Conformal Mappings Questions and Answers

Q: For the value of \(\alpha\) obtained in Q.54, the function \(g(z)\) is not conformal at a point

\(g\) is not conformal at \(\frac{\pi(1 + 3i)}{6}\).With \(\alpha = 3\) from Q.54, \(g(z) = \sinh(z - \pi/6)\). An analytic map fails to be conformal exactly where its derivative vanishes:\[ \begin{aligned} g'(z) &= \cosh(z - \pi/6) = 0 \\ z - \pi/6 &= i\left(\tfrac{\pi}{2} + k\pi\right) \end{aligned} \]since \(\cosh w = \cos(iw)\) vanishes only at \(w = i(\pi/2 + k\pi)\). With \(k = 0\), \(z = \frac{\pi}{6} + \frac{i\pi}{2} = \frac{\pi(1 + 3i)}{6}\).\(\frac{\pi(3 + i)}{6}\) swaps the real and imaginary parts, and \(\frac{2\pi}{3}\) is real, where \(\cosh\) is never zero. \(\frac{i\pi}{2}\) is the zero of \(\cosh z\) itself, forgetting the shift by \(\pi/6\).

Q: The straight lines \(L_{1} : x = 0,\ L_{2} : y = 0\) and \(L_{3} : x + y = 1\) are mapped by the transformation \(w = z^{2}\) into the curves \(C_{1}, C_{2}\) and \(C_{3}\) respectively. The angle of intersection between the curves at \(w = 0\) is

The curves meet at \(w = 0\) at an angle of \(\pi\).Only \(L_{1}\) and \(L_{2}\) pass through \(z = 0\), the point that maps to \(w = 0\); the line \(L_{3}\), \(x + y = 1\), misses the origin, so \(C_{3}\) does not pass through \(w = 0\). On \(L_{1}\), \(z = iy\) gives \(w = -y^{2}\), the negative real axis. On \(L_{2}\), \(z = x\) gives \(w = x^{2}\), the positive real axis. The two image curves leave \(w = 0\) in opposite directions, so the angle between them is \(\pi\).The map \(w = z^{2}\) is not conformal at \(z = 0\), where \(dw/dz = 2z = 0\): there it doubles angles, which is why the right angle between the axes becomes a straight angle. Answering \(\pi/2\) assumes conformality at a critical point; 0 and \(\pi/4\) do not match either image.

Q: The image of the half plane \(\mathrm{Re}(z) + \mathrm{Im}(z) \gt 0\) under the map \(w = \frac{z-1}{z+i}\) is given by

The image is the disc \(|w| \lt 1\).\(|w| = \frac{|z - 1|}{|z + i|}\), so \(|w| \lt 1\) exactly when \(z\) is nearer to 1 than to \(-i\). The points equidistant from 1 and \(-i\) form their perpendicular bisector:\[ \begin{aligned} (x-1)^{2} + y^{2} &= x^{2} + (y+1)^{2} \\ -2x &= 2y \\ x + y &= 0 \end{aligned} \]So the line \(\mathrm{Re}(z) + \mathrm{Im}(z) = 0\) maps onto the unit circle, and the half plane \(x + y \gt 0\), which contains \(z = 1\) and is on the side of 1, maps into \(|w| \lt 1\). Check: \(z = 1\) gives \(w = 0\).\(|w| \gt 1\) is the image of the other half plane. \(\mathrm{Re}(w) \gt 0\) and \(\mathrm{Im}(w) \gt 0\) are half planes, but the boundary line \(x + y = 0\) does not pass through the pole \(z = -i\), so a Mobius map sends it to a circle, not a line, and the image must be a disc or its outside.

Q: Let \(\mathbb{H} = \{z \in \mathbb{C} : \text{Im}(z) \gt 0\}\) and \(\mathbb{D} = \{z \in \mathbb{C} : |z| \lt 1\}\). Then\(\sup\left\{|f'(0)| : f\) is an analytic function from \(\mathbb{D}\) to \(\mathbb{H}\) and \(f(0) = \frac{i}{2}\right\}\)is equal to

The supremum is 1.Compose with the Cayley-type map \(\varphi(w) = \frac{w - i/2}{w + i/2}\), which sends \(\mathbb{H}\) onto \(\mathbb{D}\) and \(i/2\) to 0. Then \(\varphi\circ f\) maps \(\mathbb{D}\) into \(\mathbb{D}\) and fixes 0, so by the Schwarz lemma \(|(\varphi\circ f)'(0)| \le 1\).\[ \begin{aligned} \varphi'(w) &= \frac{i}{(w + i/2)^{2}} \\ |\varphi'(i/2)| &= \frac{1}{|i|^{2}} = 1 \end{aligned} \]So \(|f'(0)| \le 1\), with equality for \(f = \varphi^{-1}\) (times a rotation), which is analytic from \(\mathbb{D}\) onto \(\mathbb{H}\) with \(f(0) = i/2\). In general \(|f'(0)| \le 2\,\text{Im} f(0)\). The values 1/4 and 1/2 are smaller than what \(\varphi^{-1}\) attains, and 100 exceeds the bound.

Q: Let T be the Möbius transformation that maps the points 0, \(\frac{1}{2}\) and 1 conformally onto the points -3, \(\infty\) and 2, respectively, in the extended complex plane. If T maps the circle centred at 1 with radius k onto a straight line given by the equation \(\alpha x + \beta y + \gamma = 0\), then the value of \(\dfrac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma}\) is equal to

The value is \(\dfrac{1}{7}\).Since \(\tfrac{1}{2} \mapsto \infty\), write \(T(z) = (az + b)/(2z - 1)\). Then \(T(0) = -b = -3\) gives b = 3, and \(T(1) = a + 3 = 2\) gives a = -1, so \(T(z) = (3 - z)/(2z - 1)\).A circle maps to a line exactly when it passes through the point sent to \(\infty\), here 1/2; the circle centred at 1 must therefore have radius \(k = \tfrac{1}{2}\). It also passes through 3/2 and \(1 + \tfrac{i}{2}\), whose images are\[ \begin{aligned} T(\tfrac{3}{2}) &= \tfrac{3}{4} \\ T(1 + \tfrac{i}{2}) &= \tfrac{3}{4} - \tfrac{5}{4}i \end{aligned} \]so the image is the line \(x = 3/4\): \(\alpha = 1\), \(\beta = 0\), \(\gamma = -3/4\). Then\[ \begin{aligned} \frac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma} &= \frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} \\ &= \frac{1}{7} \end{aligned} \]The ratio does not depend on how the line's equation is scaled.

Q: Let \(U = \{z \in \mathbb{C} : \text{Im}(z) \gt 0\}\) and \(D = \{z \in \mathbb{C} : |z| \lt 1\}\), where Im(z) denotes the imaginary part of z. Let S be the set of all bijective analytic functions \(f: U \to D\) such that \(f(i) = 0\). Then, the value of \(\sup_{f \in S} |f(4i)|\) is

The supremum is \(\dfrac{3}{5}\).Every bijective analytic map of the upper half-plane onto the unit disc sending i to 0 has the form\[ f(z) = e^{i\theta}\,\frac{z - i}{z + i} \]for some real \(\theta\),because any two such maps differ by an automorphism of the disc fixing 0, which is a rotation. So for every f in S,\[ |f(4i)| = \left|\frac{4i - i}{4i + i}\right| = \frac{3}{5} \]The value is the same for all of them, so the supremum is 3/5. Zero would need \(f(4i) = f(i)\), impossible for an injective map, and 1/4 and 1/2 do not come from the Cayley form.

Q: Let \(D = \{z \in \mathbb{C} \mid |z| \lt 1\}\) denote the unit disc in the complex plane \(\mathbb{C}\). Let \(f : D \to D\) be an analytic function which satisfies \(f(0) = 0\). Then which one of the following is a possible value of \(f'(0)\)?

The possible value is \(f'(0) = \frac{i}{10}\).By the Schwarz lemma, an analytic map of the unit disc into itself with \(f(0) = 0\) satisfies \(|f'(0)| \le 1\). So only values in the closed unit disc are possible.\(|\frac{i}{10}| = 0.1\), and the map \(f(z) = \frac{i}{10} z\) attains it. The other values have modulus \(\frac{5}{2}\), \(\frac{3}{2}\) and \(\frac{5}{2}\), all greater than 1, so no such f can have them.

Q: The bilinear transformation :\(f(z) = \dfrac{3z + 4}{5z + 6}\) from \(\mathbf{C} - \left\{-\dfrac{6}{5}\right\}\) to \(\mathbf{C}\)assumes all complex values, except :

The omitted value is 3/5.Solve \(w = \dfrac{3z + 4}{5z + 6}\) for \(z\): \(5wz + 6w = 3z + 4\), so\[ z = \frac{4 - 6w}{5w - 3} \]This gives a valid \(z\) for every \(w\) except \(w = 3/5\). That value is the limit of \(f(z)\) as \(z \to \infty\), the ratio of the leading coefficients, and since \(\infty\) is not in the domain it is never taken.\(-6/5\) is the excluded point of the domain, not a missing value of the range. 5/3 and \(-4/3\) are both taken, at \(z = -9/4\) and \(z = -36/29\) respectively, as the formula for \(z\) shows.

Q: The number of Mobius transformations which map the real line onto the unit circle is :

There are infinitely many such Mobius transformations.One example is the Cayley-type map \(T(z) = \dfrac{z - i}{z + i}\), which sends every real \(z\) to a point with \(|T(z)| = 1\) (real \(z\) is equidistant from \(i\) and \(-i\)) and the real line onto the unit circle.Composing T with any Mobius map of the real line to itself (for example \(z \mapsto az + b\) with real \(a \ne 0\) and \(b\)), or with any rotation \(w \mapsto e^{i\theta}w\) of the circle, gives another such map. These form a three-parameter real family, so there are infinitely many, not 0, 1 or 2.

Q: Which of the following sets is mapped onto the right half plane under the map \(f(z) = \sin z\) ?

The set is \(\{z : 0 \lt \mathrm{Re}(z) \le \pi/2\}\).With \(z = x + iy\),\[ \sin z = \sin x \cosh y + i \cos x \sinh y. \]For \(0 \lt x \lt \pi/2\) the real part \(\sin x \cosh y\) is positive, and the half-strip maps one-to-one onto the right half plane except the ray \([1, \infty)\). That ray is the image of the edge \(x = \pi/2\), where \(\sin z = \cosh y \ge 1\). Including the edge \(\mathrm{Re}(z) = \pi/2\) therefore fills the right half plane exactly.The full strip \(-\pi/2 \le \mathrm{Re}(z) \le \pi/2\) also covers the left half plane, and the horizontal strips in Im(z) contain points with \(\sin x \lt 0\).

Direction for [ Question No: 1 To 1 ] :

Consider the functions \(f(z) = \frac{z^{2} + \alpha z}{(z + 1)^{2}}\) and \(g(z) = \sinh\left(z - \frac{\pi}{2\alpha}\right), \alpha \neq 0\).

For the value of \(\alpha\) obtained in Q.54, the function \(g(z)\) is not conformal at a point

A.\(\frac{\pi(1 + 3i)}{6}\)
B.\(\frac{\pi(3 + i)}{6}\)
C.\(\frac{2\pi}{3}\)
D.\(\frac{i\pi}{2}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

\(g\) is not conformal at \(\frac{\pi(1 + 3i)}{6}\).

With \(\alpha = 3\) from Q.54, \(g(z) = \sinh(z - \pi/6)\). An analytic map fails to be conformal exactly where its derivative vanishes:

\[ \begin{aligned} g'(z) &= \cosh(z - \pi/6) = 0 \\ z - \pi/6 &= i\left(\tfrac{\pi}{2} + k\pi\right) \end{aligned} \]

since \(\cosh w = \cos(iw)\) vanishes only at \(w = i(\pi/2 + k\pi)\). With \(k = 0\), \(z = \frac{\pi}{6} + \frac{i\pi}{2} = \frac{\pi(1 + 3i)}{6}\).

\(\frac{\pi(3 + i)}{6}\) swaps the real and imaginary parts, and \(\frac{2\pi}{3}\) is real, where \(\cosh\) is never zero. \(\frac{i\pi}{2}\) is the zero of \(\cosh z\) itself, forgetting the shift by \(\pi/6\).

Discuss in Forum

The straight lines \(L_{1} : x = 0,\ L_{2} : y = 0\) and \(L_{3} : x + y = 1\) are mapped by the transformation \(w = z^{2}\) into the curves \(C_{1}, C_{2}\) and \(C_{3}\) respectively. The angle of intersection between the curves at \(w = 0\) is

A.0
B.\(\pi/4\)
C.\(\pi/2\)
D.\(\pi\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The curves meet at \(w = 0\) at an angle of \(\pi\).

Only \(L_{1}\) and \(L_{2}\) pass through \(z = 0\), the point that maps to \(w = 0\); the line \(L_{3}\), \(x + y = 1\), misses the origin, so \(C_{3}\) does not pass through \(w = 0\). On \(L_{1}\), \(z = iy\) gives \(w = -y^{2}\), the negative real axis. On \(L_{2}\), \(z = x\) gives \(w = x^{2}\), the positive real axis. The two image curves leave \(w = 0\) in opposite directions, so the angle between them is \(\pi\).

The map \(w = z^{2}\) is not conformal at \(z = 0\), where \(dw/dz = 2z = 0\): there it doubles angles, which is why the right angle between the axes becomes a straight angle. Answering \(\pi/2\) assumes conformality at a critical point; 0 and \(\pi/4\) do not match either image.

Discuss in Forum

The image of the half plane \(\mathrm{Re}(z) + \mathrm{Im}(z) \gt 0\) under the map \(w = \frac{z-1}{z+i}\) is given by

A.\(\mathrm{Re}(w) \gt 0\)
B.\(\mathrm{Im}(w) \gt 0\)
C.\(|w| \gt 1\)
D.\(|w| \lt 1\)

Exam Year(s)

GATE Mathematics - 2018,

Answer: D

The image is the disc \(|w| \lt 1\).

\(|w| = \frac{|z - 1|}{|z + i|}\), so \(|w| \lt 1\) exactly when \(z\) is nearer to 1 than to \(-i\). The points equidistant from 1 and \(-i\) form their perpendicular bisector:

\[ \begin{aligned} (x-1)^{2} + y^{2} &= x^{2} + (y+1)^{2} \\ -2x &= 2y \\ x + y &= 0 \end{aligned} \]

So the line \(\mathrm{Re}(z) + \mathrm{Im}(z) = 0\) maps onto the unit circle, and the half plane \(x + y \gt 0\), which contains \(z = 1\) and is on the side of 1, maps into \(|w| \lt 1\). Check: \(z = 1\) gives \(w = 0\).

\(|w| \gt 1\) is the image of the other half plane. \(\mathrm{Re}(w) \gt 0\) and \(\mathrm{Im}(w) \gt 0\) are half planes, but the boundary line \(x + y = 0\) does not pass through the pole \(z = -i\), so a Mobius map sends it to a circle, not a line, and the image must be a disc or its outside.

Discuss in Forum

Let \(\mathbb{H} = \{z \in \mathbb{C} : \text{Im}(z) \gt 0\}\) and \(\mathbb{D} = \{z \in \mathbb{C} : |z| \lt 1\}\). Then

\(\sup\left\{|f'(0)| : f\) is an analytic function from \(\mathbb{D}\) to \(\mathbb{H}\) and \(f(0) = \frac{i}{2}\right\}\)

is equal to

A.\(\frac{1}{4}\)
B.\(\frac{1}{2}\)
C.1
D.100

Exam Year(s)

GATE Mathematics - 2024,

Answer: C

The supremum is 1.

Compose with the Cayley-type map \(\varphi(w) = \frac{w - i/2}{w + i/2}\), which sends \(\mathbb{H}\) onto \(\mathbb{D}\) and \(i/2\) to 0. Then \(\varphi\circ f\) maps \(\mathbb{D}\) into \(\mathbb{D}\) and fixes 0, so by the Schwarz lemma \(|(\varphi\circ f)'(0)| \le 1\).

\[ \begin{aligned} \varphi'(w) &= \frac{i}{(w + i/2)^{2}} \\ |\varphi'(i/2)| &= \frac{1}{|i|^{2}} = 1 \end{aligned} \]

So \(|f'(0)| \le 1\), with equality for \(f = \varphi^{-1}\) (times a rotation), which is analytic from \(\mathbb{D}\) onto \(\mathbb{H}\) with \(f(0) = i/2\). In general \(|f'(0)| \le 2\,\text{Im} f(0)\). The values 1/4 and 1/2 are smaller than what \(\varphi^{-1}\) attains, and 100 exceeds the bound.

Discuss in Forum

Let T be the Möbius transformation that maps the points 0, \(\frac{1}{2}\) and 1 conformally onto the points -3, \(\infty\) and 2, respectively, in the extended complex plane. If T maps the circle centred at 1 with radius k onto a straight line given by the equation \(\alpha x + \beta y + \gamma = 0\), then the value of \(\dfrac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma}\) is equal to

A.\(\dfrac{1}{7}\)
B.\(\dfrac{2}{7}\)
C.\(\dfrac{1}{3}\)
D.\(\dfrac{2}{3}\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: A

The value is \(\dfrac{1}{7}\).

Since \(\tfrac{1}{2} \mapsto \infty\), write \(T(z) = (az + b)/(2z - 1)\). Then \(T(0) = -b = -3\) gives b = 3, and \(T(1) = a + 3 = 2\) gives a = -1, so \(T(z) = (3 - z)/(2z - 1)\).

A circle maps to a line exactly when it passes through the point sent to \(\infty\), here 1/2; the circle centred at 1 must therefore have radius \(k = \tfrac{1}{2}\). It also passes through 3/2 and \(1 + \tfrac{i}{2}\), whose images are

\[ \begin{aligned} T(\tfrac{3}{2}) &= \tfrac{3}{4} \\ T(1 + \tfrac{i}{2}) &= \tfrac{3}{4} - \tfrac{5}{4}i \end{aligned} \]

so the image is the line \(x = 3/4\): \(\alpha = 1\), \(\beta = 0\), \(\gamma = -3/4\). Then

\[ \begin{aligned} \frac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma} &= \frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} \\ &= \frac{1}{7} \end{aligned} \]

The ratio does not depend on how the line's equation is scaled.

Discuss in Forum

Let \(U = \{z \in \mathbb{C} : \text{Im}(z) \gt 0\}\) and \(D = \{z \in \mathbb{C} : |z| \lt 1\}\), where Im(z) denotes the imaginary part of z. Let S be the set of all bijective analytic functions \(f: U \to D\) such that \(f(i) = 0\). Then, the value of \(\sup_{f \in S} |f(4i)|\) is

A.0
B.\(\dfrac{1}{4}\)
C.\(\dfrac{1}{2}\)
D.\(\dfrac{3}{5}\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: D

The supremum is \(\dfrac{3}{5}\).

Every bijective analytic map of the upper half-plane onto the unit disc sending i to 0 has the form

\[ f(z) = e^{i\theta}\,\frac{z - i}{z + i} \]

for some real \(\theta\),

because any two such maps differ by an automorphism of the disc fixing 0, which is a rotation. So for every f in S,

\[ |f(4i)| = \left|\frac{4i - i}{4i + i}\right| = \frac{3}{5} \]

The value is the same for all of them, so the supremum is 3/5. Zero would need \(f(4i) = f(i)\), impossible for an injective map, and 1/4 and 1/2 do not come from the Cayley form.

Discuss in Forum

Let \(D = \{z \in \mathbb{C} \mid |z| \lt 1\}\) denote the unit disc in the complex plane \(\mathbb{C}\). Let \(f : D \to D\) be an analytic function which satisfies \(f(0) = 0\). Then which one of the following is a possible value of \(f'(0)\)?

A.\(\frac{5}{2} i\)
B.\(\frac{i}{10}\)
C.\(\frac{3}{2}\)
D.\(-\frac{5}{2} i\)

Exam Year(s)

GATE Mathematics - 2026,

Answer: B

The possible value is \(f'(0) = \frac{i}{10}\).

By the Schwarz lemma, an analytic map of the unit disc into itself with \(f(0) = 0\) satisfies \(|f'(0)| \le 1\). So only values in the closed unit disc are possible.

\(|\frac{i}{10}| = 0.1\), and the map \(f(z) = \frac{i}{10} z\) attains it. The other values have modulus \(\frac{5}{2}\), \(\frac{3}{2}\) and \(\frac{5}{2}\), all greater than 1, so no such f can have them.

Discuss in Forum

The bilinear transformation :

\(f(z) = \dfrac{3z + 4}{5z + 6}\) from \(\mathbf{C} - \left\{-\dfrac{6}{5}\right\}\) to \(\mathbf{C}\)

assumes all complex values, except :

A.\(\dfrac{5}{3}\)
B.\(-\dfrac{4}{3}\)
C.\(-\dfrac{6}{5}\)
D.\(\dfrac{3}{5}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The omitted value is 3/5.

Solve \(w = \dfrac{3z + 4}{5z + 6}\) for \(z\): \(5wz + 6w = 3z + 4\), so

\[ z = \frac{4 - 6w}{5w - 3} \]

This gives a valid \(z\) for every \(w\) except \(w = 3/5\). That value is the limit of \(f(z)\) as \(z \to \infty\), the ratio of the leading coefficients, and since \(\infty\) is not in the domain it is never taken.

\(-6/5\) is the excluded point of the domain, not a missing value of the range. 5/3 and \(-4/3\) are both taken, at \(z = -9/4\) and \(z = -36/29\) respectively, as the formula for \(z\) shows.

Discuss in Forum

The number of Mobius transformations which map the real line onto the unit circle is :

A.0
B.1
C.2
D.Infinitely many

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

There are infinitely many such Mobius transformations.

One example is the Cayley-type map \(T(z) = \dfrac{z - i}{z + i}\), which sends every real \(z\) to a point with \(|T(z)| = 1\) (real \(z\) is equidistant from \(i\) and \(-i\)) and the real line onto the unit circle.

Composing T with any Mobius map of the real line to itself (for example \(z \mapsto az + b\) with real \(a \ne 0\) and \(b\)), or with any rotation \(w \mapsto e^{i\theta}w\) of the circle, gives another such map. These form a three-parameter real family, so there are infinitely many, not 0, 1 or 2.

Discuss in Forum

10.

Which of the following sets is mapped onto the right half plane under the map \(f(z) = \sin z\) ?

A.\(\left\{z : 0 \lt \mathrm{Re}(z) \le \dfrac{\pi}{2}\right\}\)
B.\(\left\{z : \dfrac{-\pi}{2} \le \mathrm{Im}(z) \le \dfrac{\pi}{2}\right\}\)
C.\(\left\{z : 0 \lt \mathrm{Im}(z) \le \dfrac{\pi}{2}\right\}\)
D.\(\left\{z : \dfrac{-\pi}{2} \le \mathrm{Re}(z) \le \dfrac{\pi}{2}\right\}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The set is \(\{z : 0 \lt \mathrm{Re}(z) \le \pi/2\}\).

With \(z = x + iy\),

\[ \sin z = \sin x \cosh y + i \cos x \sinh y. \]

For \(0 \lt x \lt \pi/2\) the real part \(\sin x \cosh y\) is positive, and the half-strip maps one-to-one onto the right half plane except the ray \([1, \infty)\). That ray is the image of the edge \(x = \pi/2\), where \(\sin z = \cosh y \ge 1\). Including the edge \(\mathrm{Re}(z) = \pi/2\) therefore fills the right half plane exactly.

The full strip \(-\pi/2 \le \mathrm{Re}(z) \le \pi/2\) also covers the left half plane, and the horizontal strips in Im(z) contain points with \(\sin x \lt 0\).

Conformal Mappings Questions and Answers

Direction for [ Question No: 1 To 1 ] :

Topics

Other Subjects

Tags: Conformal Mappings Questions and Answers

About Us

Conformal Mappings Questions and Answers

Direction for [ Question No: 1 To 1 ] :

Topics

Other Subjects

Tags: Conformal Mappings Questions and Answers

About Us

Stay informed about all important Dates & Notifications