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 $|P(z)|\to \mathrm{\infty }$ as $|z|\to \mathrm{\infty }$ (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 $|P(z)|\to \mathrm{\infty }$ as $|z|\to \mathrm{\infty }$, 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 $\left(\begin{array}{cc}3& -4\\ 2& 1\end{array}\right)$.   

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 :

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

$3z-4=2z^2+z$

$2z^2-2z+4=0$

$z^2-z+2=0$

⇒ $z=\frac{1\pm \sqrt{-7}}{2}=\frac{1\pm \sqrt{7}i}{2}$

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, 

$f(x_1)\ne f(x_2)\text{ whenever }{x}_1\ne x_2$

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, 

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

Explanation:

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

For $F_{\omega }(z_1)=F_{\omega }(z_2)$ , we need to show that  z₁ = z₂.

Let's assume  $F_{\omega }(z_1)=F_{\omega }(z_2)$  and then simplify:

${\displaystyle \frac{\omega -z_1}{1-\overline{\omega }{z}_1}}={\displaystyle \frac{\omega -z_2}{1-\overline{\omega }{z}_2}}$

$\Rightarrow (\omega -z_1)(1-\overline{\omega }{z}_2)=(\omega -z_2)(1-\overline{\omega }{z}_1)$

$\Rightarrow \omega -z_1-\overline{\omega }{z}_2+\overline{\omega }{z}_1{z}_2=\omega -z_2-\overline{\omega }{z}_1+\overline{\omega }{z}_1{z}_2$

$\Rightarrow -z_1+\overline{\omega }{z}_1{z}_2=-z_2+\overline{\omega }{z}_1{z}_2$

$\Rightarrow -z_1=-z_2$

$\Rightarrow z_1=z_2$

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. 

$\Rightarrow \frac{\omega -z}{1-\overline{\omega }z}=w$

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

$\Rightarrow \omega -z=w-w\overline{\omega }z$

$\Rightarrow -z+w\overline{\omega }z=w-\omega$

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

$\Rightarrow z=\frac{w-\omega }{1-w\overline{\omega }}$

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 $f(z)=\frac{az+b}{cz+d}$, 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 $f(z)=\frac{az+b}{cz+d}$ 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 $f(z)=\frac{az+b}{cz+d}$, 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) = $\frac{\mathrm{a}\mathrm{z}+\mathrm{b}}{\mathrm{c}\mathrm{z}+\mathrm{d}}$, 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)$(\bar{z}-\overline{a})$ = |a|2

⇒ z$\overline{z}$ - a$\overline{z}$ - z$\overline{a}$ - |a|2 = |a|2

⇒ $z\overline{z}=\overline{a}z+a\overline{z}$...(i)

Now, given w = f(z) = $\frac{1}{z}$ ⇒ z = $\frac{1}{w}$ ⇒ $\overline{z}=\frac{1}{\overline{w}}$

Then from (i) we get

$\frac{1}{w}.\frac{1}{\overline{w}}=\overline{a}\frac{1}{w}+a\frac{1}{\overline{w}}$

⇒ $\overline{a}\overline{w}+aw=1$ 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 $|P(z)|\to \mathrm{\infty }$ as $|z|\to \mathrm{\infty }$ (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 $|P(z)|\to \mathrm{\infty }$ as $|z|\to \mathrm{\infty }$, 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 $\left(\begin{array}{cc}3& -4\\ 2& 1\end{array}\right)$.   

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 :

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

$3z-4=2z^2+z$

$2z^2-2z+4=0$

$z^2-z+2=0$

⇒ $z=\frac{1\pm \sqrt{-7}}{2}=\frac{1\pm \sqrt{7}i}{2}$

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, 

$f(x_1)\ne f(x_2)\text{ whenever }{x}_1\ne x_2$

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, 

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

Explanation:

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

For $F_{\omega }(z_1)=F_{\omega }(z_2)$ , we need to show that  z₁ = z₂.

Let's assume  $F_{\omega }(z_1)=F_{\omega }(z_2)$  and then simplify:

${\displaystyle \frac{\omega -z_1}{1-\overline{\omega }{z}_1}}={\displaystyle \frac{\omega -z_2}{1-\overline{\omega }{z}_2}}$

$\Rightarrow (\omega -z_1)(1-\overline{\omega }{z}_2)=(\omega -z_2)(1-\overline{\omega }{z}_1)$

$\Rightarrow \omega -z_1-\overline{\omega }{z}_2+\overline{\omega }{z}_1{z}_2=\omega -z_2-\overline{\omega }{z}_1+\overline{\omega }{z}_1{z}_2$

$\Rightarrow -z_1+\overline{\omega }{z}_1{z}_2=-z_2+\overline{\omega }{z}_1{z}_2$

$\Rightarrow -z_1=-z_2$

$\Rightarrow z_1=z_2$

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. 

$\Rightarrow \frac{\omega -z}{1-\overline{\omega }z}=w$

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

$\Rightarrow \omega -z=w-w\overline{\omega }z$

$\Rightarrow -z+w\overline{\omega }z=w-\omega$

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

$\Rightarrow z=\frac{w-\omega }{1-w\overline{\omega }}$

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: $\frac{z}{2z+1}\in H$

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) = $\frac{z-\alpha }{cz+d}$

T(0) = α ⇒ - $\frac{\alpha }{d}$ = α ⇒ d = - 1

T(∞) = −α ⇒ $\frac{1}{c}$ = −α ⇒ c = $-\frac{1}{\alpha }$

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

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) ) = $\frac{(x+\frac{1}{\sqrt{2}}(1-i)}{x-1}$

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 = $\frac{z-\alpha }{-\frac{z}{\alpha }-1}$ 

⇒ $-\frac{wz}{\alpha }-w=z-\alpha$

⇒ $z(-\frac{w}{\alpha }-1)=w-\alpha$

⇒ z = $\frac{w-\alpha }{-\frac{w}{\alpha }-1}$

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

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