Back to the on-screen lesson ·
An analytic map with non-vanishing derivative is locally a rotation together with a scaling, so it preserves the angle between curves and the sense of that angle while preserving neither lengths nor areas — and where the derivative vanishes it multiplies angles instead, which is what makes it useful for changing one region into another.
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 say why an analytic map with non-zero derivative preserves angles, read the rotation and the scaling off the derivative, compute how much a map magnifies an area, say what happens at a critical point and by what factor angles are multiplied there, explain why conjugation is not conformal, and state the Riemann mapping theorem with its one exception.
The Cauchy-Riemann equations, and the observation that they force the local linear approximation of an analytic map to be a rotation together with a scaling rather than a general linear map. This unit takes that observation and reads it as geometry.
A map is conformal at a point when it preserves both the size and the sense of the angle between any two curves crossing there. A point where $f'$ vanishes is a critical point of $f$. A map is orientation preserving when it turns anticlockwise angles into anticlockwise ones. A biholomorphism between two regions is an analytic bijection with an analytic inverse; it is conformal at every point.
The theorem. If $f$ is analytic at $z_0$ and $f'(z_0) \ne 0$, then $f$ is conformal at $z_0$: any two curves through $z_0$ have images crossing at the same angle, in the same sense.
Why. Near $z_0$,
$$f(z) \approx f(z_0) + f'(z_0)(z - z_0),$$
and multiplying by the fixed complex number $f'(z_0)$ is a rotation by its argument together with a scaling by its modulus. Both tangent directions are turned by the same angle, so the angle between them is unchanged; and a rotation cannot turn anticlockwise into clockwise, so the sense survives too.
What is not preserved. Lengths are multiplied by $|f'(z_0)|$ and areas by $|f'(z_0)|^{2}$, both of which vary from point to point. Straight lines are bent. The map $z^{2}$ is conformal everywhere except the origin and turns the vertical line through $1$ into a parabola.
Where it fails. At a critical point. If $f'$ vanishes to order $k - 1$ at $z_0$, then $f$ multiplies angles at $z_0$ by $k$: the map $z^{k}$ sends $re^{i\theta}$ to $r^{k}e^{ik\theta}$, so a wedge of angle $\alpha$ at the origin opens into a wedge of angle $k\alpha$. That is a defect from the point of view of the theorem and a tool from the point of view of the next three lessons: opening a corner by a chosen factor is how an awkward region is turned into a half plane.
And conjugation is the contrast worth keeping. $\bar z$ preserves the size of every angle and reverses every sense. It is anticonformal, it is analytic nowhere, and it is the reason the theorem says and the sense rather than leaving it understood.
Another way: picture
Draw a fine square grid and push it through an analytic map. Every square comes back a square — turned, and bigger or smaller, but square. The amount of turning and the amount of resizing vary from place to place, so the grid as a whole bends into a curved net; but every intersection is still a right angle. That picture is the Cauchy-Riemann equations, drawn.
Another way: steps
Preserving angles sounds like a weak property compared with preserving distances, and it is. It is also exactly the right one, for a reason that has nothing to do with geometry.
Harmonic functions survive conformal maps. If $u$ is harmonic and $f$ is analytic, then $u \circ f$ is harmonic as well. So a solution of Laplace's equation on a complicated region can be obtained by mapping the region conformally onto a simple one — a disc or a half plane — solving there, and pulling the answer back.
A distance-preserving map of the plane is a rotation, a translation or a reflection, and there are not enough of those to be useful: no isometry sends a strip to a disc. Conformal maps are plentiful — every analytic function with a non-vanishing derivative is one — and they are exactly the maps under which Laplace's equation is unchanged. Weakening the requirement from rigid to angle-preserving buys an enormous supply of maps at the cost of nothing that the physics needed.
And the supply is as large as it could possibly be. The Riemann mapping theorem says that any simply connected open set other than the whole plane can be mapped conformally onto the open unit disc. Any of them: a square's interior, a half plane, a slit plane, a region with a fractal boundary. The exception is the plane itself, and it is excluded for a reason already familiar — such a map would be a bounded entire function, which Liouville makes constant.
The theorem promises the map exists and does not say what it is. The next three lessons are about the ones that can be written down.
Three misreadings, each of which survives a long time because the true statement is so easy to over-hear.
It is local. Conformal at $z_0$ describes the map on an arbitrarily small neighbourhood. A map conformal at every point of a region can still fail to be one-to-one on that region: $z^{2}$ is conformal everywhere except the origin and sends both $1$ and $-1$ to $1$. A conformal map between regions has to be shown to be a bijection separately; the derivative condition does not give it.
It is about angles only. Not lengths, not areas, not straightness, not shapes. A square is not sent to a square by a general conformal map — its corners stay right angles and its sides become arcs. Anybody who has seen a conformal picture of a grid has seen this and may still expect a circle to stay a circle.
The derivative has to be non-zero, and when it is not, something specific happens. A critical point is not a place where the map is badly behaved; the map is perfectly analytic there. It is a place where angles are multiplied by a definite integer, and that integer is one more than the order to which $f'$ vanishes. Treating a critical point as a mere exception loses the most useful region-changing tool in the unit.
Take $f(z) = z^{2}$ and the two curves through $z_0 = 1$: the real axis and the vertical line through $1$, crossing at a right angle.
Pick a point where the derivative is not zero.
$f'(1) = 2$, a positive real number, so the images are rotated by nothing and stretched by $2$.
The derivative gives both halves.
The real axis maps to the real axis and the vertical line maps to a parabola — and the parabola meets the real axis at a right angle. Bent, and still perpendicular.
Conformality is not about straightness.
The same map at the origin: $f'(0) = 0$, so the theorem does not apply.
Check the hypothesis before using the conclusion.
The positive real axis and the positive imaginary axis meet at $90^\circ$ there. Their images are the positive real axis and the negative real axis, meeting at $180^\circ$.
The angle doubled.
The quarter plane opened into a half plane — which is precisely what makes this map useful for changing one region into another.
The failure is the feature.
$f'(z) = 3z^{2} - 3$, so $f'(1) = 0$.
Compute the derivative and test it.
Not conformal there. The second derivative $6z$ is $6 \ne 0$, so angles at $z = 1$ are doubled.
Match each map to what it does to a small figure near the point named.
| Scales every length by $5$ and keeps every angle | Translates, changing no length and no angle | Multiplies every angle at that point by $3$ | Keeps the size of every angle and reverses its sense | |
|---|---|---|---|---|
| $f(z) = 5z$ near any point | ||||
| $f(z) = z + 5$ near any point | ||||
| $f(z) = z^{3}$ near the origin | ||||
| $f(z) = \bar z$ near any point |
$f$ is analytic at $z_0$ with $f'(z_0) \ne 0$. Is this true of $f$ near $z_0$: it keeps the sense of that angle, clockwise or anticlockwise?
Two curves cross at $36^\circ$. Fill in the table, giving every angle in degrees.
| Value | |
|---|---|
| The angle between the images under the power map, away from the origin | |
| The angle between the images under the power map, at the origin | |
| The angle the map c times z rotates by | |
| The factor by which the map c times z scales lengths | |
| The factor by which the map c times z scales areas |
Let $f(z) = z + \left(1 + 2i\right)$. Plot the images of $0$, of $1$ and of $i$, reading the first coordinate as the real part and the second as the imaginary part.
Plot your answer on the grid:
Put the steps of deciding whether a map is conformal at a point, and of describing what it does there, in order.
Number the steps in order (write the number in the box):
$f(z) = z^{2}$, and a tiny region near the real point $z = 3$ has area $A$. Approximately what is the area of its image, as a multiple of $A$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each map to what it does to a small figure near the point named.
| Scales every length by $7$ and keeps every angle | Translates, changing no length and no angle | Multiplies every angle at that point by $2$ | Keeps the size of every angle and reverses its sense | |
|---|---|---|---|---|
| $f(z) = 7z$ near any point | ||||
| $f(z) = z + 7$ near any point | ||||
| $f(z) = z^{2}$ near the origin | ||||
| $f(z) = \bar z$ near any point |
You can decide whether a map is conformal at a point and describe what it does to a small figure there. Say in your own words why preserving angles is worth more than preserving distances. Next: the Möbius transformations, which are the conformal maps of the extended plane.
9. Your turn: is $f(z) = z^{3} - 3z$ conformal at $z = 1$?, step 2