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Composing powers, exponentials, logarithms and Möbius maps carries an awkward region onto a disc or a half plane; since a harmonic function composed with an analytic one is harmonic, a boundary value problem travels with it — which is how complex analysis reaches heat, electrostatics and ideal flow.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to name what each standard map does to a region, choose the power that opens a given wedge into a half plane, compose two or three maps to carry a region onto a disc, check the direction with an interior point, describe the four moves of solving a boundary value problem by conformal mapping, and say what the Riemann mapping theorem does and does not give you.
Conformal maps and Möbius transformations, harmonic functions from unit 2, and the fact that a critical point opens a corner. This lesson puts those together and arrives at the reason the subject is used outside mathematics at all.
A Dirichlet problem asks for a function harmonic inside a region and taking prescribed values on its boundary. A sector or wedge is the set of points whose argument lies between two bounds. A strip is the set of points whose imaginary part lies between two bounds. A map is a conformal equivalence between two regions when it is analytic, one-to-one and onto, with an analytic inverse.
The fact that makes it work. If $u$ is harmonic and $f$ is analytic, then $u \circ f$ is harmonic. So a solution of Laplace's equation on one region becomes a solution on another the moment a conformal map between them exists — and the boundary goes to the boundary, so the boundary values come along too.
The method, in four moves.
The standard maps, by what they do. Nobody remembers these by their formulas; they are remembered by which region becomes which.
| Map | Region change |
|---|---|
| $z + \beta$, $\alpha z$ | move, turn and resize |
| $z^{k}$ | opens a wedge at the origin by a factor of $k$ |
| $e^{z}$ | strip of height $\pi$ to the upper half plane |
| $\log z$ | upper half plane to a strip; sector to a strip |
| $\dfrac{z - i}{z + i}$ | upper half plane to the unit disc |
| $\dfrac12\left(z + \dfrac1z\right)$ | outside of the unit circle to the plane with a slit |
A region is handled by composing two or three of them. A quarter plane becomes a half plane by squaring and then a disc by the Cayley transform; a strip becomes a half plane by the exponential and then a disc the same way.
And the guarantee behind it all. The Riemann mapping theorem: every simply connected open set other than the whole plane can be mapped conformally onto the open unit disc. So the method never fails for want of a map — though it very often fails for want of a map anybody can write down, which is the honest limitation and the reason the table above matters.
Another way: picture
Think of the region as a sheet of rubber with a temperature painted on its edge. A conformal map stretches the sheet — unevenly, but without shearing and without tearing — into a round disc, carrying the painted edge with it. The steady temperature inside the disc is something a formula knows; stretch the sheet back and the temperature comes back with it, correct at every point.
Another way: steps
This is the point in the course where complex analysis stops being about complex numbers.
Laplace's equation in two dimensions describes the steady state of anything that diffuses or flows without sources: the temperature of a plate whose edges are held fixed, the electrostatic potential in a region between conductors, the velocity potential of an incompressible irrotational fluid, the displacement of a stretched membrane, the pressure in a porous medium. In each case the geometry is given and awkward, and the boundary values are known.
The conformal method turns all of them into one problem on a disc. And the harmonic conjugate from unit 2 comes back with a physical meaning: if $u$ is the potential, the level curves of its conjugate $v$ are the field lines — the paths heat flows along, the lines of force, the streamlines of the flow. They cross the equipotentials at right angles, which is the Cauchy-Riemann equations seen in a laboratory.
Two classical applications are worth naming. The Joukowski map $\tfrac12(z + 1/z)$ turns a circle into an aerofoil-shaped curve, and the flow past the circle — which anyone can write down — becomes the flow past the aerofoil, which is how early aerodynamics computed lift. And the Schwarz-Christoffel formula writes down the map from a half plane onto the interior of any polygon, which is how field problems in rectangular geometry are solved exactly rather than numerically.
Neither is in this course. Both are the next thing, and both are this lesson's four moves with a harder map in the first one.
The rubber-sheet picture is the right one and it has to be taken seriously: the stretching is uneven. Distances change, areas change, straight boundaries become curved, and a square goes to something with four right-angled corners and four curved sides. What survives is the angle at every crossing, and — because of that — Laplace's equation. Expecting shapes to survive is the commonest source of wrong pictures here.
The second error is to treat the Riemann mapping theorem as a method. It is an existence theorem: it promises a conformal map onto the disc for every simply connected region other than the plane, and its proof is not constructive. For an arbitrary region nobody can write the map down. Every map in this unit comes from the short table of standard ones, and the skill is composing them.
Two smaller traps. Check the direction. A map that sends a boundary where you wanted often sends the region to the other side of it — the outside of the disc rather than the inside — and the algebra gives no warning. One interior point settles it. And a corner has to be at the origin before a power will open it, because a power multiplies arguments measured from the origin; a wedge elsewhere is translated first.
The first quadrant is a wedge of $90^\circ$ at the origin. Squaring doubles that to $180^\circ$: the upper half plane.
Open the corner with a power.
Then the Cayley transform $\dfrac{w - i}{w + i}$ sends the upper half plane to the unit disc.
Finish with the standard half-plane map.
So the whole map is $z \mapsto \dfrac{z^{2} - i}{z^{2} + i}$. Checking one interior point: $z = 1 + i$ gives $z^{2} = 2i$, which maps to $\dfrac{i}{3i} = \dfrac13$ — inside the disc, as intended.
One substitution confirms the direction.
The strip $0 < \operatorname{Im} z < \pi$ has horizontal lines at heights between $0$ and $\pi$.
Describe the region by what the exponential will see.
The exponential sends the line at height $c$ to the ray of argument $c$, so heights from $0$ to $\pi$ become arguments from $0$ to $\pi$: the upper half plane.
A strip of height $\pi$ opens exactly into a half plane.
A strip of any other height is rescaled first, which is a multiplication — one more standard map in front.
A power multiplies the angle, and $180 / 60 = 3$.
Choose the factor that reaches a straight angle.
So $w = z^{3}$ does it, and following with the Cayley transform would reach the disc.
Match each standard map to the change of region it makes.
| Opens a wedge at the origin by a factor of $4$ | Opens a horizontal strip into a half plane | Closes a half plane or a sector into a strip | Sends the upper half plane onto the unit disc | |
|---|---|---|---|---|
| $z^{4}$ | ||||
| $e^{z}$ | ||||
| a branch of $\log z$ | ||||
| $\dfrac{z - i}{z + i}$ |
Let $f(z) = 4z$. Plot the images of the four corners of the unit square $0$, $1$, $1 + i$, $i$, reading the first coordinate as the real part and the second as the imaginary part.
Plot your answer on the grid:
The map $w = z^{2}$ is applied to the sector at the origin of angle $50^\circ$. What is the angle of the image sector, in degrees?
Answer:
Put the steps of solving a boundary value problem for Laplace's equation by conformal mapping in order.
Number the steps in order (write the number in the box):
Select every statement that is true.
This task has no paper form; do it on a device.
A sector at the origin has angle $45^\circ$. For which whole number $k$ does $w = z^{k}$ map it onto a half plane?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each standard map to the change of region it makes.
| Opens a wedge at the origin by a factor of $2$ | Opens a horizontal strip into a half plane | Closes a half plane or a sector into a strip | Sends the upper half plane onto the unit disc | |
|---|---|---|---|---|
| $z^{2}$ | ||||
| $e^{z}$ | ||||
| a branch of $\log z$ | ||||
| $\dfrac{z - i}{z + i}$ |
You can compose standard maps to carry a region onto a disc or a half plane and can say why a boundary value problem travels with them. Say in your own words why the method needs angles preserved and not distances. That completes the course: the ideas of unit 1 about arguments and turns are the ones that have just been used to bend a region into a disc.
9. Your turn: map the sector of angle $60^\circ$ at the origin onto a half plane, step 2