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Course 9

Conformal Mapping & Laplace Transform

Explore conformal mappings, Möbius transformations, and Laplace transform theory with applications to differential equations, signal processing, and engineering problems.

Advanced5-6 hours10 Practice Questions

Learning Objectives

Understand conformal mappings and angle preservation

Master Möbius (linear fractional) transformations

Learn the Laplace transform and its properties

Compute inverse Laplace transforms using residues

Apply Laplace transforms to solve differential equations

Understand the relationship between complex analysis and transforms

Work with common conformal mappings

Apply transforms to engineering problems

Conformal Mappings

Definition 9.1: Conformal Mapping

A mapping

w = f(z)

is conformal at

z_0

if it preserves angles between curves. More precisely, if two curves intersect at

z_0

with angle

\theta

, their images intersect at

f(z_0)

with the same angle

\theta

(both magnitude and orientation).

Theorem 9.1: Conformality of Analytic Functions

f

is analytic at

z_0

and

f'(z_0) \neq 0

, then

f

is conformal at

z_0

. Conversely, if a mapping is conformal and preserves orientation, then it is analytic with non-zero derivative.

Proof of Theorem 9.1:

If $f'(z_0) \neq 0$ , then locally $f(z) - f(z_0) \approx f'(z_0)(z-z_0)$ , which is a rotation and scaling. Rotations preserve angles, and scaling preserves angles, so angles are preserved.

The converse follows from the fact that angle preservation implies the Cauchy-Riemann equations are satisfied.

∎

Example 9.1

Problem: Show that $w = z^2$ is conformal except at $z=0$ .

Solution:

$f'(z) = 2z$ , which is non-zero for $z \neq 0$ . Therefore, $w = z^2$ is conformal everywhere except at $z=0$ , where $f'(0) = 0$ .

At $z=0$ , angles are doubled (e.g., the real and imaginary axes, which meet at $90^\circ$ , map to lines meeting at $180^\circ$ ).

Möbius Transformations

Definition 9.2: Möbius Transformation

A Möbius transformation (or linear fractional transformation) is a mapping of the form:

w = \frac{az+b}{cz+d}

where

a, b, c, d \in \mathbb{C}

and

ad - bc \neq 0

. If

c = 0

, this is a linear transformation; if

c \neq 0

, it can be written as

w = \frac{a}{c} + \frac{bc-ad}{c(cz+d)}

Theorem 9.2: Properties of Möbius Transformations

Möbius transformations:

Map the extended complex plane $\hat{\mathbb{C}}$ to itself (including infinity)
Preserve generalized circles (circles and lines)
Preserve cross-ratios
Are conformal everywhere (except at the pole)
Form a group under composition

Example 9.2

Problem: Find a Möbius transformation mapping $-1, 0, 1$ to $0, 1, \infty$ respectively.

Solution:

Using the cross-ratio: $(w, 0, 1, \infty) = (z, -1, 0, 1)$

Since $(w, 0, 1, \infty) = \frac{w-0}{w-\infty} \cdot \frac{1-\infty}{1-0} = -w$ (in limit), and $(z, -1, 0, 1) = \frac{z+1}{z-1} \cdot \frac{0-1}{0+1} = -\frac{z+1}{z-1}$

Setting equal: $w = \frac{z+1}{z-1}$

Laplace Transform

Definition 9.3: Laplace Transform

The Laplace transform of a function

f(t)

(defined for

t \geq 0

) is:

\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} f(t)e^{-st}dt

where

s

is a complex variable. The transform exists for

\text{Re}(s) > \sigma_0

where

\sigma_0

is the abscissa of convergence.

Theorem 9.3: Properties of Laplace Transform

Linearity: $\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$
Derivative: $\mathcal{L}\{f'(t)\} = sF(s) - f(0)$
Integral: $\mathcal{L}\left\{\int_0^t f(\tau)d\tau\right\} = \frac{F(s)}{s}$
Shift: $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$
Convolution: $\mathcal{L}\{f * g\} = F(s)G(s)$

Example 9.3

Problem: Find the Laplace transform of $f(t) = t^2$ .

Solution:

Using integration by parts twice:

$\mathcal{L}\{t^2\} = \int_0^{\infty} t^2 e^{-st}dt = \frac{2}{s^3}$ for $\text{Re}(s) > 0$

More generally: $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$

Theorem 9.4: Inverse Laplace Transform

The inverse Laplace transform is given by the Bromwich integral:

f(t) = \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} F(s)e^{st}ds

where

\gamma

is chosen so all singularities of

F(s)

lie to the left of the line

\text{Re}(s) = \gamma

. This can be evaluated using residue theory.

Example 9.4

Problem: Solve the differential equation $y'' + y = 0$ with $y(0) = 1$ , $y'(0) = 0$ using Laplace transforms.

Solution:

Taking Laplace transform: $s^2Y(s) - sy(0) - y'(0) + Y(s) = 0$

Substituting initial conditions: $s^2Y(s) - s + Y(s) = 0$ , so $Y(s) = \frac{s}{s^2+1}$

Taking inverse transform: $y(t) = \mathcal{L}^{-1}\left\{\frac{s}{s^2+1}\right\} = \cos t$

Applications

Applications of Conformal Mappings

Fluid Dynamics

Conformal mappings transform complex flow geometries into simpler ones, making potential flow problems easier to solve. The mapping preserves streamlines and equipotential lines.

Electrostatics

In two-dimensional electrostatics, conformal mappings solve Laplace's equation by transforming regions to standard geometries where solutions are known.

Cartography

Map projections like the Mercator projection are conformal mappings that preserve angles, essential for navigation where direction accuracy is crucial.

More Conformal Mapping Examples

Example 9.5

Problem: Find a conformal map from the upper half-plane to the unit disk.

Solution:

The Möbius transformation $w = \frac{z-i}{z+i}$ maps the upper half-plane $\text{Im } z > 0$ to the unit disk $|w| < 1$ .

It sends: $i \mapsto 0$ , $\infty \mapsto 1$ , $-i \mapsto \infty$

Example 9.6

Problem: Find a conformal map from the first quadrant to the upper half-plane.

Solution:

The map $w = z^2$ sends the first quadrant $\{z: \text{Re } z > 0, \text{Im } z > 0\}$ to the upper half-plane $\text{Im } w > 0$ .

This works because squaring doubles the argument, so angles between $0$ and $\pi/2$ map to angles between $0$ and $\pi$ .

More Laplace Transform Examples

Example 9.7

Problem: Find the Laplace transform of $f(t) = t\sin t$ .

Solution:

Since $\mathcal{L}\{\sin t\} = \frac{1}{s^2+1}$ , using the derivative property:

$\mathcal{L}\{t\sin t\} = -\frac{d}{ds}\left(\frac{1}{s^2+1}\right) = \frac{2s}{(s^2+1)^2}$

Example 9.8

Problem: Solve $y'' + 4y = \sin t$ with $y(0) = 0$ , $y'(0) = 1$ .

Solution:

Taking Laplace transform: $s^2Y(s) - 1 + 4Y(s) = \frac{1}{s^2+1}$

$Y(s) = \frac{1}{s^2+4} + \frac{1}{(s^2+4)(s^2+1)} = \frac{1}{s^2+4} + \frac{1/3}{s^2+1} - \frac{1/3}{s^2+4}$

$= \frac{2/3}{s^2+4} + \frac{1/3}{s^2+1}$

Taking inverse: $y(t) = \frac{1}{3}\sin(2t) + \frac{1}{3}\sin t$

Study Tips

1. Conformal Mappings

Möbius transformations are determined by three points. Use cross-ratios to find specific transformations. Remember: they map circles and lines to circles and lines.

2. Laplace Transforms

Memorize common transforms: $1/s$ for unit step, $1/(s-a)$ for $e^{at}$ , $\omega/(s^2+\omega^2)$ for $\sin(\omega t)$ . Use properties like linearity, shifting, and convolution.

3. Inverse Laplace Transforms

Use partial fractions, complete the square for quadratics, and the Bromwich integral for complex cases. For rational functions, factor the denominator and use Heaviside cover-up method.

More Möbius Transformation Examples

Example 9.9

Problem: Find a Möbius transformation that maps the real axis to the unit circle.

Solution:

The transformation $w = \frac{z-i}{z+i}$ maps the real axis $\text{Im } z = 0$ to the unit circle $|w| = 1$ .

For $z$ real, $|w| = \left|\frac{z-i}{z+i}\right| = \frac{|z-i|}{|z+i|} = 1$ since $|z-i| = |z+i|$ for real $z$ .

Example 9.10

Problem: Find a Möbius transformation mapping the disk $|z| < 1$ to the half-plane $\text{Re } w > 0$ .

Solution:

Use $w = \frac{1+z}{1-z}$ . This maps $z = -1$ to $w = 0$ , $z = 1$ to $w = \infty$ , and $z = i$ to $w = i$ .

The boundary $|z| = 1$ maps to the imaginary axis, and the interior maps to the right half-plane.

More Laplace Transform Examples

Example 9.11

Problem: Find the Laplace transform of $f(t) = t^ne^{at}$ .

Solution:

Using the shift property and the fact that $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$ :

$\mathcal{L}\{t^ne^{at}\} = \frac{n!}{(s-a)^{n+1}}$ for $\text{Re}(s) > \text{Re}(a)$ .

Example 9.12

Problem: Solve $y'' + 2y' + y = e^{-t}$ with $y(0) = 1$ , $y'(0) = 0$ .

Solution:

Taking Laplace transform: $s^2Y(s) - s + 2(sY(s) - 1) + Y(s) = \frac{1}{s+1}$

$(s^2 + 2s + 1)Y(s) = s + 2 + \frac{1}{s+1} = \frac{(s+2)(s+1) + 1}{s+1} = \frac{s^2 + 3s + 3}{s+1}$

$Y(s) = \frac{s^2 + 3s + 3}{(s+1)^3} = \frac{1}{s+1} + \frac{1}{(s+1)^2} + \frac{1}{(s+1)^3}$

Taking inverse: $y(t) = e^{-t} + te^{-t} + \frac{t^2}{2}e^{-t} = e^{-t}\left(1 + t + \frac{t^2}{2}\right)$

Example 9.13

Problem: Find the inverse Laplace transform of $F(s) = \frac{s}{(s^2+1)^2}$ .

Solution:

Using partial fractions or recognizing the derivative of $\frac{1}{s^2+1}$ :

$\mathcal{L}^{-1}\left\{\frac{s}{(s^2+1)^2}\right\} = \frac{t\sin t}{2}$

Example 9.14

Problem: Use convolution to find the inverse Laplace transform of $F(s) = \frac{1}{s^2(s^2+1)}$ .

Solution:

Write $F(s) = \frac{1}{s^2} \cdot \frac{1}{s^2+1}$

$\mathcal{L}^{-1}\{1/s^2\} = t$ , $\mathcal{L}^{-1}\{1/(s^2+1)\}= \sin t$

By convolution: $f(t) = (t * \sin t)(t) = \int_0^t (t-\tau)\sin\tau d\tau = t - \sin t$

Example 9.15

Problem: Find the Laplace transform of $f(t) = \delta(t - a)$ where $\delta$ is the Dirac delta function.

Solution:

$\mathcal{L}\{\delta(t-a)\} = \int_0^{\infty} \delta(t-a)e^{-st}dt = e^{-sa}$ for $a \geq 0$

For $a = 0$ , $\mathcal{L}\{\delta(t)\} = 1$ .

Example 9.16

Problem: Solve $y'' + 4y = \delta(t)$ with $y(0) = 0$ , $y'(0) = 0$ .

Solution:

Taking Laplace transform: $s^2Y(s) + 4Y(s) = 1$

$Y(s) = \frac{1}{s^2+4}$

Taking inverse: $y(t) = \frac{1}{2}\sin(2t)$

Example 9.17

Problem: Find a Möbius transformation that maps the upper half-plane to itself.

Solution:

A transformation $w = \frac{az+b}{cz+d}$ maps the real axis to itself if all coefficients are real (up to a common factor).

For it to map the upper half-plane to itself, we need $ad - bc > 0$ .

Example: $w = z$ (identity) or $w = 2z + 1$ .

Example 9.18

Problem: Find the Laplace transform of $f(t) = \int_0^t g(\tau)d\tau$ in terms of $G(s) = \mathcal{L}\{g(t)\}$ .

Solution:

Using the integral property: $\mathcal{L}\left\{\int_0^t g(\tau)d\tau\right\} = \frac{G(s)}{s}$

This is useful for solving integro-differential equations.

Example 9.19

Problem: Find a Möbius transformation that maps the points $-1, 0, 1$ to $\infty, 1, 0$ respectively.

Solution:

Using cross-ratio: $(w, \infty, 1, 0) = (z, -1, 0, 1)$

Computing cross-ratios: $w = \frac{1-z}{1+z}$

Example 9.20

Problem: Find the Laplace transform of $f(t) = t^n e^{at}$ where $n$ is a positive integer.

Solution:

Using the shift property and the fact that $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$ :

$\mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s-a)^{n+1}}$ for $\text{Re}(s) > \text{Re}(a)$ .

Example 9.21

Problem: Find the Laplace transform of $f(t) = u(t-a)$ where $u$ is the unit step function.

Solution:

$\mathcal{L}\{u(t-a)\} = \int_a^{\infty} e^{-st}dt = \frac{e^{-as}}{s}$ for $\text{Re}(s) > 0$ .

Example 9.22

Problem: Solve $y'' + 3y' + 2y = e^{-t}$ with $y(0) = 0$ , $y'(0) = 1$ .

Solution:

Taking Laplace transform: $s^2Y(s) - 1 + 3sY(s) + 2Y(s) = \frac{1}{s+1}$

$Y(s) = \frac{1}{(s+1)(s^2+3s+2)} + \frac{1}{s^2+3s+2}$

Using partial fractions and taking inverse transform: $y(t) = te^{-t}$ .

Example 9.23

Problem: Find the inverse Laplace transform of $F(s) = \frac{1}{(s^2+2s+5)^2}$ .

Solution:

Complete the square: $s^2+2s+5 = (s+1)^2+4$

Using the shift property and convolution:

$\mathcal{L}^{-1}\{F(s)\} = e^{-t} \cdot \mathcal{L}^{-1}\left\{\frac{1}{(s^2+4)^2}\right\}$

Using convolution with $\sin 2t$ gives: $f(t) = \frac{e^{-t}}{16}(\sin 2t - 2t\cos 2t)$

Example 9.24

Problem: Solve $x''(t) - 2x'(t) + 2x(t) = 2e^t\cos t$ with $x(0) = x'(0) = 0$ .

Solution:

Taking Laplace transform: $s^2X(s) - 2sX(s) + 2X(s) = \frac{2(s-1)}{(s-1)^2+1}$

$X(s) = \frac{2(s-1)}{[(s-1)^2+1]^2}$

Using shift property and inverse transform: $x(t) = te^t\sin t$ .

Remark 9.1

The Laplace transform is particularly powerful for solving linear differential equations with constant coefficients, as it converts differential equations into algebraic equations in the s-domain.

Example 9.25

Problem: Find a Möbius transformation that maps the unit disk $|z| < 1$ to the upper half-plane $\text{Im}(w) > 0$ .

Solution:

We want to map three boundary points. Choose: $z_1 = -1 \to w_1 = 0$ , $z_2 = i \to w_2 = 1$ , $z_3 = 1 \to w_3 = \infty$ .

Using cross-ratio: $(w, 0, 1, \infty) = (z, -1, i, 1)$

Since $(w, 0, 1, \infty) = w$ and $(z, -1, i, 1) = \frac{z+1}{z-1} \cdot \frac{i-1}{i+1}$

Simplifying: $w = i\frac{1+z}{1-z}$

This Möbius transformation maps the unit disk to the upper half-plane, with the boundary circle mapping to the real axis.

Example 9.26

Problem: Use the Laplace transform to solve the system: $x' = x + y$ , $y' = -x + y$ with $x(0) = 1$ , $y(0) = 0$ .

Solution:

Taking Laplace transforms: $sX(s) - 1 = X(s) + Y(s)$ and $sY(s) = -X(s) + Y(s)$

Rearranging: $(s-1)X(s) - Y(s) = 1$ and $X(s) + (s-1)Y(s) = 0$

Solving the system: $X(s) = \frac{s-1}{(s-1)^2+1}$ , $Y(s) = \frac{-1}{(s-1)^2+1}$

Using inverse transforms: $x(t) = e^t\cos t$ , $y(t) = -e^t\sin t$

Practice Problems

Conformal Mapping & Laplace Transform Practice

Questions

Correct

Accuracy

A conformal mapping preserves:

Easy

Not attempted

A Möbius transformation has the form:

Easy

Not attempted

The Laplace transform of

f(t)

is defined as:

Easy

Not attempted

A function is conformal at

z_0

if:

Medium

Not attempted

The inverse Laplace transform can be computed using:

Hard

Not attempted

Möbius transformations map:

Medium

Not attempted

The Laplace transform of

e^{at}

is:

Medium

Not attempted

A Möbius transformation is determined by:

Medium

Not attempted

The Laplace transform of

f'(t)

is:

Medium

Not attempted

Conformal mappings are used in:

Easy

Not attempted

Frequently Asked Questions

What makes a mapping conformal?

A mapping is conformal at a point if it preserves angles between curves. For an analytic function $f$, this occurs at points where $f'(z) \neq 0$. The mapping preserves both the magnitude and orientation of angles.

What are Möbius transformations good for?

Möbius transformations are the most general conformal mappings of the extended complex plane. They preserve generalized circles (circles and lines) and cross-ratios, making them useful for transforming regions and solving boundary value problems.

How do I find a Möbius transformation mapping three points to three points?

Use the cross-ratio. If you want $z_1, z_2, z_3$ to map to $w_1, w_2, w_3$, set up the equation $(w, w_1, w_2, w_3) = (z, z_1, z_2, z_3)$ and solve for $w$ in terms of $z$.

What is the relationship between Laplace transform and complex analysis?

The Laplace transform $F(s)$ is a function of the complex variable $s$. The inverse transform is computed via a contour integral (Bromwich integral) using residue theory, connecting it directly to complex analysis.

Why use Laplace transforms for differential equations?

Laplace transforms convert differential equations into algebraic equations, making them easier to solve. After solving in the transform domain, we invert to get the solution in the time domain.