Conformal Mappings Questions and Answers

Q: Which of the following Mobius transformations maps the unit disc onto the right half plane ?

The map is \(f(z) = \dfrac{1 + z}{1 - z}\).For \(|z| \lt 1\):\[ \begin{aligned} \mathrm{Re}\,\frac{1 + z}{1 - z} &= \frac{\mathrm{Re}[(1 + z)(1 - \bar{z})]}{|1 - z|^{2}} \\ &= \frac{1 - |z|^{2}}{|1 - z|^{2}} \gt 0 \end{aligned} \]so the disc goes into the right half plane, and the inverse \(z = (w - 1)/(w + 1)\) shows every point with positive real part is reached. Checks: \(z = 0 \mapsto 1\), \(z = 1 \mapsto \infty\).Multiplying by \(i\) (option D) rotates the image to the upper half plane. \((z - 1)/(z + 1)\) sends the disc to the left half plane, and \((z - i)/(z + i)\) maps the upper half plane onto the disc, the reverse direction.

Q: Which of the following Möbius transformations maps the unit disk onto itself ?

The map onto the unit disk is f(z) = (2z − i)/(2 + zi).The automorphisms of the unit disk have the form \(e^{i\theta}\frac{z - a}{1 - \bar{a}z}\) with \(|a| \lt 1\). Dividing top and bottom by 2:\[ \begin{aligned} &\frac{2z - i}{2 + iz} \\ &= \frac{z - \frac{i}{2}}{1 + \frac{i}{2}z} \end{aligned} \]With \(a = \frac{i}{2}\), \(\bar{a} = -\frac{i}{2}\), so \(1 - \bar{a}z = 1 + \frac{i}{2}z\): exactly the required form.The others fail. \((2z - 1)/(2 + z)\) would need \(1 - z/2\) below. \((i - z)/(i + z)\) sends 0 to 1, on the boundary, so it cannot map the disk onto itself, and \((iz + 1)/(z + i)\) has a pole at \(-i\), on the unit circle.

Q: Suppose \(f : \mathbf{C} \to \mathbf{C}\) is an analytic one-one function. Then which of the following statements is false ?

The false statement is f'(z) = 0 for some z.An injective entire function cannot have an essential singularity at infinity (by the Casorati-Weierstrass theorem its values near infinity would overlap its values near 0), so it is a polynomial, and an injective polynomial has degree one: \(f(z) = az + b\) with \(a \neq 0\).Then \(f'(z) = a \neq 0\) everywhere, so 'f'(z) = 0 for some z' is false. The other statements are true: f is a polynomial of degree one, it is onto (solve \(az + b = w\)), and every non-constant analytic function is an open map.

Q: The image the circle\[ c = \{z \in \mathbf{C} : |z - 1| = 1\} \]under the map \(f(z) = \dfrac{1}{z}\) is :

The image is a line.The circle \(|z - 1| = 1\) passes through the origin, which \(f(z) = 1/z\) sends to infinity. A Möbius map sends circles and lines to circles and lines, and a circle through the pole becomes a line.To find it, write the circle as \(z\bar{z} = z + \bar{z}\). Dividing by \(z\bar{z}\) and putting \(w = 1/z\):\[ 1 = w + \bar{w} = 2\,\text{Re}(w) \]So the image is the vertical line \(\text{Re}(w) = \frac{1}{2}\). It would be a circle only if the original circle missed the origin; parabolas and ellipses never arise as images under \(1/z\) of a circle.

Q: Under the stereographic projection between S2 and \(\mathbf{C}_{\infty}\), the point 1 + i of C corresponds to the following point of S2.

The point is \(\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\).With projection from the north pole (0, 0, 1) of the unit sphere, the point \(z = x + iy\) corresponds to\[ \begin{aligned} &\Big(\frac{2x}{|z|^{2} + 1}, \frac{2y}{|z|^{2} + 1}, \\ &\quad \frac{|z|^{2} - 1}{|z|^{2} + 1}\Big) \end{aligned} \]For \(z = 1 + i\), \(|z|^{2} = 2\), so the point is \(\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\). It lies on the sphere: \(\frac{4 + 4 + 1}{9} = 1\).The point with \(-\frac{2}{3}\) is the image of \(1 - i\). The point with third coordinate 0 lies on the equator, which corresponds to \(|z| = 1\), and \(1 + i\) has modulus \(\sqrt{2}\). The last option has positive height but the wrong first two coordinates.

Q: Which of the following Möbius transformations maps the unit disk onto the right half plane {w / Re w > 0} ?

The map is \(f(z) = \frac{z + 1}{1 - z}\).For \(|z| \lt 1\),\[ \begin{aligned} \mathrm{Re}\, \frac{1 + z}{1 - z} &= \frac{1 - |z|^{2}}{|1 - z|^{2}} \\ &\gt 0 \end{aligned} \]and points with \(|z| = 1\) go to the imaginary axis (\(z = 1\) goes to \(\infty\)), so the disk maps onto the right half plane. For example \(0 \mapsto 1\).\(\frac{z - i}{z + i}\) is the Cayley map from the upper half plane to the disk, the reverse direction. \(\frac{z - 1}{z + 1}\) sends 0 to \(-1\), outside the right half plane. \(\frac{z + 1}{2z - 1}\) has its pole at \(z = \frac{1}{2}\), inside the disk, so it cannot map the disk onto a half plane.

Q: Which of the following functions is conformal on the unit disk ?

The conformal function is \(z e^{z}\).An analytic function is conformal wherever its derivative is non-zero. Here\[ \frac{d}{dz}\left(z e^{z}\right) = (1 + z)e^{z} \]which vanishes only at \(z = -1\), on the boundary circle and not in the open unit disk.Each other function has a critical point inside the disk. \(z^{5} + 1\) has derivative \(5z^{4}\), zero at 0. \(\cos z\) has derivative \(-\sin z\), zero at 0. \(z e^{z^{3} + 1}\) has derivative \((1 + 3z^{3})e^{z^{3} + 1}\), zero where \(z^{3} = -\frac{1}{3}\), at points of modulus \(3^{-1/3} \approx 0.69 \lt 1\).

11.

Which of the following Mobius transformations maps the unit disc onto the right half plane ?

A.\(f(z) = \dfrac{z - i}{z + i}\)
B.\(f(z) = \dfrac{z - 1}{z + 1}\)
C.\(f(z) = \dfrac{1 + z}{1 - z}\)
D.\(f(z) = i\left(\dfrac{1 + z}{1 - z}\right)\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The map is \(f(z) = \dfrac{1 + z}{1 - z}\).

For \(|z| \lt 1\):

\[ \begin{aligned} \mathrm{Re}\,\frac{1 + z}{1 - z} &= \frac{\mathrm{Re}[(1 + z)(1 - \bar{z})]}{|1 - z|^{2}} \\ &= \frac{1 - |z|^{2}}{|1 - z|^{2}} \gt 0 \end{aligned} \]

so the disc goes into the right half plane, and the inverse \(z = (w - 1)/(w + 1)\) shows every point with positive real part is reached. Checks: \(z = 0 \mapsto 1\), \(z = 1 \mapsto \infty\).

Multiplying by \(i\) (option D) rotates the image to the upper half plane. \((z - 1)/(z + 1)\) sends the disc to the left half plane, and \((z - i)/(z + i)\) maps the upper half plane onto the disc, the reverse direction.

Discuss in Forum

12.

Let \(D = \{z \in \mathbf{C} : |z| \lt 1\}\) and let \(f : D \to D\) be a bijective analytic function such that \(f\left(\frac{1}{2}\right) = 0\). Then :

A.\(f(z) = \dfrac{2z - 1}{2 - z}\)
B.\(f(z) = e^{i\theta}\left(\dfrac{2z - 1}{2 - z}\right)\) for some real number \(\theta\)
C.\(f(z) = \dfrac{2z - 1}{2 + z}\)
D.\(f(z) = \dfrac{z - \frac{1}{2}}{1 + z}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The general form is \(f(z) = e^{i\theta}\left(\frac{2z - 1}{2 - z}\right)\) for some real \(\theta\).

Every bijective analytic self-map of the unit disk is a disk automorphism \(f(z) = e^{i\theta}\dfrac{z - a}{1 - \bar{a} z}\), where \(a\) is the point sent to 0. Here \(a = 1/2\):

\[ \begin{aligned} &e^{i\theta}\frac{z - 1/2}{1 - z/2} \\ &= e^{i\theta}\frac{2z - 1}{2 - z} \end{aligned} \]

The rotation factor is not fixed by the single condition \(f(1/2) = 0\), so option A, which is the case \(\theta = 0\), is only one of these maps and cannot be asserted for every such \(f\). The map \((2z - 1)/(2 + z)\) does not send the unit circle to itself (at \(z = -1\) it gives \(-3\)), and \((z - 1/2)/(1 + z)\) has a pole at \(-1\) on the boundary.

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13.

Suppose \(f\) is a Möbius transformation on \(\mathbf{C}_{\infty}\) that has \(\infty\) as its only fixed point. Then :

A.\(f\) is a rotation
B.\(f\) is a dilation
C.\(f\) is a translation
D.\(f\) is an inversion

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

A Möbius transformation whose only fixed point is \(\infty\) is a translation.

Fixing \(\infty\) forces \(f(z) = az + b\). If \(a \neq 1\), the equation \(az + b = z\) has the finite solution \(z = b/(1 - a)\), a second fixed point. So \(\infty\) is the only fixed point exactly when \(a = 1\) and \(b \neq 0\), giving \(f(z) = z + b\), a translation.

A rotation \(e^{i\theta}z\) and a dilation \(kz\) both fix 0 as well as \(\infty\). An inversion \(1/z\) swaps 0 and \(\infty\), so it does not fix \(\infty\) at all.

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14.

Which of the following Möbius transformations maps the unit disk onto itself ?

A.\(f(z) = \dfrac{2z - i}{2 + zi}\)
B.\(f(z) = \dfrac{i - z}{i + z}\)
C.\(f(z) = \dfrac{2z - 1}{2 + z}\)
D.\(f(z) = \dfrac{iz + 1}{z + i}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The map onto the unit disk is f(z) = (2z − i)/(2 + zi).

The automorphisms of the unit disk have the form \(e^{i\theta}\frac{z - a}{1 - \bar{a}z}\) with \(|a| \lt 1\). Dividing top and bottom by 2:

\[ \begin{aligned} &\frac{2z - i}{2 + iz} \\ &= \frac{z - \frac{i}{2}}{1 + \frac{i}{2}z} \end{aligned} \]

With \(a = \frac{i}{2}\), \(\bar{a} = -\frac{i}{2}\), so \(1 - \bar{a}z = 1 + \frac{i}{2}z\): exactly the required form.

The others fail. \((2z - 1)/(2 + z)\) would need \(1 - z/2\) below. \((i - z)/(i + z)\) sends 0 to 1, on the boundary, so it cannot map the disk onto itself, and \((iz + 1)/(z + i)\) has a pole at \(-i\), on the unit circle.

Discuss in Forum

15.

Suppose \(f : \mathbf{C} \to \mathbf{C}\) is an analytic one-one function. Then which of the following statements is false ?

A.\(f\) is a polynomial of degree one
B.\(f\) is an open map
C.\(f\) is an onto map
D.\(f'(z) = 0\) for some \(z \in \mathbf{C}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The false statement is f'(z) = 0 for some z.

An injective entire function cannot have an essential singularity at infinity (by the Casorati-Weierstrass theorem its values near infinity would overlap its values near 0), so it is a polynomial, and an injective polynomial has degree one: \(f(z) = az + b\) with \(a \neq 0\).

Then \(f'(z) = a \neq 0\) everywhere, so 'f'(z) = 0 for some z' is false. The other statements are true: f is a polynomial of degree one, it is onto (solve \(az + b = w\)), and every non-constant analytic function is an open map.

Discuss in Forum

16.

The image the circle

\[ c = \{z \in \mathbf{C} : |z - 1| = 1\} \]

under the map \(f(z) = \dfrac{1}{z}\) is :

A.a circle
B.a line
C.a parabola
D.an ellipse

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The image is a line.

The circle \(|z - 1| = 1\) passes through the origin, which \(f(z) = 1/z\) sends to infinity. A Möbius map sends circles and lines to circles and lines, and a circle through the pole becomes a line.

To find it, write the circle as \(z\bar{z} = z + \bar{z}\). Dividing by \(z\bar{z}\) and putting \(w = 1/z\):

\[ 1 = w + \bar{w} = 2\,\text{Re}(w) \]

So the image is the vertical line \(\text{Re}(w) = \frac{1}{2}\). It would be a circle only if the original circle missed the origin; parabolas and ellipses never arise as images under \(1/z\) of a circle.

Discuss in Forum

17.

Under the stereographic projection between S² and \(\mathbf{C}_{\infty}\), the point 1 + i of C corresponds to the following point of S².

A.\(\left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right)\)
B.\(\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\)
C.\(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)\)
D.\(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The point is \(\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\).

With projection from the north pole (0, 0, 1) of the unit sphere, the point \(z = x + iy\) corresponds to

\[ \begin{aligned} &\Big(\frac{2x}{|z|^{2} + 1}, \frac{2y}{|z|^{2} + 1}, \\ &\quad \frac{|z|^{2} - 1}{|z|^{2} + 1}\Big) \end{aligned} \]

For \(z = 1 + i\), \(|z|^{2} = 2\), so the point is \(\left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\). It lies on the sphere: \(\frac{4 + 4 + 1}{9} = 1\).

The point with \(-\frac{2}{3}\) is the image of \(1 - i\). The point with third coordinate 0 lies on the equator, which corresponds to \(|z| = 1\), and \(1 + i\) has modulus \(\sqrt{2}\). The last option has positive height but the wrong first two coordinates.

Discuss in Forum

18.

Which of the following Möbius transformations maps the unit disk onto the right half plane {w / Re w > 0} ?

A.\(f(z) = \frac{z - i}{z + i}\)
B.\(f(z) = \frac{z - 1}{z + 1}\)
C.\(f(z) = \frac{z + 1}{2z - 1}\)
D.\(f(z) = \frac{z + 1}{1 - z}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The map is \(f(z) = \frac{z + 1}{1 - z}\).

For \(|z| \lt 1\),

\[ \begin{aligned} \mathrm{Re}\, \frac{1 + z}{1 - z} &= \frac{1 - |z|^{2}}{|1 - z|^{2}} \\ &\gt 0 \end{aligned} \]

and points with \(|z| = 1\) go to the imaginary axis (\(z = 1\) goes to \(\infty\)), so the disk maps onto the right half plane. For example \(0 \mapsto 1\).

\(\frac{z - i}{z + i}\) is the Cayley map from the upper half plane to the disk, the reverse direction. \(\frac{z - 1}{z + 1}\) sends 0 to \(-1\), outside the right half plane. \(\frac{z + 1}{2z - 1}\) has its pole at \(z = \frac{1}{2}\), inside the disk, so it cannot map the disk onto a half plane.

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19.

Which of the following functions is conformal on the unit disk ?

A.\(z^{5} + 1\)
B.\(\cos z\)
C.\(z e^{z^{3} + 1}\)
D.\(z e^{z}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The conformal function is \(z e^{z}\).

An analytic function is conformal wherever its derivative is non-zero. Here

\[ \frac{d}{dz}\left(z e^{z}\right) = (1 + z)e^{z} \]

which vanishes only at \(z = -1\), on the boundary circle and not in the open unit disk.

Each other function has a critical point inside the disk. \(z^{5} + 1\) has derivative \(5z^{4}\), zero at 0. \(\cos z\) has derivative \(-\sin z\), zero at 0. \(z e^{z^{3} + 1}\) has derivative \((1 + 3z^{3})e^{z^{3} + 1}\), zero where \(z^{3} = -\frac{1}{3}\), at points of modulus \(3^{-1/3} \approx 0.69 \lt 1\).

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