Juan Carlos Ponce Campuzano
Independent Mathematics Educator
Visualization of Complex Functions
Domain Coloring
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
0
0
2π
2\pi
f(z)=z2+z+1z−1
f(z)=\dfrac{z-1}{z^2+z+1}
θ=phase
Complex functions are not easy to visualize
f:R→R
f:\mathbb R \to \mathbb R
compared to functions of real variable:
R= Set of real numbers
f:R→R
f:\mathbb R \to \mathbb R
x0
x_0
y0=f(x0)
y_0=f(x_0)
Input: x0∈R
\text{Input: }x_0\in\mathbb R
Output: y0=f(x0)∈R
\text{Output: }y_0=f(x_0)\in \mathbb R
C={x+iyx,y real numbers &i=−1}
\mathbb C = \bigg\{x+iy\;\big|\;x,y \,\text{ real numbers \&\;} i = \sqrt{-1}\bigg\}
f:C→C
f:\mathbb C \to \mathbb C
For z∈C,
then z=x+iy.
Real
component
Imaginary
component
f:C→C
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∈C,
then z=x+iy.
real and imaginary components.
has also
=u+iv
∈C
f:C→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:R→R
f:\mathbb R \to \mathbb R
f:R2→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
z=x+iy
x
x
y
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=reiθ
z=x+iy= r e^{i\theta}
θ
\theta
r
r
r=x2+y2=∣z∣
r = \sqrt{x^2+y^2}=|z|
θ=arctan(y/x)
\theta = \arctan(y/x)
θ∈[
\theta \in [
0
0
,
,
2π
2\pi
)
)
phase/argument
\text{phase/argument}
modulus
\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=reiθ
z=x+iy= r e^{i\theta}
θ
\theta
r
r
θ→
\theta\rightarrow
0
0
2π
2\pi
θ
\theta
Hue
phase
\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=reiθ
z= r e^{i\theta}
r
r
θ→
\theta\rightarrow
0
0
2π
2\pi
θ
\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=reiθ
z= r e^{i\theta}
r
r
θ→
\theta\rightarrow
0
0
2π
2\pi
θ
\theta
How does domain coloring work?
0
0
2π
2\pi
How does domain coloring work?
The identity function
Color represents
θ=phase of z
f(z)=z
f(z)=z
0
0
2π
2\pi
Assign a color to every point in the complex plane.
Domain coloring
The identity function
Color represents
θ=phase of z
Phase Portraits
f(z)=z
f(z)=z
0
0
2π
2\pi
Basic examples: Phase Portraits
[−2,2]×[−2,2]
[-2,2] \times [-2,2]
f(z)=z
f(z)=z
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
Basic examples: Phase Portraits
f(z)=z1
f(z)=\dfrac{1}{z}
[−2,2]×[−2,2]
[-2,2] \times [-2,2]
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
Basic examples: Phase Portraits
f(z)=1/z
f(z)=1/z
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
f(z)=z
f(z)=z
Basic examples: Phase Portraits
[−2,2]×[−2,2]
[-2,2] \times [-2,2]
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
f(z)=z2+z+1z−1
f(z)=\dfrac{z-1}{z^2+z+1}
Rational function
Basic examples: Phase Portraits
f(z)=z2+z+1z−1
f(z)=\dfrac{z-1}{z^2+z+1}
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
Note that f(1)=0.
z0=1 is a zero/root of f.
Also f is undefined at
z1=2−1+i3,z2=2−1−i3.
z0
z_0
z1
z_1
z2
z_2
z1,z2 are poles of f.
Basic examples: Phase Portraits
f(z)=z2+z+1z−1
f(z)=\dfrac{z-1}{z^2+z+1}
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
z0
z_0
z1
z_1
z2
z_2
To locate Zeros and Poles
we just need to look at where all
the colors meet!
Basic examples: Phase Portraits
f(z)=z2+z+1z−1
f(z)=\dfrac{z-1}{z^2+z+1}
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
z0
z_0
z1
z_1
z2
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
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
z0
z_0
z1
z_1
z2
z_2
We can solve this issue by introducing the level curves of the modulus ∣f(z)∣.
Enhanced Phase Portraits
Recall we used Hue ↔ Phase
0
π/2
π
3π/2
2π
Enhanced Phase Portraits
Recall we used Hue → Phase
0
π/2
π
3π/2
0
π/2
π
3π/2
2π
The color wheel
Enhanced Phase Portraits
Recall we used Hue → Phase
0
π/2
π
3π/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
logf−⌊log∣f∣⌋
, Saturation
- S = 1
- B =
HSB scheme
& Brightness
Enhanced Phase Portraits
f(z)=z
f(z)=z
Plain
Level curves: modulus
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
0
0
0
0
i
i
2i
2i
−1
-1
2
2
1
1
−2i
-2i
−i
-i
Enhanced Phase Portraits
f(z)=z
f(z)=z
Level curves: Phase
Plain
B = logf−⌊log∣f∣⌋
B = Phase−⌊ Phase⌋
Level curves: modulus
Enhanced Phase Portraits
f(z)=z
f(z)=z
Combine
B = logf−⌊log∣f∣⌋
B = Phase−⌊ Phase⌋
Enhanced Phase Portraits
f(z)=z
f(z)=z
f(z)=1/z
f(z)=1/z
Level curves: Modulus
Zero at z0=0
Pole at z0=0
Lighter
Enhanced Phase Portraits
f(z)=z
f(z)=z
f(z)=1/z
f(z)=1/z
Level curves: Modulus
Zero at z0=0
Pole at z0=0
Darker
Enhanced Phase Portraits
f(z)=z
f(z)=z
f(z)=1/z
f(z)=1/z
Level curves: Modulus
Zero at z0=0
Pole at z0=0
Darker
Enhanced Phase Portraits
f(z)=z
f(z)=z
f(z)=1/z
f(z)=1/z
Level curves: Modulus
Zero at z0=0
Pole at z0=0
Lighter
Enhanced Phase Portraits
Plain
f(z)=z2+z+1z−1
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)=z2+z+1z−1
f(z)=\dfrac{z-1}{z^2+z+1}
This is a zero
These are poles
Enhanced Phase Portraits
Level curves: Modulus
z2+2+2i(z2−1)(z−2−i)2
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
f(z)=
Enhanced Phase Portraits
Level curves: Modulus
z2+2+2i(z2−1)(z−2−i)2
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
f(z)=
Zeros
Enhanced Phase Portraits
Level curves: Modulus
z2+2+2i(z2−1)(z−2−i)2
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
f(z)=
Poles
Enhanced Phase Portraits
Level curves: Modulus
z2+2+2i(z2−1)(z−2−i)2
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
f(z)=
0
0
0
0
1.25i
1.25i
2.5i
2.5i
−1.25
-1.25
2.5
2.5
1.25
1.25
−2.5i
-2.5i
−1.25i
-1.25i
Enhanced Phase Portraits
Level curves: Modulus
z2+2+2i(z2−1)(z−2−i)2
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
f(z)=
3 zeros
0
0
0
0
1.25i
1.25i
2.5i
2.5i
−1.25
-1.25
2.5
2.5
1.25
1.25
−2.5i
-2.5i
−1.25i
-1.25i
Enhanced Phase Portraits
Level curves: Modulus
z2+2+2i(z2−1)(z−2−i)2
\dfrac{\left(z^2-1\right)\left(z-2-i\right)^2}{z^2+2+2i}
f(z)=
f(z)=
2 poles
0
0
0
0
1.25i
1.25i
2.5i
2.5i
−1.25
-1.25
2.5
2.5
1.25
1.25
−2.5i
-2.5i
−1.25i
-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)=anzn+an−1zn−1+⋯+a1z+a0
p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0
- Roots of unity
zn−1=0
z^n-1=0
and more...
n=1∑201−znzn
\displaystyle \sum_{n=1}^{20}\dfrac{z^n}{1-z^n}
Endless possibilities!
Mathematical Art
1/(iz)−1(1/(iz))18−(1/(iz))
\dfrac{(1/(iz))^{18}-(1/(iz))}{1/(iz)-1}
Finite Blaschke product
\text{Finite Blaschke product}
f(z)=0.926(z+0.073857z5+0.0045458z9)
RGB
B&W
Discrete HSB
Mathematical Art
Other color schemes
Level curves of the real and imaginary components
Mathematical Art
Other color schemes
Hydrodynamics
Joukowski’s Airfoil
\text{Joukowski's Airfoil}
Electric potential
log(z+1z−1)
\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.
Visualization of Complex Functions Domain Coloring 0 0 0 0 i i 2 i 2i − 1 -1 2 2 1 1 − 2 i -2i − i -i 0 0 2 π 2\pi f ( z ) = z − 1 z 2 + z + 1 f(z)=\dfrac{z-1}{z^2+z+1} θ = p h a s e
Domain Coloring
By Juan Carlos Ponce Campuzano
Domain Coloring
Visualization of complex functions using domain coloring.
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