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.

  1. Assign a color to every point in the complex plane.
  2. 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.