y^2}\)

Since y > 0, the imaginary part of w is negative. Thus,\( \frac{-z}{z+1}\) does not map z in H to another point in H.

Option 4: \(\frac{z}{2 z+1} \in H\)

Let \(w = \frac{z}{2z+1} = \frac{x+iy}{2(x+iy)+1}\)

\(\Rightarrow w = \frac{(x+iy)(2x-2iy+1)}{(2x+1)^2 + (2y)^2}\)

\(\Rightarrow \text{Im}(w) = \frac{2xy - 2xy + y}{(2x+1)^2 + 4y^2} = \frac{y}{(2x+1)^2 + 4y^2}\)

Since y > 0, the imaginary part of w is positive, implying that \( \frac{z}{2z+1}\) 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) = \(z-\alpha\over cz+d\)

T(0) = α ⇒ - \(\alpha\over d\) = α ⇒ d = - 1

T(∞) = −α ⇒ \(\frac1c\) = −α ⇒ c = \(-\frac1α \)

Hence T(z) = \(z-\alpha\over -\frac z{\alpha}-1\) 

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) ) = \((x+\frac1{\sqrt2}(1-i)\over x-1\)

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 = \(z-\alpha\over -\frac z{\alpha}-1\) 

⇒ \(-\frac {wz}{\alpha}-w={z-\alpha}\)

⇒ \(z(-\frac {w}{\alpha}-1)={w-\alpha}\)

⇒ z = \(w-\alpha\over -\frac w{\alpha}-1\)

⇒ T^-1(z) = \(z-\alpha\over -\frac z{\alpha}-1\)

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