Komplekse tall og funksjoner

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Transcript Komplekse tall og funksjoner

Complex numbers and function
- a historic journey
(From Wikipedia, the free encyclopedia)
Contents
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Complex numbers
Diophantus
Italian rennaissance mathematicians
Rene Descartes
Abraham de Moivre
Leonhard Euler
Caspar Wessel
Jean-Robert Argand
Carl Friedrich Gauss
Contents (cont.)
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Complex functions
Augustin Louis Cauchy
Georg F. B. Riemann
Cauchy – Riemann equation
The use of complex numbers today
Discussion???
Diophantus of Alexandria
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Circa 200/214 - circa 284/298
An ancient Greek mathematician
He lived in Alexandria
Diophantine equations
Diophantus was probably a Hellenized
Babylonian.
Area and perimeter problems
• Collection of taxes
• Right angled triangle
• Perimeter = 12 units
• Area = 7 square units
?
Can you find such a triangle?
• The hypotenuse must be (after some
calculations) 29/6 units
• Then the other sides must have sum =
43/6, and product like 14 square units.
• You can’t find such numbers!!!!!
Italian rennaissance
mathematicians
• They put the quadric equations into three
groups (they didn’t know the number 0):
• ax² + b x = c
• ax² = b x + c
• ax² + c = bx
Italian rennaissance
mathematicians
• Del Ferro (1465 – 1526)
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Found sollutions to: x³ + bx = c
• Antonio Fior
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Not that smart – but ambitious
• Tartaglia (1499 - 1557)
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Re-discovered the method – defeated Fior
• Gerolamo Cardano (1501 – 1576)
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Managed to solve all kinds of cubic equations+ equations of degree four.
• Ferrari
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Defeated Tartaglia in 1548
Cardano’s formula
x  px  q  0
3
x
3
p 3
q 2 q 3
( ) ( )  
3
2
2
p 3
q 2 q
( ) ( ) 
3
2
2
Rafael Bombelli
Made translations of
Diophantus’ books
Calculated with
negative numbers
Rules for addition,
subtraction and
multiplication of complex
numbers
A classical example using
Cardano’s formula
x  15x  4  0
3
Lets try to put in the number 4 for x
64 – 60 – 4 = 0
We see that 4 has to be the root (the positive root)
Cardano’s formula gives:
x
3
 121  2 
3
(Cont.)
 121  2
Bombelli found that:
3
 121  2   1  2
3
 121  2   1  2
WHY????
(Cont.)
(i  2)  i  3  i  2  3  i  2  2  i  6i  12i  8
3
3
2
i  1
2
3
3
2
i  1
2
(i  2)  i  6  12i  8  11i  2
3
11i  11  1   11   121
2
(i  2)   121  2
3
3
 121  2  (i  2)  i  2   1  2
3
3
Rene Descartes (1596 – 1650)
• Cartesian coordinate
system
• a + ib
• i is the imaginary unit
• i² = -1
Abraham de Moivre (1667 - 1754)
• (cosx + i sinx)^n = cos(nx) + i sin(nx)
• z^n= 1
• Newton knew this formula in 1676
• Poor – earned money playing chess
Leonhard Euler 1707 - 1783
• Swiss mathematician
• Collected works fills
75 volumes
• Completely blind the
last 17 years of his
life
Euler's formula in complex
analysis
Caspar Wessel (1745 – 1818)
• The sixth of fourteen children
• Studied in Copenhagen for a law degree
• Caspar Wessel's elder brother, Johan Herman
Wessel was a major name in Norwegian and
Danish literature
• Related to Peter Wessel Tordenskiold
Wessels work as a surveyor
• Assistant to his brother Ole Christopher
• Employed by the Royal Danish Academy
• Innovator in finding new methods and
techniques
• Continued study for his law degree
• Achieved it 15 years later
• Finished the triangulation of Denmark in
1796
Om directionens analytiske
betegning
On the analytic representation of direction
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Published in 1799
First to be written by a non-member of the RDA
Geometrical interpretation of complex numbers
Re – discovered by Juel in 1895 !!!!!
Norwegian mathematicians (UiO) will rename
the Argand diagram the Wessel diagram
Wessel diagram / plane
Om directionens analytiske
betegning
• Vector addition
Om directionens analytiske
betegning
• Vector multiplication
An example:
11 
11  

z1  8 cos
 i sin

12
12 

1
19  
19   
z 2   cos
 i sin

2
12
12 
The modulus is:
1
8  4
2
(Cont.)
The argument is:
11   19   30   5  




 2 
12
12
12
2
2
Then (by Wessels discovery):



z1 z 2  4 cos  i sin   4(0  i)  4i
2
2

Jean-Robert Argand (1768-1822)
• Non – professional mathematician
• Published the idea of geometrical
interpretation of complex numbers in 1806
• Complex numbers as a natural extension
to negative numbers along the real line.
Carl Friedrich Gauss (1777-1855)
• Gauss had a
profound influence in
many fields of
mathematics and
science
• Ranked beside Euler,
Newton and
Archimedes as one of
history's greatest
mathematicians.
The fundamental theorem of
algebra (1799)
Every complex polynomial of degree n
has exactly n roots (zeros), counted with
multiplicity.
If:
(where the coefficients a0, ..., an−1 can be
real or complex numbers), then there exist
complex numbers z1, ..., zn such that
Complex functions
• Gauss began the development of the
theory of complex functions in the
second decade of the 19th century
• He defined the integral of a complex
function between two points in the
complex plane as an infinite sum of the
values ø(x) dx, as x moves along a curve
connecting the two points
• Today this is known as Cauchy’s integral
theorem
Augustin Louis Cauchy (1789-1857)
• French mathematician
• an early pioneer of
analysis
• gave several
important theorems in
complex analysis
Cauchy integral theorem
• Says that if two different paths connect the
same two points, and a function is holomorphic
everywhere "in between" the two paths, then the
two path integrals of the function will be the
same.
• A complex function is holomorphic if and only if
it satisfies the Cauchy-Riemann equations.
The theorem is usually formulated for closed
paths as follows: let U be an open subset of C
which is simply connected, let f : U -> C be a
holomorphic function, and let γ be a path in U
whose start point is equal to its end
point. Then
Georg Friedrich Bernhard
Riemann (1826-1866)
• German
mathematician who
made important
contributions to
analysis and
differential geometry
Cauchy-Riemann equations
Let f(x + iy) = u + iv
Then f is holomorphic if and only if u and
v are differentiable and their partial
derivatives satisfy the Cauchy-Riemann
equations
and
The use of complex numbers today
In physics:
Electronic
Resistance
Impedance
Quantum Mechanics
…….
hz   3x  2x  3 y  1  i6xy  2 y
2
2
u = 3x  2 x  3 y  1
2
V=
6 xy  2 y 
2