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Elementary complex functions
Working with complex functions we will need less elementary functions than in the real
case. Our two basic functions are the exponential function, and its “inverse”, i.e. the
logarithm.
i z x iy re
q = + = - p < q £ p
The exponential function is defined by its series expansion:
(^1 2 ) 1 2! 3!
z e = + z + z + z + ◊◊◊ ,
which by use of Eulers formula leads to
(cos sin )
z x iy x e = e e = e y + i y.
The logarithm is given by:
ln z = ln r + i q + i k 2 p ,
where k is an integer (positive or negative). Thus, the logarithm is multivalued, with a
distinct “branch” for each value of k.
General power:
c c ln z z = e ,
where c can be any number (complex or real).
n’th root:
1 1 ln (^2 ) cos sin
z n n n k^ k z e r i n n
È^ q^ +^ p^ q^ + p ˘ = = + Í ˙ Î ˚
with n distinct values for k = 0,1, ◊◊◊, n - 1.
The trigonometric, hyperbolic and inverse trigonometric functions can be expressed in
terms of the exponential- and logarithmic functions. Thus, the number of basic
elementary functions is reduced in the complex case.
Complex power series
The complex power series
0 0
n n n
a z z
=
Â^ -
is according to the ratio criterion for convergence convergent if
1 1 0 0 0 1
1 , or ( )
n n n n n n
a z z a z z a z z a
Now, assume that
1
when
n
n
a R n a (^) +
Æ Æ •.
R is called the radius of convergence for the power series, and we have convergence for
z - z 0 < R ,
i.e. for all z within a radius R around z 0.
Analytic functions
A complex function f(z) is analytic in a point z 0 if it has a derivative in z 0 ,and in all
points in a neighborhood around z 0.
Neighborhood: All points z so that z - z 0 < r , i.e. all points within a radius r around z 0.
Domain(D): Open, connected set of points in the complex plane.
Cauchy-Riemann equations:
Assume that f(z)=u(x,y)+iv(x,y) is analytic in a domain D. Then we have the following
relations between its real and imaginary parts:
( , ) ( , ) ( , ) ( , ) ,
u x y v x y u x y v x y
x y y x
From these equations follows that
2 2
2 2
u x y u x y
x y
and similarly for v(x,y). This means that the real or imaginary part of any analytic
function is a solution of the Laplace equation in two dimensions.
Singular point (singularity) of a function f(z):
A point in the complex plane where f(z) is not analytic.
Isolated singular point z 0 :
f(z) is analytic in a neighbourhood around z 0 , but not in z 0 itself.
Example: f(z)=1/(1-z), z 0 =1.
We may encounter two types of isolated singularities:
Poles: Assume that
0 0
lim ( ) ( )
n z z f z z z
È - ˘
Î ˚
Æ
exists, and is different from zero. Then f(z) has a pole of order n in z 0.
Example: The function
3
2
z z f z z
has poles of order one in z=i and z=-i.
Essential singularity: The limit discussed above does not exist for any finite n.
Example:
1
( )
z f z = e. Essential singularity in z=0.
Finally, we may also have removable singularities: Assume that f(z 0 ) does not exist, but
f(z)Æa as zÆz 0. Then we set f(z 0 )=a, and the singularity is removed.
Noen kommentarer ang. Laurent-rekka, nullpunkter og poler.
Teorem: Enhver funksjon f(z) som er analytisk for R 1 <|z-z 0 |<R 2 kan utvikles i en rekke
av typen
0 1 0
n n n n n (^) C
f z f z a z z a dz p i z z
=-•
 (^) Ú
C er en enkeltlukket integrasjonsvei i området R 1 <|z-z 0 |<R2. Merker oss at for n>0 er an
gitt ved
( ) 0
( ) (Taylorrekka). !
n an f z n
Dette generelle uttrykket for an er imidlertid ofte lite nyttig for å bestemme rekka. Vi
fortsøker i stedet å omskrive f(z) ved hjelp av kjente rekker, spesielt
0 0 0 0 0
0
z z z z z n
n
n
n
e e e e z z n
z
z b b z b b
b
=
=
Â
Â
Av rekka for e
z følger også rekker for de trigonometriske og hyperbolske funksjonene.
For på finne rekker for de inverse trigonometriske og hyperbolske funksjonene må vi ha
en rekke for logaritmen lnz. Dette er som kjent en flertydig funksjon , og vi gjør kanskje
best i å unngå dette problemet!
Den siste rekka ovenfor er nyttig når vi skal finne rekker for rasjonale funksjoner av
typen
1
( ) ( N ) ( )
g z g z f z h z z z z z
der polynomet h(z) antas faktorisert. Delbrøkoppspalting gir så funksjoner av typen
F(z)/(z-b), der F(z) er et polynom. Ønsker vi en rekke om et annet punkt enn origo, setter
vi
0
n
n
z a
z b z a c z^ a c c c c
=
Â
Dermed har vi fått en rekke om a (b=a+c) som er konvergent for |z-a|<|c|. Tegn figur!
En rekke som er konvergent for |z-a|>|c| finner vi på denne måten:
1 0 0
n n n n n
c c
z b z a c z a c z a z a z a
z a
= =
 Â
Eksempel:
0
( ). Finn Laurent-rekka om z =1. 3
f z z
0
n
n
z f z z z
=
Â
og rekka er konvergent om z=1 for |z-1|<2. Dette betyr at vi med denne rekka om z=
ikke kan passere singulariteten for f(z) ved z=3. Tegn figur!
For å finne en rekke som konvergerer for |z-1|>2 setter vi
1 0 0
n n n n n
f z z z z z z
z
= =
 Â
Vi ser da at f(z) i dette tilfelle har en Taylorrekke om z=1 som er riktig ut til
singulariteten ved z=3. Utenfor punktet z=3 finner vi en mer generell Laurent-type rekke
med bare negative eksponenter. Den delen av Laurentrekka som har negative eksponenter
kaller vi gjerne hoveddelen, eller prinsipaldelen.
Nullpunkter og poler
Vi antar at f(z) har en Laurentrekke som vi skriver
0 (^0 1 )
n (^) n n n n n
b f z a z z z z
= =
 Â
f(z) har da et nullpunkt av orden m for z=z 0 hvis alle bn=0 (Taylorrekke), an=0 for n<m,
og am π 0. Rekka for f(z) om z 0 har da formen
n n n m
f z a z z m
=
= (^) Â - >
Som regel vil vi forsøke å finne orden av et nullpunkt uten først å bestemme
rekkeutviklingen. Vi merker oss da at hvis f(z) har et nullpunkt av orden m, så vil vi ha
0 0
lim( ) ( )
k m
k m f z z z a k m z z k m
Ï^ <
Ô
Æ = Ì =
Ô
^ >
Vi ser at vi må forsøke oss litt frem med ulike valg av k til finner en endelig grense som
ikke er lik null.
Eksempel:
3
3
sin ( ) ( ).
z f z z
p
f(z) har nullpunkter for z=n, der n kan være alle hele positive og negative tall. For z=
har f(z) grenseverdien p
3 , dvs. ikke nullpunkt. Etter litt regning vil vi finne:
3
3
3 3
3
( ) sin ( ) lim( ) lim( 0) ( ) ( ) ( )
sin ( ) =( 1) lim( 0) ( 1) 3 ( ) ( )
k k
n k n k
f z t n z n t z n t z n t n t
k
t t k t n t n
k
p p
p p p p
Æ = Æ - =
Ï <
Ô
Ô Ê ˆ
- Æ = Ì - Á ˜ =
+ Ë ¯
Ô
ﱥ >
Ó
Altså har f(z) nullpunkt av orden 3 for z=n (n (^) π 0).
1 1
( ) 2 Re ( ) Re ( )
n m
k l k l
P f x dx pi sf z pi sf z
-• =^ =
= (^) Â + Â Ú
where zk are singular points (poles) in the upper half plane ( Im zk > 0 ), and zl are simple
poles on the real axis.
Fourier series and Fourier transforms
A function f(x) is expanded in a Fourier series in the region [-L,L]:
L
n x b L
n x a
a f x
n
n n
n
p p cos sin 2
1 1
0 Â Â
=
=
dx L
n x f x L
dx b L
n x f x L
a
L
L
n
L
L
n (^) Ú Ú
p p ( )sin
( )cos ,
We notice that if f(x) is odd, i.e. f(-x)=-f(x), then an=0 for all n, whereas an even function
f(-x)=f(x) implies bn=0 for all n.
We also note that the Fourier series is periodic with the period 2L, i.e. f(x+2L)=f(x).
Dirichlets conditions : Sufficient conditions for the Fourier series associated with f(x) to
be equal to f(x) (converge to f(x)) is that f(x) is only allowed to have a finite number of
finite discontinuities, and a finite number of maxima and minima in the interval [-L,L].
If f(x) has a (finite) discontinuity at x=x 0 , the Fourier series gives the result
[ ( ) ( )] 2
lim( e Æ 0 ) f x 0 - e + f x 0 + e for f(x) at x=x 0.
Assume that we just know f(x) in the region [0,L], and want to expand it in a Fourier
series that is valid in this region. Then we may make an arbitrary extension of f(x) to the
interval [-L,0], and we still obtain a Fourier series that is valid in the region [0,L], maybe
except for a discontinuity at x=0. The simplest choice for the region [-L,0] is an even
function, then there will be no discontinuity at the origin, and the series will have bn=0.
The Fourier series may be integrated term by term, and also differentiated term by term if
the derivative of f(x) satisfies the Dirichlet conditions.
If we want a Fourier series for f(x) in a general interval [a,b], we change the variable x to
t according to the relation
x = a t + b ,
so that x=a gives t=-L, and x=b gives t=L , i.e. a=(b-a)/2L, b=(a+b)/2. Then we can
expand the function f(x)=f(at+b)=g(t) in the interval tŒ[-L,L], and finally insert
t=(x-b)/a for t in the series. We notice that the Fourier series for a general unsymmetric
interval is not very nice!
Complex Fourier series:
/ ( ) ,
in x L n n
f x c e
p
=-•
= (^) Â
L in x L n L
c f x e dx L
Ú
We recall the orthogonality relation
/ /
L in x L im x L
L
m n e e dx L m n
p - p
Ï^ π = Ì
Ó =
Ú
If the variable is the time t, i.e. x=t, it is convenient to set L=T/2, wT=2p, and npx/L=nwt
The Fourier transform:
The Fourier series expansion is valid only for a finite interval [-L,L], and is periodic with
period 2L. If we want an “expansion” valid for the whole real axis, i.e. [-•,•], we have to
change to the Fourier integral transform
i t f t F e d
w w w p
-•
Ú
F (^) [ f t ( ) (^) ] =
i t F f t e dt
w w p
-•
Ú
We notice that the variable is t, which indicate time, since the Fourier transform is most
frequently used in connection with time and frequency problems.
Furthermore, we notice that if f(t) is an odd function of t, we have a sine transform, and if
f(t) is even the result is a cosine transform (cf. textbook, 7.12).
The transform of the Gauss curve is a frequent example:
2 ( ) , 0 , with the result
t f t ke
a a
= >
2
4 ( ) , 2
k F e
w a w a
=
which means that the transform of a Gauss curve in t is another Gauss curve in w.
Fourier transform of the derivative:
F (^) [ f '( ) t (^) ] =i w F [ f t ( ) (^) ] ,
F (^) [ f ''( ) t (^) ] = - w^2 F (^) [ f t ( ) (^) ].
The Fourier transform of the derivative is in particular useful for solution of differential
equations.
Example: y ''+ ay ' + by = r t ( ).
By making a Fourier transform of the equation we have
2
- w Y ( w ) + iw aY ( w ) + bY ( w ) = R ( w ) ,
with the result 2
( ) , and ( ) ( ). 2
R i t Y y t Y e d i a b
w w w w w w w p
-•
Ú
If the Laplace transform takes the form of a product F(s)G(s), then we know according to
the convolution theorem that we have the transform of a function h(t) given by
Ú
t
ht f gt d
0
() ( t ) ( t ) t ,
where F(s) and G(s) are respectively the transforms of f(t) and g(t). In particular we have
the simple case
Ú
t
gt ht f d s
Gs
0
, i.e. () 1 ,andconsequently () ( )
() t t.
Frequently F(s) takes the form
( )
Qs
Ps F s = , where P(s) and Q(s) are polynomials (The
degree of Q(s) will be higher than that of P(s)). Partial fraction decomposition is then
useful/necessary.
Example:
2 ( 2 )
2
ss s ss s s s s
F s ,
which gives
t t f t e e
2
= - + , from our knowledge that L [ ]
s a
e
at
, also
for a=0.
To solve differential equations we need the Laplace transform of derivatives:
L [ f ' ( t )] = sL [ f ( t )] - f ( 0 ), L [ ]
2
f ' '( t ) = s L [ f ( t )] - sf ( 0 )- f '( 0 )
Example:
y ' '+ y = 2 t
Laplace transform:
2
s L [ y ] - sy ( 0 )- y '( 0 )+ L [ y ] = L [ ]
2
s
t = ,
which gives L [ ] '( 0 )
2 2 2 2
y s
y s
s
s s
y
= , and after some
manipulations (partial fractions)
y ( t )= 2 t +( y '( 0 )- 2 )sin t + y ( 0 )cos t.
We see that we have obtained a unique solution determined by the values of y ' ( 0 )and
y(0).
The Heaviside step-funksjon:
Ó
Ì
Ï
t a
t a H t a 0 for
1 for ( )
L [ ]
as
a
st e s
H t a e dt
Ú
( ) and
L [ ( ) ( )] ( ) ( ) ()
0
( ) H t a f t a e f t adt e f d e Fs
s a sa
a
st -
Ú Ú
t t
t .
Thereby we have the following result: If the Laplace transform takes the form e F ( s )
then this expression is the transform of the function H(t-a)f(t-a).
Example: The Laplace transform is ()
3
3 e F s s
e
which means that
2
2
f ( t )= t , and the transformed function is
2 ( 3 )( 3 ) 2
H t - t -.
The derivative of Laplace transforms:
From the definition
Ú
=
0
F ( s ) e f ( t ) dt
st follows by taking the derivative under the
integral sign that
Ú
=- =-
0
F ( s ) te f ( t ) dt ds
d (^) st
L [ tf ( t )], and in general
n n
n
F s ds
d
( )= (- 1 ) L [ t f ( t )]
n .
Example: f t kt s k
k F ( s ) , () sin 2 2
= , and
2 2 2 ( )
s k
ks
ds
dF s
L [ t sin kt ].
This expression for the derivative may be used to solve some problems that might seem
to be very difficult. For instance the inverse of the transform
t
kt f t k
s F s Arc
sin ( )= cot ,issimply ()=.
Example:
Differential equation y ''( ) t + aty '( ) t + y t ( ) = r t ( ).
We assume for simplicity the conditions y(0)=y’(t)=0. Laplace transformed:
L [ ] [ ]
d dY s ty t sY s y Y s s ds ds
dY s s Y s aY s as Y s R s ds
- = , where Y(s) and R(s) are the Laplace transforms
of y(t) and r(t) respectively. Thus, we first have to obtain Y(s) from a first order diff.
equation, and then invert the solution to get y(t).
This technique of Fourier expansion may obviously be applied to other partial differential
equations as long as the boundary conditions are the same, u(0,t)=u(L,t)=0, for instance
the diffusion equation.
A wave equation in two spatial dimensions x and y takes the form
2 2 2
2 2 2 2
u x y t ( , , ) u x y t ( , , ) 1 u x y t ( , , )
x y c t
Separation of variables, i.e. a solution of the form F(x)G(y)T(t) with boundary conditions
u(0,y,t)=u(L,y,t)=u(x,0,t)=u(x,L,t)=0 yields after a little piece of calculation a solution of
the form
nm ( ,^ , )^ sin^ sin^ [^ nm cos(^ )^ nm sin(^ )]
n x m y u x y t A ct B ct L L
p p = m + m ,
with (^) ( )
2 2 2 2 2
n m L
p m = + , n=1,2,3,,, m=1,2,3,,,
A general solution is then given by the superposition
, 1
( , , ) (^) nm ( , , )
n m
u x y t u x y t
=
= (^) Â.
For a given initial condition u(x,y,0) we have
, 1
( , ,0) (^) nm sin sin
n m
n x m y u x y A L L
p p
=
= (^) Â ,
which is a “double” Fourier series. The coefficients Anm are now obtained from a double
integral (work it out, remember the orthogonality relations for the sines)
2
0 0
( , , 0) sin sin
L L
nm
n x m y A u x y dxdy L L L
Ê ˆ^ p^ p = Á ˜ Ë ¯
Ú Ú
For the initial velocity v(x,y,0) we have
v
, 1
( , , 0) (^) nm sin sin
n m
n x m y x y B c L L
p p m
=
= (^) Â , and the coefficients Bnm are
determined by the double integral
2
0 0
( , , 0)sin sin
L L
nm
n x m y B c v x y dxdy L L L
p p m
Ê ˆ
= Á ˜
Ë ¯
Ú Ú
Thus, once more we have a unique solution for the given boundary and initial conditions.
The Laplace equation in spherical coordinates
In spherical coordinates the Laplace equation reads
2 2 2 2 2 2 2 2
( , , ) ( ) (sin ) ( , , ) 0 sin sin
u r r u r r r r r r
q f q q f q q q q f
È ∂ ∂ ∂ ∂ ∂ ˘
Í ˙
Î ∂^ ∂^ ∂^ ∂^ ∂ ˚
We shall look for a solution that is independent of the angle f, that is 0
u
f
Separation of variables: u r ( , q ) = F r G ( ) ( ) q yields
2 2 2
d F r dF r r r n n F r dr dr
where the separation constant is denoted n(n+1), but n is so far an arbitrary real number.
This differential equation for F(r) is an Euler-Cauchy equation, with solution
1
n Fn r A rn Bn (^) n r
For G(q) one obtains the equation
1 ( ) (sin ) ( 1) ( ) sin
d dG n n G d d
q q q q q q
The solution of this equation is a bit more tricky (cf. lecture). One makes the substitution
x=cosq , and the equation is transformed into the Legendre equation
2 2 2
d g x dg x x x n n g x dx dx
where g(x)=G(q). The Legendre equation is solved in terms of series expansion, but the
series will be divergent for x = ±1 ( q = 0 or p )unless n is an integer, n=0,1,2,3,,,,. The
solution is then a Legendre polynomial Pn(x)=Pn(cosq), and we have a solution
1
( , ) (cos )
n un r A rn Bn (^) n Pn r
q q
È ˘
Í ˙
Î ˚
The most general solution we can obtain by this method is the superposition
1 0 0
( , ) ( , ) (cos )
n n n n (^) n n n n
u r u r A r B P r
q q q
= =
È ˘
Í ˙
Î ˚
 Â.
The unknown coefficients An and Bn have to be determined from boundary conditions.
Example: Assume that we want to find the potential outside a sphere of radius R, where
the potential u(R,q) on the sphere is given. As the total charge on the sphere is expected
to be finite, we must have u(r,q)Æ0 as rÆ•. This requires An=0 all n. Thus
1 0
( , ) (cos )
n n n n
B
u r P r
q q
=
= (^) Â , and in particular 1 0
( , ) (cos )
n n n n
B
u R P R
q q
=
= (^) Â.
To obtain the coefficients Bn we need the orthogonality relation for the Legendre
polynomials (Textbook, chapter 12, eq.(8.4)):
1
1 0
( ) ( ) (cos ) (cos ) sin 2 1
P n x Pm x dx Pn Pm d (^) nm n
p
q q q q d
Ú Ú
This relation yields
1 1 0 0
( , ) (cos ) sin 2 1 2 1
n m m (^) n nm m n
B B
u R P d R m R m
p
q q q q d
=
Ú Â.
Thus, we se that all coefficients Bm are determined from the boundary condition u(R,q),
and there is a unique solution.
If we want the potential inside a spherical shell with radius R and boundary condition
u(R,q), we first notice that in this case we must have Bn=0 all n to avoid a singularity in
the origin (r=0). Hence, the general solution is
0
( , ) (cos )
n n n n
u r q A r P q
=
= (^) Â.
Fourier transform solution of the wave equation
If we make a Fourier transform of the wave equation
2 2
2 2 2
u x t ( , ) 1 u x t ( , )
x c t
we get the the following second order differential equation for the Fourier transform
F ( w , ) t of the solution u(x,t):
2 2 2
d F t c F t dt
w
By the Fourier transform we have taken x as a variable, and t as a parameter, so that
i x F t u x t e dx
w w
p
-•
Ú
The solution for the Fourier transform F ( w , ) t is
i ct i ct F t A e B e
w w w w w
= + , and u(x,t) is given by
( ) ( )
i x
i x ct i x ct
u x t F t e d
A e d B e d
w
w w
w w p
w w w w p p
-•
-• -•
Ú
Ú Ú
Now, it remains to determine the integration constants A(w) and B(w) from the initial
conditions. To make a Fourier transform of u(x,t) we have assumed that u(x,t)Æ0 as
|x|Æ• , which is no problem, as infinity might mean the “border” of the universe.
Given initial conditions could be u(x,0) and v(x,0), where v(x,0) means
v (^0)
( , 0) | t
u x t x t
=
Thus, we have
i x F u x e dx A B
w w w w p
-•
Ú
which is one of two equations we need to obtain separate determinations of A(w) and
B(w). The other equation is obtained in a similar way from v(x,0). First we notice that
dF t u x t (^) i x i x e dx v x t e dx dt t
w (^) w w
p p
-• -•
Ú Ú
. From this relation we
get for t=0 the second equation we need:
0
i x t
dF t v x e dx i cA i cB dt
w w w w w w p
= -•
Ú
Hence, we have obtained two equations that yield a unique solution for A(w) and B(w) in
terms of integrals of the given initial positions u(x,0) and velocities v(x,0). This also
leads to a unique solution u(x,t) in terms of the integrals over A(w) and B(w) given above.
