Complex Variables 2010/2011EEM2046 Engineering Mathematics IV1 Complex Variables Learning Objective: Throughout this chapter, we will study thebasic notion of complex functions of a complex variable(in Section 2) and limits and continuity (in Section 3); analyticity, the Cauchy-Riemann Equations, harmonicity and harmonic conjugacy (in Section 4). Integration in the complex plane, the fundamental theorem of calculus of contour integral, the principle of contour independence, Cauchy’s theorem, and the Cauchy’s integral formula ( in Section 5);Sequences of complex number, series of complex numbers, test ofconvergence, power series, Taylor series, and Laurent series (in Section 6); Residue theory, Cauchy residue thorem, and discuss some applications of the residue calculus in evaluation of improper integrals (in Section 7). Teaching Outcome. A set of tutorial questions and exercises will be provided to extend the understanding of t he basic not ion presented in this chapter. This will provide training to acquire and apply fundamental principles ofmathematical sciences. A mid-term test will be given for testing the students understanding ofthe mathematical concept that have been covered. The test wil l train the students to identify, formulate and apply mathematical principles in problem solving.
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A subset S of is said to be closed if cS is open. For example, the empty set
and the complex plane are open and closed. For any ,r and any fixed
,w the set : A z z w r is closed because its complement is open.
We next introduce the notion of connectedness. Generally, we say that a subset
S of is connected if it is “all in one piece”, that is, if it is not the union of two
nonempty subsets that do not touch each other. More precisely, we said that a subset
S of is said to be disconnected if it is union of two nonempty subsets 1S and 2S ,
neither of which intersects the clouse of the other one. The set S is connected if it is
not disconnected.
There is another notion of connectedness that is important in many situations. A
subset S of is called pathwise connected if any two points in S can be joined by a
polygonal line, which is contiunous curve, consisting of a finite number of straight
line segments jointed end to end, and that lies entirely in .S
pathwise connected not pathwise connected
In fact, if a subset S of is open, then S is pathwise connected if and only if S is connecteed. A nonempty subset S of is called a region if the set S is open and
connected. Consequently, a region is an open set which is pathwise connected.
Here are some useful examples of regions.
The complex plane is a region.
For any ,r and any fixed ,w r -neighborhood of w and deleted r -
neighborhood of w are regions.
For any ,r and any fixed ,w the set : z z w r is a region.
For any fixed ,w and any 1 2,r r with 2 1,r r the open annulus
centered at ,w 1 2( : , ) A w r r is a region.
2. Complex Functions
2.1 Complex Functions of a Complex Variable
Let S and B be nonempty subsets of . A complex function f from S to , B
denoted by : , f S B is a rule which assigns to each element z S and only one
element ,w B we write ( )w f z and call w the image of z under f . The set S is
the domain of , f and the set B is the codomain of . f The set of all images
The elementary algebra of complex functions of a complex variable does not
differ radically from that of real-valued functions of a real variable. Let A and B be
nonempty subsets of . Presented with a pair of complex functions : f A and
: g B , we multiply f by a complex number c to obtain the function ,cf add f
and g to form , g f multiply these functions to produce g f , and take the quotient
. g f Here, we have their definitions:
)())(( z cf z cf
for all , z A
)()())((
),()())((
z g z f z g f
z g z f z g f
for all , z A B and
( / )( ) ( ) ( ) f g z f z g z
with the domain-set : ( ) 0 . z A B g z
Example Let ( ) , f z z ( ) z g z e and ( ) sinhh z z for all . z Determine the
formula of the following function. (i) 4 , f and (ii) . f g h
Solution (i) 2 2( ) ( ) ( ) , f z f z f z z z z and 3 2( ) ( ) ( ) ( ) ( ) ( ) f z f z f z f z f z f z 2 3 , z z z and 4 3 4( ) ( ) ( ) . f z f z f z z
(ii) ( )( ) f g h z ( ) ( ) ( ) sinh z f z g z h z z e z for all . z
A more interesting way of combining functions to manufacture a new function is
the composite function.
Definition 2.1.1 Let A and B be nonempty subsets of . Let : f A and
: g B be complex functions, if the range of f is contained in the domain of ; g
i.e., ( ) , f A B then the composite function of g with , f : g f A is well
defined , and is defined by
))(())(( z f g z f g for all . z A
Example Let : \ 0 f and : g be complex functions defined as
( ) 1 f z z for all \ 0 , z and 2( ) g z z for all . z It is clear that
( \ 0 ) \ 0 , f and since 2 2( ) 2 g z x y i xy for all , z x iy we have
( ) . g Hence ( \ 0 ) , f and g f is well defined and ( )( ) g f x ( ( )) g f x
21 z for all \ 0 . z Find and explain your answer for . f g
2.2 Transformations and Mappings
Properties of real functions of a real variable are often exhibited and visualized by thegraph of the function in 2-dimensitonal Cartesian plane. When w z f )( , where z
and w are complex numbers, no such convenient graphical representation of the
complex function f is available because each of the numbers z and w is located in the
complex plane , rather than on the real line . However one can display some
information about the function by indicating pairs of corresponding points ),( y x z
in the domain of , f and ( , )w u v in the range of . f To do this, it is generally
simpler to draw the z and w planes separately. When a function f is thought of in this
way, it is often referred to as a transformation or mapping . Let us see some examples.
Example
To illustrate some of these ideas, consider ,)( 2 z z f defined on the first quadrant of
the z- plane: ,0 x .0 y Then ( ) ( , ) ( , ) f z u x y iv x y2 2 2 ; x y i xy
i.e.,22),( y x y xu and .2),( xy y xv
Since x0 and ,0 y it follows from the form of u and v that
u and .0 v Since we cannot draw a graph of ,)( 2 z z f and we
do not wish to plot the surface ,),( 22 y x y xu .2),( xy y xv A more appealing
and commonly used device is to display the images of representative curve. For
instance, the image of y x 0 ,1 is given parametrically by ,1 2 yu
yv 2 or eliminating the parameter y, by the parabola .0 ,41 2 vvu
Similarly, the image of x y 0 ,1 is given by the parabola ,122
vu
v0 , and their images are shown.
3. Limits and Continuity
3.1 Limits of Complex Functions
In calculus of one real variable, the reader learned the notion of the limit of a function
as well as the definition of continuity as applied to real variables. These concepts
apply with some modification to complex functions of a complex variable. Let
0 , z and let be an open set containing 0 . z Let : f be a complex
function defined on , except possibly at 0 z itself. We say, roughly speaking, thatthe number 0w is the limit of the function )( z f as z approaches 0 z if )( z f stays
Example For any 0 , z show that ( ) sin f z z is continuous at 0 . z
Solution Let , z x iy , , x y and let 0 0 0 . z x iy Then ( ) sin( ) f z x iy
sin cosh (cos sinh ). x y i x y Clearly, Re sin cosh f x y and Im cos sinh f x y are
continuous functions at 0 0( , ). x y Therefore, ( ) sin f z z is continuous at 0 . z Similarly, we can show that cos z is continuous at any 0 . z
Theorem 3.3.2 Let 1 and 2 be open subsets of containing ,0 z and let .c
If both functions 1: f and 2: g are continuous at ,0 z then
,cf , g f , g f , g f ,Re f ,Im f , f and , f
are continuous at 0 z provided that 0( ) 0 g z in the case of . g f
Example Show that 2 3 3( ) sin 5 z z z z is continuous for all . z
Solution Clearly 2( ) f z z , ( ) sin g z z and 3( )h z z are continuous for all . z
Thus, 2( )( ( )) 5 ( ) f z g z h z 2 3 3sin 5 z z z is continous for all . z
Theorem 3.3.3 Let 1 be an open set containing ,0 z and let 2 be a set . If
1: f is continuous at 0 z , and if 2: g is continuous at )( 00 z f w with
1 2( ) , f then ( )( ) g f z is continuous at 0 ; z i.e., the composition of two
continuous functions is continuous function.
Example Show that 3( ) sin(5 2) z z is continuous for all . z
Solution Clearly, 3( ) 5 2 f z z and ( ) sin g z z are continuous for all z , and
( ) . g f D Thus, 3( ) ( )( ) sin(5 2) z g f z z is continous for all . z
As a side remark, if a function that is not continuous at a given point 0 z is said to
be discontinuous at 0 , z that means, either ( ) f z is not well defined at the point 0 z z ,
or limit does not exist at 0 , z or the limit exists at 0 z but is different from 0( ) f z , or the
limit exists at 0 z but ( ) f z is not well defined at the point 0 . z z For the third and
final case, it is called a removable discontinuity, because if we simply redefine thefunction at 0 z to equal its limit then we will have a continuous function there. If the
discontinuity at a point cannot be removed, then it is called a nonremovable
discontinuity.
Example
Consider the complex function 2 1( ) , z i
z f z \ , . z i i It is easy to verify that
2
1( )( ) 21
lim ( ) lim lim lim . z i z i i z i z i z i z z i z i z i z i
Thus both limits exist, and since the point 0 z is arbitrary chosen, so we conclude that
'( ) 1 g z and '( ) 2 f z z for all . z
Example Show that the function ( ) f z z is not differentiable at 0. z
Solution Firstly, we consider z approaches the origin along the real axis:
0 ( ,0) (0,0) 0
( ) (0) 0lim lim lim 1.
0 0 z x x
f z f x i x
z x i x
We next consider an approach toward origin along the imaginary axis:
0 (0, ) (0,0) 0
( ) (0) 0
lim lim lim 1.0 0 z y y
f z f iy y
z iy y
Since the limit is unique, we must conclude that limit does not exist. So, ( ) f z z is
not differentiable at 0. z
Theorem 4.1.2 Let 1 and 2 be open subsets of containing a point ,0 z and let
.c Let 1: f and 2: g be complex functions. Then we have the
following properties:
(i) If f is differentiable at 0 z then it is continuous at .0 z
(ii) If f and g are differentiable at 0 z , then their ,cf sum , g f product , g f and quotient , g f where ,0)( 0 z g are also differentiable at ,0 z and
0 0
0 0 0
0 0 0 0 0
0 0 0 0
0 02
0
( ) '( ) '( ),
( ) '( ) '( ) '( ),
( ) '( ) '( ) ( ) ( ) '( ),
'( ) ( ) ( ) '( )( ) , ( ) 0.
( ( ))
cf z cf z
f g z f z g z
f g z f z g z f z g z
f z g z f z g z f z provided g z
g g z
Furthermore, we obtain
1 0 0 0
0
0 02
0
( ) ( ) '( ) 2,
'( )1( ) ( ) 0.
( )
n n f z n f z f z for any positive integer n
f z z provided f z
f f z
Example Show that for any ,n and ,c 1n ndcz dz
cnz for all . z
Solution We give a proof by induction. The case 1n was already shown in
previous example Suppose as an induction hypothesis holds true for 1,n k and
we will prove that it holds for .n k Consider ( ) .k h z z Then we can write it as1( ) k h z z z . Applying the induction hypothesis and Theorem 4.1.2, we have
The logarithm z log , and complex exponent c z and z c are multi-valued. When
principal branches are used, these relations become single-valued complex functions
and analytic.
(a) ,
1
Log z z dz
d
where \ 0 , z arg . z
(b) ,1cc cz z dz
d where \ 0 , z arg . z
(c) ,log cccdz
d z z where c and a value clog is specified.
The inverse complex trigonometric and the inverse complex hyperbolic are
multiple-valued. When principal branches of the square root and logarithmic are
used, all these multiple-valued relations becomes single-valued complex functions
and analytic because they are the composition of analytic functions.
(a) 1
2
1
2
1Sin .(1 )
d z dz z
(b) 1
2
1
2
1Cos .
(1 )
d z
dz z
(c) 1
2
1Tan .
1
d z
dz z
(d) 1
2
1
2
1Sinh .
(1 )
d z
dz z
(e) 1
2
1
2
1Cosh .
( 1)
d z dz z
(f) 1
2
1Tanh .
1
d z
dz z
4.5 Harmonic Functions and Laplace’s Equation
Let be an open subset of . A complex function : f is said to be
harmonic in if, throughout this open set, it has continuous partial derivatives of thefirst and second order and satisfies the partial differential equation
02
2
2
2
y
f
x
f (4.5.1)
on . The equation (4.5.1) is known as Laplace’s equation. The operator
2
2
2
2
y x
is called the Laplace operator , or Laplacian, and sometimes it is denoted by the
Theorem 4.6.1 A complex function ),(),()( y xiv y xu z f is analytic in a simply
connected region if and only if v is a harmonic conjugate of u on .
Example Show that 2 2( , )u x y x y x is harmonic in the entire plane, and find a
harmoinc conjugate.
Solution Evidently, that u is harmomic in the entire plane follows from 2 xxu and
2. yyu To find a harmonic conjugate ,v we make use of Cauchy-Riemann
equations as follows. Since we want f u iv to be analytic throughout the entire
plane, it follows that v must satisfy the Cauchy-Riemann equations
x yu v and . y xu v
Since 2 1 xu x , it follows that 2 1. yv x To obtain v we integrate both sides of
this equation with respect to . y However, because v is a function of x and y , the
constant if integration may a function of , x thus we yield
2 1 yv dy x dy implies ( , ) (2 1) ( ),v x y x y c x
where ( )c x is a function of . x Plugging this into the equation , y xu v we have
2 2 ( ) ,d y x dx
y u v y c x
whence
( ) 0d dx
c x implies ( )c x ,C
where C is a real constant. Thus ( , ) 2 .v x y xy y C By choosing 0,C ( , )v x y 2 xy y is a harmonic conjugate of ( , ) ,u x y and ( ) ( , ) ( , ) f z u x y iv x y
2 2 (2 ) x y x i xy y C is an entire function.
For more numerical examples on harmonic conjuage, see Chapter 7: Complex
y as functions of third real variable .t Generally, a parametric form of a curve
( ), y f x is a representation of the curve by a pair of equations ( ) x x t and
( ), y y t where t ranges over a set of real numbers, usually a parametric closed
interval [ , ]a b . Each value [ , ] ,t a b determines a point ( ) ( ( ), ( )),t x t y t which
traces the curve as t varies. When t varies from a to b the point ( )t traces the
curve in a specific direction, starting from ( )a , the initial point of , and
terminating at ( ),b the terminal point of . As usual, for continuous curve, this
direction is denoted by an arrow on the curve . In the complex plane , as
, z x iy it makes sense to adopt the notation ( ) ( ) ( ).t x t iy t Therefore, it is
natural for us to adopt the parametric form of a curve , which lies in the complex
plane, using the complex notation as
( ) ( ) ( ),t x t iy t [ , ] .t a b
Consequently, we can think of the curve ( ) ( ) ( ),t x t iy t as a complex-valued
function of a real variable .t ( ) y t
( )b
t ( ) x t
a b ( )a
A curve ( ) ( ) ( )t x t iy t is smooth on [ , ]a b if ( )t and '( )t continuous
throughout the entire interval [ , ]a b ; i.e., ( )t and '( )t continuous on ( , )a b and both
limits lim '( )t a
t and lim '( )t b
t exist. By a contour ( )t we mean a continuous
piecewise smooth curve ( )t over a parametric closed interval [ , ] .a b More precisely,
a contour is the curve consisting of a finite number of smooth curve joined end to
end; i.e., there exists a finite subdivision, 0 1 1i i na t t t t t b of the
parametric closed interval [ , ]a b such that ( )t is smooth on 1,i it t for all1,2, , .i n A contour is called simple if )()( 21 t t whenever 1 2t t . The
contour on [ , ]a b is closed if the initial point is joined to the terminal point; i.e.,
( ) ( ).a b Furthermore, if the contour is simple except for the initial and
terminal points of , we say that is a simple closed contour .
Suppose that 0 z and 1 z are two fixed points in a region , and that 1 and 2
are contours joining 0 z to 1 z . We say that 1 is continuously deformable into 2
relative to if we can continuously move 1 over 2 while keeping the two end
points fixed at 0 z to 1 z without leaving .
Suppose that 1 and 2 are two closed contours in a region . We say that 1 is
continuously deformable into 2 relative to if we can continuously move 1 without leaving , in such a way that 1 overlaps with 2 in direction and position.
If is a closed contour in a region , we say that is continuously deformable
into a point 0 z relative to if we can continuously shrink into the point 0 z in
without leaving . Indeed, this is the same situation as the continuously deformable
of the closed contour 1 into another closed contour 2 , where, in this case, 2 is
degenerated into a point; i.e., 2 0( )t z for all .t
Two contours 1 and 2 are said to be mutually deformable in a region if 1 is
continuously deformable into 2 relative to , and also, 2 is continuously
deformable into 1 relative to .
Definition 5.5.1 A region is called simply connected if all contours joining any
two fixed points 0 z and 1 z in are mutually continuously deformable relative to .
If a region is not simply connected, then it is called multiply connected .
Theorem 5.6.5 Let and i , ,,,2,1 ni be the simple closed contours such that
(i) is a simple closed contour, described in the counterclockwise direction;
(ii) i , 1, , ,i n are simple closed contours, all described in the counterclockwise
directions, that are interior to and whose interiors have no points in common. If a function ( ) f z is analytic throughout the closed region consisting of all
points within and on except for points interior any i , then
1
( ) ( )i
n
i
f z dz f z dz .
Example Let be the positively oriented simple closed contour , and let 0 z be a
point not on . Show that
0
00
0 ,1
2 .
if z is in the exterior of dz
i if z is in the interior of z z
Solution If 0 z is the exterior of , then the function 0( ) 1 ( ) f z z z is analytic on
and inside , and hence, by Cauchy’s theorem, its integral along is zero. For the
second case, where 0 z is the interior of , then we cannot claim that the function
0( ) 1 ( ) f z z z is analytic on and inside . So, we cannot apply Cauchy’s theorem,
however by previous examples, we conclude that
0
12 .dz i
z z
5.7 Cauchy’s Integral Formula and Derivatives of Analytic
Functions
Theorem 5.7.1 (Cauchy’s Integral Formula) Let ( ) f z be analytic throughout within
and on a simple closed positively oriented contour . If 0 z is any point interior to ,
then
.)(
2
1)(
0
0 dz z z
z f
i z f (5.7.1)
Note that formula (5.7.1) tells us that if a complex function ( ) f z is to be analytic
within and on a simple closed contour , then its values everywhere interior to a