Visualization of Complex Functions
Domain Coloring
0
0
i
2i
-1
2
1
-2i
-i
0
2\pi
f(z)=\dfrac{z-1}{z^2+z+1}
\(\theta=\text{phase}\)
Complex functions are not easy to visualize
f:\mathbb R \to \mathbb R
compared to functions of real variable:
\(\mathbb R=\) Set of real numbers
f:\mathbb R \to \mathbb R
x_0
y_0=f(x_0)
\text{Input: }x_0\in\mathbb R
\text{Output: }y_0=f(x_0)\in \mathbb R
\mathbb C = \bigg\{x+iy\;\big|\;x,y \,\text{ real numbers \&\;} i = \sqrt{-1}\bigg\}
f:\mathbb C \to \mathbb C
For \(z\in \mathbb C,\)
then \(z=x+iy.\)
Real
component
Imaginary
component
f:\mathbb C \to \mathbb C
So the plot of a complex function \(f\) lives in a four-dimensional space.
The output \(\,f(z)\)
For \(z\in \mathbb C,\)
then \(z=x+iy.\)
real and imaginary components.
has also
\(=u+iv\)
\(\in \mathbb C\,\)
f:\mathbb C \to \mathbb C
So the plot of a complex function \(f\) lives in a four-dimensional space.
We can use what we know about plotting functions of real variables:
f:\mathbb R \to \mathbb R
f:\mathbb R^2 \to \mathbb R
Good News!
Methods to visualize complex functions
Plotting Real and Imaginary components
-
Analytic landscapes
-
Mappings
-
Domain coloring
-
Methods to visualize complex functions
Plotting Real and Imaginary components
-
Analytic landscapes
-
Mappings
-
-
Domain coloring
What is domain coloring?
A technique for visualizing complex functions by assigning a color to each point of the complex plane.
Proposed by Frank Farris in the 90s as an alternative way to explore geometric properties of complex functions.
- Assign a color to every point in the complex plane.
- Color the domain of \(f\) by painting the location \(z\) with the color determined by the value \(f(z)\).
How does domain coloring work?
Color the domain of \(f\) by painting the location \(z\) with the color determined by the value \(f(z)\).
z=x+iy
x
y
How does domain coloring work?
Color the domain of \(f\) by painting the location \(z\) with the color determined by the value \(f(z)\).
z=x+iy= r e^{i\theta}
\theta
r
r = \sqrt{x^2+y^2}=|z|
\theta = \arctan(y/x)
\theta \in [
0
,
2\pi
)
\text{phase/argument}
\text{modulus}
How does domain coloring work?
Color the domain of \(f\) by painting the location \(z\) with the color determined by the value \(f(z)\).
z=x+iy= r e^{i\theta}
\theta
r
\theta\rightarrow
0
2\pi
\theta
Hue
\text{phase}
How does domain coloring work?
Color the domain of \(f\) by painting the location \(z\) with the color determined by the value \(f(z)\).
z= r e^{i\theta}
r
\theta\rightarrow
0
2\pi
\theta
How does domain coloring work?
Color the domain of \(f\) by painting the location \(z\) with the color determined by the value \(f(z)\).
z= r e^{i\theta}
r
\theta\rightarrow
0
2\pi
\theta
How does domain coloring work?
0
2\pi
How does domain coloring work?
The identity function
Color represents
\(\theta=\text{phase of }z\)
f(z)=z
0
2\pi
Assign a color to every point in the complex plane.
Domain coloring
The identity function
Color represents
\(\theta=\text{phase of }z\)
Phase Portraits
f(z)=z
0
2\pi
Basic examples: Phase Portraits
[-2,2] \times [-2,2]
f(z)=z
0
0
i
2i
-1
2
1
-2i
-i
Basic examples: Phase Portraits
f(z)=\dfrac{1}{z}
[-2,2] \times [-2,2]
0
0
i
2i
-1
2
1
-2i
-i
Basic examples: Phase Portraits
f(z)=1/z
0
0
i
2i
-1
2
1
-2i
-i
0
0
i
2i
-1
2
1
-2i
-i
f(z)=z
Basic examples: Phase Portraits
[-2,2] \times [-2,2]
0
0
i
2i
-1
2
1
-2i
-i
f(z)=\dfrac{z-1}{z^2+z+1}
Rational function
Basic examples: Phase Portraits
f(z)=\dfrac{z-1}{z^2+z+1}
0
0
i
2i
-1
2
1
-2i
-i
Note that \(f(1)=0.\)
\(z_0=1\) is a zero/root of \(f.\)
Also \(f\) is undefined at
\[z_1=\frac{1+i\sqrt{3}}{2},\;z_2=\frac{1-i\sqrt{3}}{2}.\]
z_0
z_1
z_2
\(z_1,z_2\;\) are poles of \(f.\)
Basic examples: Phase Portraits
f(z)=\dfrac{z-1}{z^2+z+1}
0
0
i
2i
-1
2
1
-2i
-i
z_0
z_1
z_2
To locate Zeros and Poles
we just need to look at where all
the colors meet!
Basic examples: Phase Portraits
f(z)=\dfrac{z-1}{z^2+z+1}
0
0
i
2i
-1
2
1
-2i
-i
z_0
z_1
z_2
it becomes difficult to determine which points are Zeros or Poles.
If we do not know the expression defining the function \(f,\)
Enhanced Phase Portraits
0
0
i
2i
-1
2
1
-2i
-i
z_0
z_1
z_2
We can solve this issue by introducing the level curves of the modulus \(|f(z)|.\)
Enhanced Phase Portraits
Recall we used Hue \(\leftrightarrow\) Phase
\(0\)
\(\pi/2\)
\(\pi\)
\(3\pi/2\)
\(2\pi\)
Enhanced Phase Portraits
Recall we used Hue \(\rightarrow\) Phase
\(0\)
\(\pi/2\)
\(\pi\)
\(3\pi/2\)
\(0\)
\(\pi/2\)
\(\pi\)
\(3\pi/2\)
\(2\pi\)
The color wheel
Enhanced Phase Portraits
Recall we used Hue \(\rightarrow\) Phase
\(0\)
\(\pi/2\)
\(\pi\)
\(3\pi/2\)
The color wheel
HSB scheme
- H = Phase
Hue
Enhanced Phase Portraits
, Saturation
- B = 1
HSB scheme
& Brightness
- S = 1
- H = Phase
Hue
Enhanced Phase Portraits
\(\log\big|f\big|- \lfloor \log |f| \rfloor\)
, Saturation
- S = 1
- B =
HSB scheme
& Brightness
Enhanced Phase Portraits
f(z)=z
Plain
Level curves: modulus
0
0
i
2i
-1
2
1
-2i
-i
0
0
i
2i
-1
2
1
-2i
-i
Enhanced Phase Portraits
f(z)=z
Level curves: Phase
Plain
B = \(\log\big|f\big|- \lfloor \log |f| \rfloor\)
B = \(\text{Phase}-\lfloor \text{Phase} \rfloor\)
Level curves: modulus
Enhanced Phase Portraits
f(z)=z
Combine
B = \(\log\big|f\big|- \lfloor \log |f| \rfloor\)
B = \(\text{Phase}-\lfloor \text{Phase} \rfloor\)
Enhanced Phase Portraits
f(z)=z
f(z)=1/z
Level curves: Modulus
Zero at \(z_0=0\)
Pole at \(z_0=0\)
Lighter
Enhanced Phase Portraits
f(z)=z
f(z)=1/z
Level curves: Modulus
Zero at \(z_0=0\)
Pole at \(z_0=0\)
Darker
Enhanced Phase Portraits
f(z)=z
f(z)=1/z
Level curves: Modulus
Zero at \(z_0=0\)
Pole at \(z_0=0\)
Darker
Enhanced Phase Portraits
f(z)=z
f(z)=1/z
Level curves: Modulus
Zero at \(z_0=0\)
Pole at \(z_0=0\)
Lighter
Enhanced Phase Portraits
Plain
f(z)=\dfrac{z-1}{z^2+z+1}
There are three points where to colors meet!
Enhanced Phase Portraits
Level curves: Modulus
f(z)=\dfrac{z-1}{z^2+z+1}
This is a zero
These are poles
Enhanced Phase Portraits
Level curves: Modulus
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
Enhanced Phase Portraits
Level curves: Modulus
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
Zeros
Enhanced Phase Portraits
Level curves: Modulus
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
Poles
Enhanced Phase Portraits
Level curves: Modulus
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
0
0
1.25i
2.5i
-1.25
2.5
1.25
-2.5i
-1.25i
Enhanced Phase Portraits
Level curves: Modulus
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
3 zeros
0
0
1.25i
2.5i
-1.25
2.5
1.25
-2.5i
-1.25i
Enhanced Phase Portraits
Level curves: Modulus
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
2 poles
0
0
1.25i
2.5i
-1.25
2.5
1.25
-2.5i
-1.25i
Domain coloring
Visualizing
complex functions
- Zeros & Poles of functions
Enhanced Phase Portraits
- Multiplicity of zeros
- Classification of singularities
- The Fundamental Theorem of Algebra
p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0
- Roots of unity
z^n-1=0
and more...
\displaystyle \sum_{n=1}^{20}\dfrac{z^n}{1-z^n}
Endless possibilities!
Mathematical Art
\dfrac{(1/(iz))^{18}-(1/(iz))}{1/(iz)-1}
\text{Finite Blaschke product}
\(f(z)=0.926(z+0.073857 z^5+0.0045458 z^9)\)
RGB
B&W
Discrete HSB
Mathematical Art
Other color schemes
Level curves of the real and imaginary components
Mathematical Art
Other color schemes
Hydrodynamics
\text{Joukowski's Airfoil}
Electric potential
\log\left(\dfrac{z-1}{z+1}\right)
Online tools
www.dynamicmath.xyz
Online tools
www.dynamicmath.xyz
Complex Analysis
A visual and Interactive Introduction
complex-analysis.com
Thanks for
watching!
Patreons:
Ruan Ramon, Miguel Díaz, bleh, Dennis Watson, Doug Kuhlmann, mirror, Newnome Beauton, Adam Parrott, Sophia Wood (Fractal Kitty), pmben, Abei, Edward Huff.
Thanks for
watching!
Patreons:
Ruan Ramon, Miguel Díaz, bleh, Dennis Watson, Doug Kuhlmann, mirror, Newnome Beauton, Adam Parrott, Sophia Wood (Fractal Kitty), pmben, Abei, Edward Huff.
Domain Coloring
By Juan Carlos Ponce Campuzano
Domain Coloring
Visualization of complex functions using domain coloring.
