Introduction
Colorful
Functions
Series
Colorful visualization of complex functions
Levente Lócsi
Department of Numerical Analysis, Faculty of Informatics,
Eötvös Loránd University, Budapest, Hungary
NuHAG Seminar
Vienna, April 6, 2011
Conclusions
Introduction
Colorful
Functions
Series
Motivation – What are these?
Conclusions
Introduction
Colorful
Functions
Series
Motivation – History & Possible future work
Conclusions
Introduction
Colorful
Functions
Table of Contents
Introduction
’Colorful’ plots
Discussion of some functions
Function series
Conclusions
Series
Conclusions
Introduction
Colorful
Functions
Table of Contents
Introduction
’Colorful’ plots
Discussion of some functions
Function series
Conclusions
Series
Conclusions
Introduction
Colorful
Functions
Series
Plotting R → R functions
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
Complex numbers & arithmetic
Im
Im
z = x + iy
y
z1 + z2
z2
r
ϕ
z1 · z2Im
z1
x
Re
Re
Im
ϕ1 + ϕ2
z2
ϕ2z1
ϕ1
Re
z
Re
z
Introduction
Colorful
Functions
Series
Complex plots – Two planes
f (z) = z 2
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – Two planes
f (z) = exp z = e z = e x +iy = e x · e iy = e x (cos y + i sin y )
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – Vectorfields
f (z) = z 2 − 1
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – Vectorfields
f (z) = exp z
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – 3D (2 × R2 → R)
f1 (x, y ) = x 2 − y 2
f2 (x, y ) = 2xy
f (z) = z 2
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – 3D (2 × R2 → R)
f1 (x, y ) = e x cos y
f2 (x, y ) = e x sin y
f (z) = exp z
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – One image (complexplot)
f (z) = z 2
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – One image (complexplot)
f (z) = exp z
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – One image (complexplot)
f (z) = sin z
Conclusions
Introduction
Colorful
Functions
Series
Complex plots – Fractal coloring
The Mandelbrot set
Conclusions
Introduction
Colorful
Functions
Table of Contents
Introduction
’Colorful’ plots
Discussion of some functions
Function series
Conclusions
Series
Conclusions
Introduction
Colorful
Functions
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. . .
Introduction
Colorful
Functions
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. . .
Introduction
Colorful
Functions
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. . .
Introduction
Colorful
Functions
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
Series
Conclusions
Introduction
Colorful
Functions
Colorings
• </=, magnitude / argument
Series
Conclusions
Introduction
Colorful
Functions
Series
Advantages & disadvantages
• Concise, perspicuous (’übersichtlich’)
• Beautyful
• ’Waste of paint’
Conclusions
Introduction
Colorful
Functions
Table of Contents
Introduction
’Colorful’ plots
Discussion of some functions
Function series
Conclusions
Series
Conclusions
Introduction
Colorful
Functions
Functions
• f (z)
Series
Conclusions
Introduction
Colorful
Functions
Constant
• f (z) = 0
Series
Conclusions
Introduction
Colorful
Functions
Identity
• f (z) = z
Series
Conclusions
Introduction
Colorful
Functions
Conjugate – 1
• f (z) = z
Series
Conclusions
Introduction
Colorful
Functions
Conjugate – 2
• f (z) = z
Series
Conclusions
Introduction
Colorful
Functions
Linear
• f (z) = (2 + i)z + 2
Series
Conclusions
Introduction
Colorful
Functions
Square – 1
• f (z) = z 2
Series
Conclusions
Introduction
Colorful
Functions
Square – 2
• f (z) = z 2
Series
Conclusions
Introduction
Colorful
Functions
Square – 3
• f (z) = z 2
Series
Conclusions
Introduction
Colorful
Functions
Series
A polynomial of degree two
• f (z) = z 2 − 1
Conclusions
Introduction
Colorful
Functions
Series
Polynomials of degree two – animation
• Let f (z) = (z − t0 )(z + t0 ), where
• t0 = e iϕ ,
ϕ ∈ [0..π].
Im
t0
−t0
Re
Conclusions
Introduction
Colorful
Functions
Series
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
Conclusions
Introduction
Colorful
Functions
Series
A polynomial of degree three – 1
• f (z) = (z − 2)(z + i)(z + 2 − i)
Conclusions
Introduction
Colorful
Functions
Series
A polynomial of degree three – 2
• f (z) = (z − 2)(z + i)(z + 2 − i)
Conclusions
Introduction
Colorful
Functions
And yet another one
• f (z) = (z − 2)(z + 1)2
Series
Conclusions
Introduction
Colorful
Functions
Exponential – 1
• f (z) = exp z
Series
Conclusions
Introduction
Colorful
Functions
Exponential – 2
• f (z) = exp z
Series
Conclusions
Introduction
Colorful
Functions
Sine
• f (z) = sin z
Series
Conclusions
Introduction
Colorful
Functions
Square root – 1
• f (z) =
√
z (principal branch)
Series
Conclusions
Introduction
Colorful
Functions
Square root – 2
• f (z) =
√
z (principal branch)
Series
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
Square root branches – animation
• Invert the square function restricted to different domains
• Where is the (branch) cut / the jump?
Im
Im
Re
Re
Introduction
Colorful
Functions
Series
Square root branches – animation
Please download video by clicking here.
Or use this url:
http://locsi.web.elte.hu/complex/doc/k_video2_gyokagak.avi
Conclusions
Introduction
Colorful
Functions
Logarithm
• f (z) = log z (principal branch)
Series
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
Logarithm branches – animation
• Invert the exponential function restricted to different domains
• Where is the cut / jump?
Im
Im
Re
Re
Introduction
Colorful
Functions
Series
Logarithm branches – animation
Please download video by clicking here.
Or use this url:
http://locsi.web.elte.hu/complex/doc/k_video3_logagak.avi
Conclusions
Introduction
Colorful
Functions
Inversion – 1
• f (z) = 1/z
Series
Conclusions
Introduction
Colorful
Functions
Inversion – 2
• f (z) = 1/z
Series
Conclusions
Introduction
Colorful
Functions
Reciprocal
• f (z) = 1/z = (1/z)
Series
Conclusions
Introduction
Colorful
Functions
A pole of order two – 1
• f (z) = 1/z 2
Series
Conclusions
Introduction
Colorful
Functions
A pole of order two – 2
• f (z) = 1/z 2
Series
Conclusions
Introduction
Colorful
Functions
Series
Of singularities
Laurent series
+∞
X
k=−∞
ck (z − a)k
Order of the pole ∼ smallest (negative) index of terms with
non-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 .
Conclusions
Introduction
Colorful
Functions
Series
Of singularities
Laurent series
+∞
X
k=−∞
ck (z − a)k
Order of the pole ∼ smallest (negative) index of terms with
non-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 .
Conclusions
Introduction
Colorful
Functions
An essential singularity
• f (z) = cos
1
z
Series
Conclusions
Introduction
Colorful
Functions
A linear fraction
• f (z) = (z − 2)/(z + 2) (Zhukovsky)
Series
Conclusions
Introduction
Colorful
Functions
Table of Contents
Introduction
’Colorful’ plots
Discussion of some functions
Function series
Conclusions
Series
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
n=
zk
k!
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=0
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=1
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=2
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=3
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=4
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=5
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=6
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=7
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=8
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n=9
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n = 10
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of exp
• Tn (z) =
n
P
k=0
zk
k!
n = 11
= 1 + z + 12 z 2 +
1 3
3! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=0
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=1
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=2
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=3
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=4
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=5
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=6
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Taylor series of sin
• Tn (z) =
n
P
k=0
(2k+1)
z
=z−
(−1)k (2k+1)!
n=7
1 3
3! z
+
1 5
5! z
+ ...
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Series
Another series of functions
• M0 (z) = z
Mn+1 (z) = (Mn (z))2 + z
Conclusions
Introduction
Colorful
Functions
Table of Contents
Introduction
’Colorful’ plots
Discussion of some functions
Function series
Conclusions
Series
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
• A small (non-complete) discussion of complex functions
• Idea and validity of ’colorful’ visualization
• No implementation known with different colorings available
and easy to use
• Matlab implementation, Gabor transforms
• Exam ,
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
• A small (non-complete) discussion of complex functions
• Idea and validity of ’colorful’ visualization
• No implementation known with different colorings available
and easy to use
• Matlab implementation, Gabor transforms
• Exam ,
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
• A small (non-complete) discussion of complex functions
• Idea and validity of ’colorful’ visualization
• No implementation known with different colorings available
and easy to use
• Matlab implementation, Gabor transforms
• Exam ,
Conclusions
Introduction
Colorful
Functions
Exam – Question 1
Series
Conclusions
Introduction
Colorful
Functions
Exam – Question 1
Logarithm
f (z) = log z
Series
Conclusions
Introduction
Colorful
Functions
Exam – Question 2
Series
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
Exam – Question 2
A polinomial of degree five with one root of multiplicity two and
three roots of multiplicity one
3
3
i)(z + 1 − 10
i)
f (z) = i(z − i)2 (z + i)(z − 1 − 10
Introduction
Colorful
Functions
Exam – Question 3
Series
Conclusions
Introduction
Colorful
Functions
Exam – Question 3
Tangent
f (z) = sin z/ cos z
Series
Conclusions
Introduction
Colorful
Functions
Exam – Question 4
Series
Conclusions
Introduction
Colorful
Functions
Series
Exam – Question 4
A polinomial√of degree three
√
f (z) = (z − i)(z − 23 + 12 i)(z + 23 + 12 i)
Conclusions
Introduction
Colorful
Functions
Exam – Question 5
Series
Conclusions
Introduction
Colorful
Functions
Series
Conclusions
Exam – Question 5
A function with a zero of multiplicity one and a pole of order two
f (z) = 1/z 2 − iz
Introduction
Colorful
Functions
Exam – Question 6
Series
Conclusions
Introduction
Colorful
Functions
Exam – Question 6
f (z) = exp cos z
Series
Conclusions
Introduction
Colorful
Functions
Exam – Question 6
f (z) = exp cos z
Series
Conclusions
Introduction
Colorful
Functions
What are these?
Series
Conclusions
Introduction
Colorful
Functions
Series
References
• See WWW
• References on my homepage
(as soon as it’s translated and broken links are fixed)
Conclusions
Introduction
Colorful
Functions
Series
Colorful visualization of complex functions
Author:
Occasion:
Web:
E-mail:
Levente Lócsi
NuHAG Seminar
Vienna, April 8, 2011
http://locsi.web.elte.hu/complex/
[email protected]
Conclusions