Colorful visualization of complex functions · 2011-04-14 · Introduction Colorful Functions Series Conclusions The idea of ’colorful’ plotting • Assign unique colors to complex

Introduction Colorful Functions Series Conclusions

Colorful visualization of complex functions

Levente Lócsi

Department of Numerical Analysis, Faculty of Informatics,Eötvös Loránd University, Budapest, Hungary

NuHAG SeminarVienna, April 6, 2011


Motivation – What are these?


Motivation – History & Possible future work


Table of Contents

Introduction

’Colorful’ plots

Discussion of some functions

Function series

Conclusions


Table of Contents

Introduction



Function series

Conclusions


Plotting R → R functions


Complex numbers & arithmetic

Re

Im

ϕ

yz = x + iy

x

r

Re

Im

z1

z2

z1 + z2

Re

Im

ϕ1

z1ϕ2z2

ϕ1 + ϕ2

z1 · z2

Re

Im

z

z


Complex plots – Two planes

f (z) = z2


Complex plots – Two planes

f (z) = exp z = ez = ex+iy = ex · eiy = ex (cos y + i sin y)


Complex plots – Vectorfields

f (z) = z2 − 1


Complex plots – Vectorfields

f (z) = exp z


Complex plots – 3D (2 × R2 → R)

f1(x , y) = x2 − y2 f2(x , y) = 2xy

f (z) = z2


Complex plots – 3D (2 × R2 → R)

f1(x , y) = ex cos y f2(x , y) = ex sin yf (z) = exp z


Complex plots – One image (complexplot)

f (z) = z2



f (z) = exp z



f (z) = sin z


Complex plots – Fractal coloring

The Mandelbrot set


Table of Contents

Introduction



Function series

Conclusions


The idea of ’colorful’ plotting

• Assign unique colors to complex numbers

• C → B3, with B = [0..255], RGB

• Plot f : C → C by painting the pixels on a plane (on a square/ interval) the color assigned to the value f(z)

• Different colorings. . .




• C → B3, with B = [0..255], RGB






• C → B3, with B = [0..255], RGB




Example coloring: ImRe

• Treat the real and imaginary part separately

• R → B (e.g. red and blue)

• In other words:

• Note: we already have the plot of f (z) = z


Colorings

• </=, magnitude / argument


Advantages & disadvantages

• Concise, perspicuous (’übersichtlich’)

• Beautyful

• ’Waste of paint’


Table of Contents

Introduction



Function series

Conclusions


Functions

• f (z)


Constant

• f (z) = 0


Identity

• f (z) = z


Conjugate – 1

• f (z) = z


Conjugate – 2

• f (z) = z


Linear

• f (z) = (2 + i)z + 2


Square – 1

• f (z) = z2


Square – 2

• f (z) = z2


Square – 3

• f (z) = z2


A polynomial of degree two

• f (z) = z2 − 1


Polynomials of degree two – animation

• Let f (z) = (z − t0)(z + t0), where

• t0 = eiϕ, ϕ ∈ [0..π].

Re

Im

t0

−t0


Polynomials of degree two – animation

Please download video by clicking here.

Or use this url:http://locsi.web.elte.hu/complex/doc/k_video1_negyzetes.avi

http://locsi.web.elte.hu/complex/doc/k_video1_negyzetes.avi


A polynomial of degree three – 1

• f (z) = (z − 2)(z + i)(z + 2 − i)


A polynomial of degree three – 2

• f (z) = (z − 2)(z + i)(z + 2 − i)


And yet another one

• f (z) = (z − 2)(z + 1)2


Exponential – 1

• f (z) = exp z


Exponential – 2

• f (z) = exp z


Sine

• f (z) = sin z


Square root – 1

• f (z) =√

z (principal branch)


Square root – 2

• f (z) =√

z (principal branch)


Square root branches – animation

• Invert the square function restricted to different domains

• Where is the (branch) cut / the jump?

Re

Im

Re

Im


Square root branches – animation


Or use this url:http://locsi.web.elte.hu/complex/doc/k_video2_gyokagak.avi

http://locsi.web.elte.hu/complex/doc/k_video2_gyokagak.avi


Logarithm

• f (z) = log z (principal branch)


Logarithm branches – animation

• Invert the exponential function restricted to different domains

• Where is the cut / jump?

Re

Im

Re

Im


Logarithm branches – animation


Or use this url:http://locsi.web.elte.hu/complex/doc/k_video3_logagak.avi

http://locsi.web.elte.hu/complex/doc/k_video3_logagak.avi


Inversion – 1

• f (z) = 1/z


Inversion – 2

• f (z) = 1/z


Reciprocal

• f (z) = 1/z = (1/z)


A pole of order two – 1

• f (z) = 1/z2


A pole of order two – 2

• f (z) = 1/z2


Of singularities

Laurent series

+∞∑

k=−∞

ck(z − a)k

Order of the pole ∼ smallest (negative) index of terms withnon-zero coefficient (if there are finitely many of such)Essential singularity ∼ infinite number of such terms exist

Picard’s theorem

May f : C → C have an essential singularity at a ∈ C. Then:∃w0 ∈ C : ∀ε > 0 : ∀w ∈ C \ {w0} : ∃z ∈ kε(a) : f (z) = w .


Of singularities

Laurent series

+∞∑

k=−∞

ck(z − a)k

Order of the pole ∼ smallest (negative) index of terms withnon-zero coefficient (if there are finitely many of such)Essential singularity ∼ infinite number of such terms exist

Picard’s theorem

May f : C → C have an essential singularity at a ∈ C. Then:∃w0 ∈ C : ∀ε > 0 : ∀w ∈ C \ {w0} : ∃z ∈ kε(a) : f (z) = w .


An essential singularity

• f (z) = cos 1z


A linear fraction

• f (z) = (z − 2)/(z + 2) (Zhukovsky)


Table of Contents

Introduction



Function series

Conclusions


Taylor series of exp

• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n =



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 0



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 1



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 2



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 3



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 4



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 5



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 6



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 7



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 8



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 9



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 10



• Tn(z) =n∑

k=0

zk

k! = 1 + z + 12z2 + 1

3!z3 + . . .

n = 11


Taylor series of sin

• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n =



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 0



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 1



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 2



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 3



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 4



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 5



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 6



• Tn(z) =n∑

k=0(−1)k z (2k+1)

(2k+1)! = z − 13!z

3 + 15!z

5 + . . .

n = 7


Another series of functions

• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z



• M0(z) = zMn+1(z) = (Mn(z))

2 + z


Table of Contents

Introduction



Function series

Conclusions


Conclusions

• A small (non-complete) discussion of complex functions

• Idea and validity of ’colorful’ visualization

• No implementation known with different colorings availableand easy to use

• Matlab implementation, Gabor transforms

• Exam ,


Conclusions





• Exam ,


Conclusions





• Exam ,


Exam – Question 1


Exam – Question 1

Logarithmf (z) = log z


Exam – Question 2


Exam – Question 2

A polinomial of degree five with one root of multiplicity two andthree roots of multiplicity one

f (z) = i(z − i)2(z + i)(z − 1 − 310 i)(z + 1 − 3

10 i)


Exam – Question 3


Exam – Question 3

Tangentf (z) = sin z/ cos z


Exam – Question 4


Exam – Question 4

A polinomial of degree three

f (z) = (z − i)(z −√

32 + 1

2 i)(z +√

32 + 1

2 i)


Exam – Question 5


Exam – Question 5

A function with a zero of multiplicity one and a pole of order twof (z) = 1/z2 − iz


Exam – Question 6


Exam – Question 6

f (z) = exp cos z


Exam – Question 6

f (z) = exp cos z


What are these?


References

• See WWW

• References on my homepage(as soon as it’s translated and broken links are fixed)


Colorful visualization of complex functions

Author: Levente Lócsi

Occasion: NuHAG SeminarVienna, April 8, 2011

Web: http://locsi.web.elte.hu/complex/E-mail: [email protected]

Documents

Colorful visualization of complex functions · 2011-04-14 · Introduction Colorful Functions Series Conclusions The idea of ’colorful’ plotting • Assign unique colors to complex