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An Elementary Note: Contour & Virtual Integrals of Implicit Complex Functions - Images
John Gill July 2015
Consider the following problem: Evaluate ( )I z dzγ
ω= ∫ where both ( )z t and ( )zω are
defined implicitly, and :0 1t → . Assume the functions here are differentiable with respect to t
and (1) ( ), 0z t =F and (2) ( ) ( )f zω =G .
Solve for ( )0,0 0z =F and
0 0( ) ( )f zω =G using traditional technique.
Differentiate both expressions with respect to t , producing
(1a) ( , )z
dzz t
dtϕ= and (2a) ( ), ,
dz t
dtω
ωϕ ω= . Now convert to discrete formats:
(1b) ( )1, 1, 1,
, kk n k n n z k n nz z zη ϕ −
− −= + ⋅ and (2b) ( )1, 1, 1, 1,, , kk n k n n k n k n n
zωω ω η ϕ ω −− − −= + ⋅ , 1 k n≤ ≤ .
Define (3) ( ), ,
1
1( , ) ,
n
kn n k n z k n nn
k
S zn
η ω ϕ η=
= ⋅ ⋅ =∑ , Thus limn
nI S
→∞= .
In a force field : ( )F f f z= the work done along a contour γ is given by
1
0
Re ( ( )) '( )W f dz f z t z t dtγ
= =∫ ∫�
With the formula holding when both ( )f z and ( )z t are given implicitly.
The Zeno contours , 1, 1, 1,
( , )kk n k n n z k n nz z zη ϕ −
− −= + ⋅ , where ( , )z
d d dzz t
dt dt dt
ζϕ= ⇒ =
F, can be
expanded linearly:
1
0 1 1, 0 0 0 0 0 0
0
( , ) ( , ) ( , ) ( , )nn n n n nn n nz z z z z z z t dtη ψ η ψ η ψ ψ−= + + + + ≈ + ∫� , with
0 ,
1( , ) : ( , ) , k k
z k n nn nz z
nψ ϕ η= = . I call this a virtual integral [1] due to the fact that the integrand
normally cannot be precisely stated although its value can be easily computed. However,
when ( )z t can be derived in closed form, 0( , ) ( ( ), )z t z t tψ ϕ= and
( )1 1
0 0
( ) ( , ) ( ), (1) (0)z z t dt z t t dt z zλ ψ ϕ= = = −∫ ∫
Example 1 : 2( ) zz z e t it= + = +F ,
21( )
2i zω ω ω= − + =G .
2( , )
1z z
t iz t
eϕ
+=
+,
2
( ) (1 )zt i
i eωϕ
ω
+=
− + ⋅ +.
1
0
( ) ( ( )) ( ) .1373 .2264I z dz z t z t dt iγ
ω ω ′= = ⋅ ≈ − +∫ ∫
The ( )zF contour is in green. Integral value is the red vector. And the blue contour is the
following: 0
( ) ( ( )) ( )
t
C t z z dω τ τ τ′= ∫ , :0 1t → .
Example 2 : ( ) 2 2, ( ) 0 , ( ) 4 2 1z t z t t t itζ ζ= − = = + +F ,
( ) ( 2 ) (2 1) 0 , ( )u v i u z x iy u ivω ω= + + + − = + = +G . 0 0
1 , .5 .75z iω= = − + .
1
0
( ) ( ( )) ( ) .7803 1.0858I z dz z t z t dt iγ
ω ω ′= = ⋅ ≈ − +∫ ∫
Example 3 : 2 2( , ) (2 1) 0z t z t it= − − + =F , ( ) 0e i zωω ω ω= + − =G .
( ) .9713 1.5080z dz iγ
ω ≈ − −∫
Example 4 : 2:F i zω ω+ = and
2( , ) 0 , .6742 .0725zz t ze c t c i= + − = ≈ −F , giving
0 0.5 , .5 .5z i iω= − + = + ⇒ .3612W f dz
γ
= ≈ −∫ � . The red VF is the force field F , the green
VF is the field of contours.
The following are simple topographical images of ( )zλ where ( )z t is unavailable or tedious to
unscramble.
Example 5 : 3 2( , ) ( ) 0z t iz t it cΦ = − + + = ,
2
2( , )
3
t iz t
izϕ
+= .
1
0
( ) ( , )z z t dtλ ψ= ∫ :
(The z in the integral is actually 0z )
-1.2<Re(z)<1.2 , -1.2<Im(z)<1.2
Example 6 :2( , ) ( ) ( ) 0z t Sin zt iz t it cΦ = − − + + = ,
2 ( )( , )
( )
t i zCos ztz t
tCos zt iϕ
+ −=
−.
1
0
( ) ( , )z z t dtλ ψ= ∫ :
-15<Re(z)<15 , -15<Im(z)<15
Example 7 : 2 2( , ) ( ) 0zz t zte z t it cΦ = − − + + = ,
2( , )
(1 ) 2
z
z
t i zez t
te z zϕ
+ −=
+ −.
1
0
( ) ( , )z z t dtλ ψ= ∫ :
-15<Re(z)<10, -15<Im(z)<15
Example 8 : 3 2( , ) ( ) 0z t z t iz t it cΦ = − − + + = ,
3
2
2( , )
3
t i zz t
z t iϕ
+ −=
−.
1
0
( ) ( , )z z t dtλ ψ= ∫ :
-5<Re(z)<5, -5<Im(z)<5
Example 9 : 2 2( , ) ( ( 1)) ( ) 0z t Ln z t z t it cΦ = + − − + + = ,
2
2 1 /( 1)( , )
1 2
t i tz t z
zϕ
+ − += ⋅
−.
1
0
( ) ( , )z z t dtλ ψ= ∫ : -2<Re(z)<2, -2<Im(z)<2
Example 10 : ( ) ( ) ( ) 2, ; ( ) ( ) ( ) 0x y t x Sin y i y Cos x t it cΦ = + + − − + + = ,
2 ( ) 1 2 ( )( , )
1 ( ) ( ) 1 ( ) ( )
t Cos y tSin xz t i
Sin x Cos y Sin x Cos yϕ
− −= + ⋅
− −.
1
0
( ) ( , )z z t dtλ ψ= ∫ :
-10<Re(z)<10, -10<Im(z)<10
Example 11 : ( ) 2, ; ( ) ( ) ( ) 0x y t xCos y iySin x t it cΦ = + − + + = ,
1
0
( ) ( , )z z t dtλ ψ= ∫ :
10<Re(z)<10, -10<Im(z)<10
Compositions of Sets of Implicit Contours
Suppose the set of Zeno contours arising from an implicit function is designated:
[ ] ( ): , 0 ( , )m m m
dzz t z t
dtγ ϕΦ = ⇒ =
Then compose sets of contours in the following manner:
[ ] [ ] ( ) *
1 2 1 2: ( , ), ( , )dz
z t t z tdt
γ γ ϕ ϕ ϕ= =�
This leads to virtual integrals
1
* *
0
( ) ( , )z z t dtλ ψ= ∫
Example 12 : [ ] ( ) 3 2
1 1: , ( ) 0z t z t itγ ζΦ = − + + = , [ ] ( ) 2
2 2: , ( ) 0zz t te z t itγ ζΦ = + − + + =
1 22
2 2 ,
3 1
z
z
t i t i e
z teϕ ϕ
+ + −⇒ = =
+.
1
* *
0
( ) ( , )z z t dtλ ψ= ∫ :
-.5<Re(z)<2 , -1<Im(z)<2.5
Example 13 : [ ] ( ) 2 2
1 1: , ( ) 0z t tz z t it cγ Φ = + − + + = , [ ] ( ) 2 2
2 2: , ( ) 0ztz t e iz t it cγ Φ = − − + + =
2
1 2
2 2 ( , ) , ( , )
2 1 2
tz
tz
t i z t i zez t z t
tz te ziϕ ϕ
+ − + −⇒ = =
+ −.
1
* *
0
( ) ( , )z z t dtλ ψ= ∫ :
-3<Re(z)<8 , -10<Im(z)<10
Integrals of Contours
Suppose for an implicit function ( ), 0z tΦ = involving an undetermined constant c we consider
the set of contours { ( )z t } . Individual contours are determined by 0
( )c zζ ζ= = so that we
may graph
1
0
( ) ( , )z t dtν ζ ζ= ∫ topographically over a part of the ζ -plane (which lies
superimposed over the z-plane).
Example 14 : 3 2( , ) ( ) 0z t iz t it cΦ = − + + = (example 5).
1
0
( ) ( , )z t dtν ζ ζ= ∫ :
-1.2<Re(z)<1.2 , -1.2<Im(z)<1.2
Example 15 : 3 2( , ) ( ) 0z t z t iz t it cΦ = − − + + = (example 8).
1
0
( ) ( , )z t dtν ζ ζ= ∫ :
-5<Re(z)<5, -5<Im(z)<5
Implicit Contour Integrals
We wish to evaluate the integral ( )f z dzγ∫ if we are given ( ) ( )f z f x iy= + but the contour γ
lies implicit (and unrevealed) in ( ), 0z tΦ = :
Example 16 : Given ( ) ( ) sin( )f x iy xCos y iy x+ = + and the contour derived from
2( ) 0zte z t it ζΦ = + − + + = , 0
(0)z z= , evaluate I = ( )f z dzγ∫ .
The last condition exists when 0zζ = . We have
( ) ( )(1) 1
0
( ) ( ) ( ) ( ) ( ), ( )
z
f x iy dz f x t iy t x t y t dtζ
ϑ ζ ϕ= + = +∫ ∫ where 2
1
z
z
t i e
teϕ
+ −=
+. A simple
topographical image results from graphing ( )ϑ ζ over part of the complex plane.
-10<Re(z)<10, -10<Im(z)<10
[1] J. Gill, Informal Notes: Zeno Contours, Parametric Forms, & Integrals , Comm. Anal. Th. Cont. Frac., Vol XX
(2014)