Visualising
Complex Functions
Enhanced phase portraits
- What domain colouring is
- How it can be implemented in the computer
- How we can use it to explore complex functions
Complex functions
f:C→C
f:\mathbb C \rightarrow \mathbb C
live in a 4-dimensional space
Methods to visualize complex functions
- Real and Imaginary components
- Analytic Landscapes
- Mappings
- Domain colouring
Real and Imaginary components
f(z)=Re(z)+iIm(z)
f(z) = \text{Re}(z) + i \, \text{Im}(z)
f(z)=z2
f(z) = z^2
f(z)=x2−y2+2xyi
f(z) = x^2-y^2 + 2xy\,i
f(z)=x2−y2+2xyi
f(z) = x^2-y^2 + 2xy\,i
Re(z2)
\text{Re}(z^2)
Im(z2)
\text{Im}(z^2)
Analytic Landscapes
f(z)
\big| f(z) \big|
A historical analytic landscape of Γ(z) from 1909
Funktionentafeln mit Formeln und Kurven by Eugene Jahnke & Fritz Emde
Mappings
Mappings
Domain colouring
Phase portraits
Domain colouring
- Assign a colour to every point in the complex plane.
- Colour the domain of f by painting the location z with the colour determined by the value f(z).
The colour wheel
- H = Phase
-
S = 1
- B = 1
Hue , Saturation & Brightness
(HSB)
Implementation in the computer
- Mathematica
- MATLAB
- Python
- Java
- C++
- GeoGebra
- JavaScript
- CindyJS
- p5.js
Implementation in the computer
Basic Examples
Phase portraits
Phase portrait
f(z)=z[−2,2]×[−2,2]
f(z)=z\\
[-2,2] \times [-2,2]
Phase portrait
f(z)=z1[−2,2]×[−2,2]
f(z)=\dfrac{1}{z}\\
[-2,2] \times [-2,2]
Phase portrait
f(z)=z2+z+1z−1[−2,2]×[−2,2]
f(z)=\dfrac{z-1}{z^2+z+1}\\
[-2,2] \times [-2,2]
Enhanced phase portraits
- H = Phase
-
S = 1
- B = logf−⌊log∣f∣⌋
Elias Wegert's work from 2012
=Phase−⌊Phase⌋
=\text{Phase}-\lfloor \text{Phase} \rfloor
Enhanced phase portraits: Level curves
f(z)=z
f(z)=z
Phase
Modulus
Enhanced phase portraits: Level curves
f(z)=z
f(z)=z
Phase
Modulus
Combined
Enhanced phase portraits: Modulus
f(z)=z
f(z)=z
f(z)=1/z
f(z)=1/z
Enhanced phase portraits: f(z)=z2+z+1z−1
More examples...
Roots of unity: zn−1,n=2,3,…,10
Multiplicity of zeros & Order of poles
f(z)=(z+1)z2(z−1)3
f(z)=(z+1)z^2(z-1)^3
g(z)=1/f(z)
g(z)=1/f(z)
Analytic vs Non-Analytic functions
f(z)=23z(1+iz)
f(z)=\frac{3}{2}z(1+iz)
g(z)=23z(1−iz)
g(z)=\frac{3}{2}z(1-i\overline{z})
Laurent series
f(z)=n=0∑∞an(z−z0)n+n=1∑∞(z−z0)nbn,
f(z)=\displaystyle\sum_{n=0}^{\infty} a_n(z-z_0)^n+\displaystyle\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n},
R1<∣z−z0∣<R2.
R_1 <|z-z_0|< R_2.
an=2πi1∮C(z−z0)n+1f(z)dz(n=0,1,2,…)
a_n=\frac{1}{2\pi i}\displaystyle \oint_C \frac{f(z)dz}{(z-z_0)^{n+1}}\quad (n=0,1,2,\ldots)
bn=2πi1∮C(z−z0)−n+1f(z)dz(n=1,2,…)
b_n=\frac{1}{2\pi i}\displaystyle \oint_C \frac{f(z)dz}{(z-z_0)^{-n+1}}\quad (n=1,2,\ldots)
z31−cosh(z)
\dfrac{1-\cosh(z)}{z^3}
Poles of order m: ∃m≥1,bm=0 and bk=0 for k>m.
(z−1)2exp(2z)
\dfrac{\exp(2z)}{(z-1)^2}
z4sinh(z)
\dfrac{\sinh(z)}{z^4}
Removable singularities: If bn=0,∀n
zsin(z)
\dfrac{\sin(z)}{z}
ez−1z
\dfrac{z}{e^z-1}
Essential singularities: If bk=0 for infinitely many k
sin(z1)
\sin \left(\dfrac{1}{z}\right)
exp(z1)
\exp \left(\dfrac{1}{z}\right)
Zooming in
f(z)=exp(z1)
Other colour schemes
f(z)=0.926(z+0.073857z5+0.0045458z9)
Discrete HSV
RGB
B&W
Thank you!
Online resources:
https://www.dynamicmath.xyz/domain-coloring/
Contact:
j.ponce@uq.edu.au
Slides: reveal.js
Visualising Complex Functions Enhanced phase portraits