Including Complex Dynamics in Complex Analysis Courses Mandelbrot and Julia sets

Benefits: Introduces “modern” topics in mathematics into the course Illustrates different ways to use complex analysis “tools”

Topics: Mandelbrot set (complex square roots) Attracting cycles (Schwarz Lemma) Understanding chaos (rational maps) Exponential dynamics (geometry of exp(z))

Begin by recalling iteration Start with a function: 2 x + constant and a seed: x0

Then iterate: x = x + constant x = x + constant x = x + constant 2 x = x + constant 1 2 x = x + constant 2 1 Orbit of x 2 x = x + constant 3 2 2 x = x + constant 4 3 etc. Goal: understand the fate of orbits.

Example: x + 1 Seed 0 x = 0 x = 1 x = 2 x = 5 x = 26 x = big x = 1 1 x = 2 2 x = 5 “Orbit tends to infinity” 3 x = 26 4 x = big 5 x = BIGGER 6

Example: x + 0 Seed 0 x = 0 x = 0 x = 0 x = 0 x = 0 x = 0 x = 0 2 Example: x + 0 Seed 0 x = 0 x = 0 1 x = 0 2 “A fixed point” x = 0 3 x = 0 4 x = 0 5 x = 0 6

Example: x - 1 Seed 0 x = 0 x = -1 x = 0 x = -1 x = 0 x = -1 x = 0 2 Example: x - 1 Seed 0 x = 0 x = -1 1 x = 0 2 x = -1 “A two- cycle” 3 x = 0 4 x = -1 5 x = 0 6

Example: x - 1.1 Seed 0 x = 0 x = -1.1 x = 0.11 x = x = x = x = 2 Example: x - 1.1 Seed 0 x = 0 x = -1.1 1 x = 0.11 2 x = 3 time for the computer! x = 4 x = 5 x = 6

Observation Sometimes orbit of 0 goes to infinity, other times it does not.

Big Question: Answer: How do we understand the chaotic behavior that occurs? Answer: We move to the complex plane and use tools from complex analysis

Complex Iteration 2 Iterate z + c complex numbers

Example: z + i Seed 0 z = 0 z = i z = -1 + i z = -i z = -1 + i z = -i 2 Example: z + i Seed 0 z = 0 z = i 1 z = -1 + i 2 z = -i 3 z = -1 + i 4 z = -i 5 2-cycle z = -1 + i 6

Example: z + 2i Seed 0 z = 0 z = 2i z = -4 + 2i z = 12 - 14i z = 2i 1 z = -4 + 2i 2 Off to infinity z = 12 - 14i 3 z = -52 + 336i 4 z = big 5 z = BIGGER 6

Same observation Sometimes orbit of 0 goes to infinity, other times it does not.

The Mandelbrot Set: All c-values for which orbit of 0 does NOT go to infinity. Why the heck do we care about the orbit of 0?

Why use the orbit of 0? Answer: 0 is a “critical point” So spend some time talking about behavior of complex functions near critical points (not one-to-one) and near other points (always one-to-one)

The Filled Julia Set: The Julia Set: Fix a c-value. The filled Julia set is all of the complex seeds whose orbits do NOT go to infinity. The Julia Set: The boundary of the filled Julia set

Example: z Seed: In Filled Julia set? Yes 1 Yes -1 Yes i Yes 2i No 5 Yes 1 Yes -1 Yes i Yes 2i No 5 I doubt it...

The filled Julia set of z2 is the unit disk A little complex analysis: |z| < 1: orbit of z 0 |z| > 1: orbit of z infinity |z| = 1: orbit of z remains on the unit circle

The filled Julia set of z2 is the unit disk The Julia set is the unit circle

Chaos occurs on the Julia set

Chaos occurs on the Julia set 3 nearby seeds Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Chaos occurs on the Julia set

Nearby orbits on the Julia set have vastly different behavior

i.e., the map just doubles angles On the Julia set the map is ei e2i i.e., the map just doubles angles

On the Julia set the map is ei e2i 1/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 4/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.

Rational angles on the Julia set are eventually periodic. 1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

But irrational angles on the Julia set are very different.

Irrational angles have orbits that are dense on the circle.

So why do we use the orbit of 0 to plot the Mandelbrot set? The orbit of the critical point knows “everything!”

So why do we use the orbit of 0 to plot the Mandelbrot set? The orbit of the critical point knows “everything!” To see this, we need the geometry of complex square roots

The square root of a circle (s.c.c.) that does not surround the origin.... i 1

is a pair of s.c.c.’s that do not The square root of a circle (s.c.c.) that does not surround the origin.... i z z2 1 is a pair of s.c.c.’s that do not surround the origin

The square root of a circle (s.c.c.) that touches the origin.... 1

The square root of a circle (s.c.c.) that touches the origin.... z z2 1 is a figure 8 that passes through the origin

The square root of a circle (s.c.c.) that surrounds the origin.... 1

The square root of a circle (s.c.c.) that surrounds the origin.... z z2 1 is one s.c.c. that now surrounds the origin

To find the preimages of a s.c.c. under z2 + c, we solve: so that:

To find the preimages of a s.c.c. under z2 + c, we solve: so that: So to find the preimage of a s.c.c., we translate the curve by -c, then take the complex square root....

So now suppose the orbit of 0 escapes to infinity c

This large circle is mapped outside itself c

This large circle is mapped outside itself c

so all blue points escape c

so all blue points escape To find the preimage of this curve, subtract c and take the square root... c

so all blue points escape To find the preimage of this curve, subtract c and take the square root... but subtracting c means the curve still encircles 0 c

so all blue points escape so this is the preimage c

so all blue points escape so these points also escape c

so all blue points escape so these points also escape c

so all blue points escape so these points also escape as do these points c

so all blue points escape so these points also escape as do these points c

so all blue points escape so these points also escape as do these points c

8 so all blue points escape so these points also escape as do these points c 8

Eventually 0 is in a preimage of the See what’s happening? Eventually 0 is in a preimage of the circle, so we get a figure 8 preimage. And then..... c = .4975

So the filled Julia set consists of infinitely many distinct components if 0 escapes (in fact it is a Cantor set). But when 0 does not escape, we never get a figure 8, so the filled Julia set is a connected set. That’s why the critical point “knows it all.”

Attracting cycles Understanding chaos Dynamics of the exponential

If F is complex analytic and has an attracting cycle, then there must be a critical point inside its immediate basin of attraction. So F(z) = z2 + c can have at most one attracting cycle, and the orbit of 0 must find it.

Reason: suppose F has an attracting fixed point p:

Then there is a neighborhood U of p mapped 1-1 inside itself.

Then there is a neighborhood U of p mapped 1-1 inside itself. So we have an inverse map F-1: F(U) U with a repelling fixed point at p. F-1 p U

Since there are no critical points in the basin, we can continue F-1 p U

Since there are no critical points in the basin, and on and on ..... F-1 p U

Eventually get an open disk V that is mapped 1-1 to itself by F-1 F-1 ... p U V

Eventually get an open disk V that is mapped 1-1 to itself by F-1 F-1 p V

But there is a repelling fixed point in V, so this can’t happen by the Schwarz Lemma F-1 p V

So this begins to describe the structure of the Mandelbrot set: each “bulb” consists of parameters for which there is an attracting cycle of some given period

The eventual orbit of 0 Eventual orbit

The eventual orbit of 0 Eventual orbit

The eventual orbit of 0 3-cycle

The eventual orbit of 0 3-cycle

The eventual orbit of 0 3-cycle

The eventual orbit of 0 3-cycle

The eventual orbit of 0

The eventual orbit of 0

The eventual orbit of 0 4-cycle

The eventual orbit of 0 4-cycle

The eventual orbit of 0 4-cycle

The eventual orbit of 0 4-cycle

The eventual orbit of 0 4-cycle

The eventual orbit of 0

The eventual orbit of 0

The eventual orbit of 0 5-cycle

The eventual orbit of 0 5-cycle

The eventual orbit of 0 5-cycle

The eventual orbit of 0 5-cycle

The eventual orbit of 0 5-cycle

The eventual orbit of 0 5-cycle

The eventual orbit of 0 5-cycle

The eventual orbit of 0 2-cycle

The eventual orbit of 0 2-cycle

The eventual orbit of 0 2-cycle

The eventual orbit of 0 fixed point

The eventual orbit of 0 fixed point

The eventual orbit of 0 fixed point

How understand the periods of the bulbs?

How understand the periods of the bulbs?

junction point three spokes attached

junction point three spokes attached Period 3 bulb

Period 4 bulb

Period 5 bulb

Period 7 bulb

Period 13 bulb

Complex exponential dynamics Let E (z) = exp(z) where > 0. Differences: no critical points, but one “asymptotic value” at z = 0.

Complex exponential dynamics Let E (z) = exp(z) where > 0. Differences: no critical points, but one “asymptotic value” at z = 0. H is wrapped infinitely often around 0, which plays the role of the critical value H

Complex exponential dynamics Let E (z) = exp(z) where > 0. Differences: no critical points, but one “asymptotic value” at z = 0. The Julia set is now the closure of the set of escaping points, not the boundary of this set

When 0 < < 1/e, the Julia set is a “Cantor bouquet,” i.e., infinitely many curves extending to infinity in the right half plane.

0 < < 1/e

attracting fixed point q

q p repelling fixed point

q x0 p

So where is J?

So where is J?

So where is J? Green points lie in the basin of q, so not in the Julia set

So where is J? Green points lie in the basin of q, so not in the Julia set

So where is J? Green points lie in the basin of q, so not in the Julia set

So where is J? Green points lie in the basin of q, so not in the Julia set

So where is J? Green points lie in the basin of q, so not in the Julia set

The Julia set is a collection of curves (hairs) in the right half plane, each with an endpoint and a stem. hairs endpoints stems

A “Cantor bouquet” q p

Colored points escape to and so are in the Julia set. q p

When = 1/e, E undergoes a “saddle node bifurcation” but much more happens...

When = 1/e, E undergoes a “saddle node bifurcation” The two fixed points on the real line disappear and the orbit of the asymptotic value now escapes....

And, because of this, the Julia instantaneously becomes the entire complex plane!

Replay

How understand the chaotic behavior? z2 z2 -.12+.75i

so use that information How understand the chaotic behavior? We understand what’s going on here, so use that information to understand what’s going on here.

How understand the chaotic behavior? These angles are just rotated by z2

How understand the chaotic behavior? These angles are just rotated by z2 So we can find a “conjugacy” that creates angles here that map in the same way

Example: z2 - 2 z2 z2 - 2 How to map the exterior of the circle to the exterior of [-2,2]? -2 2 z2 z2 - 2

Consider H(z) = z + 1/z -2 2 z2 z2 - 2

Consider H(z) = z + 1/z H takes rays to hyperbolas -2 2 z2 z2 - 2

Consider H(z) = z + 1/z Well, usually that’s the case -2 2 z2 z2 - 2

Consider H(z) = z + 1/z z2 z2 - 2 And H takes the circle to [-2, 2] -2

Consider H(z) = z + 1/z z2 z2 - 2 And H takes the circle to [-2, 2] since eit + e-it = 2 cos(t) -2 2 z2 z2 - 2

On the circle we have: H H On [-2,2] we have: What is this function???

On the circle we have: H H + 2 - 2

On the circle we have: H H + 2 - 2

On the circle we have: H H + 2 - 2 So this function is w w2 - 2

So the orbits of F(z) = z2 on the unit circle (which we completely understand are taken to orbits of G(z) = z2 - 2 by the “conjugacy” H(z) = z + 1/z. z F(z) F2(z) F3(z) F4(z) ... H H H H H w G(w) G2(w) G2(w) G4(z) ... And that’s how we understand the chaotic behavior on the Julia sets.