Complex Variables
Complex numbers are really two numbers packaged
into one entity (much like matrices). The two
“numbers” are the real and imaginary portions of the
complex number:
z  x  jy.
x  Re{z}.
y  Im{z}.
We may plot complex numbers in a complex plane:
the horizontal axis corresponds to the real part and
the vertical axis corresponds to the imaginary part.
Im{z}
z = x + jy
y
Re{z}
x
Often, we wish to use polar coordinates to specify
the complex number. Instead of horizontal x and
vertical y, we have radius r and angle q.
Im{z}
z = x + jy
y
r
q
Re{z}
x
The best way to express a complex number in polar
coordinates is to use Euler’s identity:
jq
e  cosq  j sin q .
So,
z  re
and
jq
 r cosq  jr sin q ,
x  r cos q .
y  r sin q .
We also have
x  y  r cos q  r sin q  r .
2
2
2
2
2
2
2
A summary of the complex relationships is on the
following slide.
Im{z}
re jq  r cosq  jr sin q .
z  x  jy.
r  x2  y2 .
y
tan q  .
x
y = r sin q
r
q
x = r cos q
Re{z}
The magnitude of a complex number is the squareroot of the sum of the squares of the real and
imaginary parts:
z  x  jy  x  y .
2
2
If we set the magnitude of a complex number equal
to a constant, we have
z  x  jy  x  y  c,
2
2
or,
z x y c .
2
2
2
2
This is the equation of a circle, centered at the
origin, of radius c.
Im{z}
c2 = |z|2 = x2 + y2
z = x + jy
y
c
Re{z}
x
Suppose we wish to find the region corresponding to
z x y c .
2
2
2
2
This would be a disk, centered at the origin, of
radius c.
Im{z}
x2 + y2 = |z|2 < c2
y
c
Re{z}
x
Suppose we wish to find the region corresponding to
z  z0  c .
2
2
This would be a disk, centered at z0, of radius c.
z  z0  ( x  x0 )  ( y  y0 )  c .
2
2
2
2
Im{z}
(x-x0)2 + (y-y0)2 = |z-z0|2 < c2
= |z-z0|2
y0
c
z0
Re{z}
x0
Functions of Complex
Variables
Suppose we had a function of a complex variable,
say
w  f ( z)  z 2 .
Since z is a complex number, w will be a complex
number. Since z has real and imaginary parts, w will
have real and imaginary parts.
w  f ( z)  z 2
 ( x  jy) 2
 x 2  2 jxy  y 2


 x 2  y 2  j2 xy.
The standard notation for the real and imaginary
parts of z are x and y respectively.
The standard notation for the real and imaginary
parts of w are u and v respectively.
w  x

 y  j2 xy
 u  jv,
2
2
where


u  x2  y 2 .
v  2 xy.
Both u and v are functions of x and y.
So a complex function of one complex variable is
really two real functions of two real variables.
w  f ( z)  u( x, y)  jv( x, y).
Exercise: Find u(x,y) and v(x,y) for each of the
following complex functions:
f ( z)  z 3.
f ( z)  e .
z
f ( z )  e jz .
f ( z )  cos z.
f ( z)  z .
f ( z)  z .
z
Continuity of Complex
Functions
In order to perform operations such as differentiation
and integration of complex functions, we must be
able to verify of the complex function is continuous.
A complex function
f (z )
is said to be continuous at a point z0 if as z
approaches z0 (from any direction) then f(z) can be
made arbitrarily close to f(z0).
A more mathematical definition of continuity would
be for any , we can make
f ( z)  f ( z0 )  
for some d such that
z  z0  d .
Since we are dealing with complex numbers, the
geometric interpretation of this statement is different
from that of real numbers.
The region |z-z0| < d defines a disk in the complex
plane of radius d centered about z0.
d
Im{z}
z0
Re{z}
So, if we wish |f(z)-f(z0)| <  we must find a d to
make this so.

Im{w}
f(z0)
Re{w}
Example: Suppose
f ( z)  z .
2
Find d such that
f ( z)  f ( z0 )    0.5
for
z0  1  j 2.
Solution:
f ( z)  x  jy   x  y  j 2xy.
2
2
2
f ( z0 )  1 j 2  3  j 4.
2
f ( z )  f ( z0 ) 
x
2

 y  3  2 xy  4
2
2
2
All we need to do is to find a value of d such that if
z  z0 
x  12   y  22  d ,
then
x
2

 y  3  2 xy  4  0.5.
2
2
2
We can do some calculations on a spreadsheet
(continuity.xls).
x
1.100
1.050
1.050
0.950
0.950
1.010
y
2.100
2.050
1.950
2.050
1.950
2.010
x12 y 22
0.1414
0.0707
0.0707
0.0707
0.0707
0.0141
A value of d < 0.1 seems to do it.
x
2

 y 2  3  2 xy  4
2
2
0.6515
0.3210
0.3147
0.3178
0.3115
0.0634
A MATLAB plot (by continuity.m) of the previous
example is shown on the following slide.
5
4
|z2 - z20| <  = 0.5
Im{z}
3
|z - z0| < d = 0.1
2
1
0
-1
-4
-3
-2
-1
Re{z}
0
1
2
Differentiation of Complex
Functions
How do we take derivatives of complex functions
with respect to complex variables?
If
w  f (z ),
what is
dw df ( z )

?
dz
dz
The differential dz can vary in one of two ways:
along the real axis (dx) or along the imaginary axis
(dy).
Im{z}
y+dy
y
dy
dx
Re{z}
x
x+dx
As z varies in either direction, the derivative must be
the same.
x direction
dw w u
v


j .
dz x x
x
y direction
dw w
w
u v

j
j
 .
dz jy
y
y y
So, we must have
u v
 .
x y
u
v
 .
y
x
These last two conditions
u v
 .
x y
u
v
 .
y
x
are called the Cauchy-Riemann equations. These
equations are the criteria for a complex function to
be differentiable (with respect to z = x + jy).
Example: Show that the function
w  f ( z)  z .
2
is differentiable
Solution: We have shown that


u  x2  y 2 .
v  2 xy.
u
 2 x.
x
v
 2 x.
y
u
 2 y.
y
v
 2 y.
x

u v
 .
x y

u
v
 .
y
x
Now that we have determined that this function is
differentiable, the derivative can be found using
dw u
v

j .
dz x
x
or
dw v
u
 j .
dz y
y
If we apply these formulas to
w  f ( z)  z  u( x, y)  jv( x, y).
2
where


u  x2  y 2 .
v  2 xy.
we have


dw  x  y
2 xy

j
dz
x
x
 2 x  j 2 y.
or
2
2


dw 2 xy
 x2  y2

j
dz
y
y
 2 x  j (2 y)  2( x  jy).
We see that the derivative in both cases is
dw
 2( x  jy )  2 z.
dz
The answer is what we would expect to get if z were
treated as a real variable.
As it turns out, for most well-behaved complex
functions, the derivative can be found by treating
z as if it were a real variable.
Example: Show that the function
w  f ( z )  Re{z}.
is not differentiable
Solution: We have shown that
u  x.
v  0.
u
 1.
x
v
 0.
y
u
 0.
y
v
 0.
x

u v
 .
x y

u
v
 .
y
x
Exercise: Is
w  f ( z )  Im{z}
differentiable?
Definition: A function
w  f (z ).
is said to be analytic if it is differentiable throughout
a region in the complex plane.
Integration of Complex
Functions
What happens when we try to take the integral of a
complex function along some path in the complex
plane?

C
f ( z ) dz .
A complex integral is like a line integral in two
dimensions.

C
f ( z ) dz   u ( x, y )  jv( x, y )dx  jdy
C
  u ( x, y ) dx   v( x, y ) dy
C

C

 j  v( x, y ) dx   u ( x, y ) dy .
C
C
The real and the imaginary parts of the integral are
nearly identical to classic line integrals.
Example: Integrate
w  f ( z)  z 2
over the real interval z = 0 + j0 to z = 2 + j0.
Solution: We have shown that


u  x2  y 2 .
v  2 xy.

C
f ( z ) dz   u ( x, y ) dx   v( x, y ) dy
C

C

 j  v( x, y ) dx   u ( x, y ) dy
C
  x
C

C
2

 y dx   2 xy dy
2
C



 j  2 xy dx   x 2  y 2 dy .
C
C
Since we are integrating along the real (x) axis, all
integrals with respect to dy are zero. In addition
y=0. So,
 
2
2
z
dz

x

 dx
C
C
2
  x dx
2
0
3 2
x

3
0
 83 .
The result is exactly what we would expect to get if
we simply integrated a real variable from 0 to 2.
Example: Integrate
w  f ( z)  z 2
over the imaginary interval z = 0 + j0 to z = 0 + j2.
Solution: The integral becomes

C
  
2
f ( z ) dz  j   y dy   j  y 2 dy   j 83.
2
C
0
The result is exactly what we would expect to get if
we simply integrated
 z dz
2
C
where C = jy and where y=[0,2] :


2
2

   j 8.
(
jy
)
djy

j

y
dy
3
C
 0

2
Example: Integrate
w  f ( z)  z 2
over the complex path z = 0 + j2 to z = 2 + j2.
Im{z}
2
2
Re{z}
Solution:

C




f ( z ) dz   x 2  y 2 dx  j  2 xy dx
C
2


C
2
  x  y dx  j  2 xy dx.
0
2
2
0
The value of y is that of the path: y=2.

C
2


2
f ( z) dz   x  (2) dx  j  2 x(2) dx.
0
2
2
0

C
2


2
f ( z ) dz   x  (2) dx  j  2 x(2) dx
0
2
2
2
0
x

2 2
   4 x   j 2 x   163  j8.
0
3

0
3
Example: Integrate
w  f ( z)  z 2
over the complex path z = 2 + j0 to z = 2 + j2.
Im{z}
2
2
Re{z}
Solution:

C



f ( z ) dz    2 xy dy  j  x  y dy
C
2
2
C


2
2
2

   2(2) y dy  j  (2)  y dy
 0

0
2
2

y 
 2 y  j 4 y    8  j 163 .
0
3 0

2 2
3
Example: Integrate
w  f ( z)  z 2
over the complex path z = 0 + j0 to z = 2 + j2.
Im{z}
2
2
Re{z}
Solution: The path of integration is a line z = x + jy
where x = y = t = [0,2].
Solution: The integral is more complicated.

C


f ( z ) dz   x 2  y 2 dx   2 xy dy
C

C



 j  2 xy dx   x 2  y 2 dy
2

C
C

2
  t  t dt   2tt dt
0
2
2
0


2
2
2
2

 j  2tt dt   t  t dt.
 0

0

C
2
2

f ( z ) dz    2t dt  j  2t dt
 0

0
2
2
3 2
2t

3
3 2
2t
j
3
0
 2 83  1  j .
0
This result is the same as the sum of the integral
from 0+j0 to 0+j2 with the integral from 0+j2 to 2+j2.
Im{z}
2
2
Re{z}
This result also is the same as the sum of the
integral from 0+j0 to 2+j0 with the integral from 2+j0
to 2+j2.
Im{z}
2
2
Re{z}
So it seems that it does not matter what path is
taken as long as the endpoints are the same.
Example: Integrate
w  f ( z)  z 2 .
over two paths:
(1) a semicircle z = ejq, where q = [0,p].
(2) a semicircle z = e-jq, where q = [0,p].
Show that the two integrals are the same.
Im{z}
C1
Re{z}
q=0
q=p
C2
Solution: This integration is best handled using
polar coordinates:
 
f ( z)  z  re
2
jq
jq 2
r e
2
jq
j 2q
e
jq
j 2q
. (r  1.)
dz  d (re )  jre dq  je dq . (r  1.)
The integral around curve C1 is

C1
p
f ( z )dz   e j 2q je jq dq
0
p
j j 3q p
 j  e dq  e
0
0
j3
1 j 3p
1 1
j0
 e e 
  23 .
3
3

j 3q

The integral around curve C2 is

C2
p
f ( z )dz   e j 2q je jq dq
0
p
j jq p
 j  e dq  e
0
j3 0
1  jp
1 1
j0
 e e 
  23 .
3
3

jq

If we were to integrate around the whole circle C =
ejq for q = [0, 2p], we would get

C
2p
f ( z )dz   e
0
2p
j 2q
jq
je dq
j j 3q 2p
 j  e dq  e
0
0
j3
1 j 6p
11
j0
 e e 
 0.
3
3

j 3q

The curve C can be thought of as C1 + (-C2 ).
Cauchy’s Integral Theorem: If a function f(z) is
analytic over a region R enclosed by a (closed) path
C, then

C
f ( z )dz  0.
Im{z}
C
R
Re{z}
Simple Proof:

C
f ( z )dz   udx  vdy   j  udy  vdx .
C
C
Both integrals are line integrals around a closed
curve C. We can apply Green’s theorem (a special
case of Stoke’s theorem) to these line integrals

C
 v u 
 u v 
f ( z )dz     dxdy  j    dxdy.
R x
R x
y 
y 


If f(z) is analytic, then the Cauchy-Riemann
equations apply:
u v
 .
x y
u
v
 .
y
x
If these are true, then both integrands of

C
 v u 
 u v 
f ( z )dz     dxdy  j    dxdy
R x
R x
y 
y 


are zero and the theorem is proved.
If

C
f ( z )dz  0,
and C = C1 + C2, then

C
f ( z )dz   f ( z )dz   f ( z )dz  0.
C1
C2
Im{z}
C1
C2
Re{z}
We also have

C1
f ( z )dz    f ( z )dz
C2

C2
f ( z )dz.
Im{z}
C1
-C2
Re{z}
So it does not matter what path that you take so long
as the endpoints are the same provided f(z) is
analytic between any of the two paths.
If f(z) is not analytic at some point between two
paths, then the path does matter.
Example: Integrate
1
w  f ( z) 
z
over a unit circle z = ejq, where q = [0,2p].
Im{z}
Re{z}
Solution: As with the previous example, this
integration is best handled using polar coordinates:
1
1
1  jq
f ( z )   jq  e  e  jq . (r  1.)
z re
r
dz  d (re jq )  jre jq dq  je jq dq . (r  1.)

C
2p
f ( z )dz   e
0
 jq
2p
je dq  j  dq  j 2p .
jq
0
This integral is not zero because there is a
discontinuity (actually a pole) at z = 0.
Example: Integrate
1
w  f ( z) 
z 1
over a unit circle z = 1 + ejq, where q = [0,2p].
Im{z}
Re{z}
Solution: Let us first write the integral.

C
dz
.
z 1
f ( z )dz  
C
To carry-out this integration, we first perform a
substitution of variables. Let z = z-1.

C
f ( z )dz  
C
dz
'
z
.
The path C’ is equal to C minus one:
Im{z}
C’
C
Re{z}
We see that

C
f ( z )dz  
C
dz
'
z
 2p j.
This integral is not zero because there is a
discontinuity (a pole) at z = 1 or z = 0.
Exercise: Show that no matter what z0 is, if C is a
circular path (of any radius) around z0 we will have

C
dz
 2p j.
z  z0
Cauchy’s Integral Formula: Let f(z) be analytic
over a region R enclosed by a closed path C. If z0 is
a point within R, then

C
f ( z)
dz  2p j f ( z0 ).
z  z0
Note that while f(z) is analytic throughout R,
f(z)/(z-z0 ) is not analytic (z0 is a pole).
Im{z}
R
C
z0
Re{z}
Proof: We add and subtract f(z0) to the numerator of
the integrand so as to split-up the integral into two
terms:

C
f ( z )  f ( z0 )  f ( z0 )
f ( z)
dz  
dz
C
z  z0
z  z0

C
f ( z )  f ( z0 )
dz
dz  f ( z0 ) 
.
C zz
z  z0
0
If f(z) is analytic within the region R, then it is also
continuous. So, for any , we can find a d such that
for
z  z0  d ,
we have
f ( z)  f ( z0 )   .
Let us choose a r  d such that z  R, i.e., the disk
|z-z0| < r is totally within R. Let denote the path |zz0| = r by the symbol C’ .
r
Im{z}
R
C
z0
C’
Re{z}
So if
z  z0  r  d ,
f ( z)  f ( z0 )   ,
and
f ( z )  f ( z0 )




z  z0
z  z0 r
(on thepathC' ).
So for appropriate values of d and r, the integrand in

C'
f ( z )  f ( z0 )
dz
z  z0
can be made arbitrarily small. Now since the
integrand is analytic except at z = z0, we have

C'
f ( z )  f ( z0 )
f ( z )  f ( z0 )
dz  
dz.
C
z  z0
z  z0
The integral

C
dz
z  z0
is equal to 2pj. So,

C
f ( z)
dz  0  f ( z0 )2p j ,
z  z0
and the theorem is proved.
Example: Evaluate the integral

C
z 1
dz,
z 1
where C is a closed curve around z = 1.
Solution:
z0  1.
f ( z )  z  1.
So,

C
z 1
dz  f ( z0 )2p j  (1  1)2p j  4p j.
z 1
The formula
f ( z0 ) 
1
2p j C
f ( z)
dz
z  z0
makes calculating derivatives with respect to z0
relatively easy. We do not have to worry about z: it
is independent of z0.
1
f ( z)
f ' ( z0 ) 
dz
.
2
2p j C z  z0 
In general it can be shown that
f
(n)
n!
f ( z)
( z0 ) 
dz.
n

1

2p j C z  z0 
This formula is very useful in deriving the Taylor
series.
Example: Evaluate the integral
z 1
C z  12 dz,
where C is a closed curve around z = 1.
Solution:
z0  1.
f ( z )  z  1. n  1.
So,
z 1
(1)
C z  12 dz  f ( z0 )2p j  (1)2p j  2p j.
Example: Evaluate the integral
z 1
C z  13 dz,
where C is a closed curve around z = 1.
Solution:
z0  1.
f ( z )  z  1. n  2.
So,
z 1
2p j
( 2)
C z  13 dz  f ( z0 ) 2!  (0)p j  0.
Example: Evaluate the integral
z3 1
C z  23 dz,
where C is a closed curve around z = 2.
Solution:
z0  2.
f ( z )  z 3  1. n  2.
f ( 2) ( z )  6 z.
z3 1
2p j
( 2)
C z  23 dz  f ( z0 ) 2!  (6)(2)p j  12p j.
Exercise: Evaluate the integral
cos z
C z  12 dz,
where C is a closed curve around z = 1.
Inverse Laplace Transforms
We used a formula to calculate the forward Laplace
transform, but we did not use a formula to calculate
the inverse Laplace transform. Such a formula
exists!
The forward Laplace transform was found using

L{x(t )}  X (s)   e x(t )dt.

 st
The inverse Laplace transform can be found using a
complex inversion integral formula:
L { X ( s)}  x(t ) 
-1
1

2p j


st
e X ( s)ds.
We can evaluate the inverse Laplace transform
using Cauchy’s integral formula.
Cauchy’s integral formula is for an integration
around a closed loop. The inverse Laplace
transform formula is an integral along an infinite line.
This infinite line integral can actually be thought of
as a loop.
Let us construct a closed curve C consisting of a line
along the imaginary axis and a semicircle in the lefthalf plane.
Im{s}
s  +j
r
s  -
Re{s}
C
s  -j
As the radius r of the semicircle approaches infinity,
the closed loop approaches an infinite line (from s =
-j to +j). As r approaches infinity, both the
semicicular curve and the infinite line pass through
something called the point at infinity.
The point at infinity can be reached from either the
positive or negative half of the real or the imaginary
axis. The limits s  +j, s  -j and s  - are all
the same.
Example: Find the inverse Laplace transform of
1
X (s) 
.
s 1
Solution:
L { X ( s)}  x(t ) 
-1
1
1

2p j



st
e X ( s)ds
1

e
ds
.
2p j  s  1
st
s0  1.
f ( s)  e st .
1

1
L { X ( s )} 
e
ds

2p j  s  1
-1
 2p j
t
e .
st
1
2p j
e
s0 t
s0  1
Example: Find the inverse Laplace transform of
1
X ( s)  2 .
s 1
Solution:
L { X ( s)}  x(t ) 
-1
1
1

2p j



st
e X ( s)ds
1

e 2 ds.

2p j  s  1
st
The integral can be evaluated in the complex plane
about a curve C:
1

1
L { X ( s)} 
e 2 ds

2p j  s  1
-1
1
st
1

e 2 ds

2p j C s  1
st
The curve C is an extension of the s= -j to s = +j
line.
Im{s}
Re{s}
C
There are actually two points of discontinuity:
s0  { j, j}.
We can evaluate the complex inversion integral by
evaluating the integral around these two points of
discontinuity. Let us call the paths around these two
points C1 and C2.
1
1
1
st
L { X ( s)} 
e 2 ds 
e 2 ds.


2p j C1 s  1
2p j C2 s  1
-1
1
st
Im{s}
C1
s0 = +j
Re{s}
C
C2
s0 = -j
L { X ( s )} 
-1
1
2p j 
C1
e
st
1
s  j s  j 
1
 2p j
e st
2p j s  j 
ds 
1
2p j 
C2
e
st
1
s  j s  j 
1
 2p j
e st
s j
2p j s  j 


s  j
1
1
1 jt  jt
jt
 jt

e 
e 
e  e  sin t.
 j  j
 j  j 
2j
ds
Exercise: Using the complex inversion integral, find
the inverse Laplace transforms of the following
functions:
s
X ( s)  2 .
s 1
1
X ( s) 
.
2
( s  1)  1
(First find the points of discontinuity and then
evaluate the integral in paths around these points.)
Example: Find the inverse Laplace transform of
cos s
X (s) 
.
s p
Solution:
L { X ( s)}  x(t ) 
-1
1
1

2p j



st
e X ( s)ds
cos s

e
ds.

2p j  s  p
st
There is one point of discontinuity:
s0  p .
1
cos s
L { X ( s)} 
e
ds

2p j C1 s  p
-1

1
2p j
st
2p j e st cos s
s  p
 e pt .
Sequences and Series
Consider the sequence of values
1, 12 , 13 , 14 ,
each term in this sequence can be represented by
zn  1n .
What happens as n goes to infinity? For this
example zn goes to zero.
How about the sequence of values
1, 32 ,1 34 , 1 78 ,
each term in this sequence can be represented by
zn  2[1  
 ].
1 n
2
This sequence is said to converge to two (2): as n
goes to infinity, zn goes to two.
How about the sequence of values
1, 32 , 2, 52 ,
This series does not converge to any value, but
rather it is said to diverge.
This sequence {zn } is said to converge to a value c
if zn can be made arbitrarily close to c for a large
enough value of n.
A more formal definition of convergence of series
would be for any positive value , we can find an
integer N such that for
nN
we must have
zn  c   .
The expression
zn  c   .
is for a complex number zn and defines a disk in the
complex plane.

Im{z}
c
Re{z}
Example: Plot the sequence of values in
2pn
1 j 20
zn  e
n
in the complex plane
Solution: The radius of zn is 1/n and the angle of zn
is 2pn/20. The plot is performed using MATLAB
(sequence.m) and is shown on the following slide
page.
1
0.8
Im{z}
0.6
0.4
0.2
0
-0.2
-0.2
0
0.2
0.4
Re{z}
0.6
0.8
This sequence converges to zero. The relationship
between the sequence index n and distance to the
limit is rather easy in this case.
For
n  1 ,
we must have
zn  c   .
where 1/ corresponds to the smallest integer
greater than 1/.
Similarly, for
n  N,
we must have
1
zn  c  .
N
A necessary (but not sufficient) condition for a
sequence to converge is that it be bounded. A
bounded sequence is {zn } is one that for all n we
have
zn  B
for some finite value B. If a sequence is not
bounded, it will diverge.
Just because a sequence is bounded does not
mean it converges. Many sequences which are
bounded do not converge. For example,
zn   1, 1, 1, 1,.
For this sequence
zn  1
n
for n  0,1,.
zn
n
1 2 3 4 5
While this sequence does not have a limit, it does
have an upper bound (+1) and a lower bound (-1) as
n   . These “bounds” are called the supremum
and the infimum respectively. The supremum is the
smallest upper bound (+1) and the infimum is the
largest lower bound (-1). These bounds are
abbreviated sup and inf respectively.
lim sup zn  1.
n 
lim inf zn  1.
n 
Note that while {zn } does not converge, there are
subsequences
z1n   1, 1,.
z2n    1,  1,.
that do converge.
Series
Suppose we were to add the numbers in a
sequence:
n
sn   z k .
k 1
The term sn is called the partial sum of the series
{zn }. The summation is a series.
If we were to take the limit as n   , we would get
an infinite series:

s  lim sn   zk .
n 
k 1
If the sequence of partial sums converges, we say
that the series converges.
A necessary (but not sufficient) condition for the
series {sn } to converge is that the sequence {zn }
converges. The series is generally “wilder” that the
sequence. If the series converges, the sequence
must necessarily converge. If the series diverges,
the sequence may or may not converge.
Example: Consider the (convergent) sequence
1
zn  .
n
The corresponding series does not converge (as n
goes to infinity):
n
n
1
sn   z k   .
k 0
k 0 k
Now how do we find sufficient conditions for a series
to converge? There are five (5) standard tests for
series convergence:
(1) Comparison Test: compare series term-by-term
against a known convergent series.
(2) Geometric Series Test: a geometric series
converges if each geometric term is less than one
(3) Ratio Test: Take the limit of the ratio of one term
to the previous term. If the limit is less than one, the
series converges
(4) Root Test: Take the k-th root of the k-th term. If
the limit is less than one, the series converges.
(5) Integral Test: compare the sequence to an
integrand of a known integral.
(1) Comparison Test: compare series term-by-term
against a known convergent series.
If we have a known convergent series
n
sn   z k ,
k 1
then any series
n
rn   wk
k 1
such that
wk  zk
also converges.
(2) Geometric Series Test: a geometric series
converges if each geometric term is less than one
A geometric series is of the form
n
sn   q .
k
k 1
We can use this form to find a closed-form
expression for the geometric series
n
sn   q .
k
k 0
n
qsn   q
n 1
 q .
k 1
k
k 0
k 1
n 1
n
sn  qsn   q   q
k
k 0
 1 q
k 1
n 1
.
k
So,
sn  sn q  1  q
n 1
.
n 1
1 q
sn 
.
1 q
If the geometric term q is such that qn goes to zero
as n goes to infinity, then
1
lim sn 
.
n 
1 q
In order for qn to go to zero, we must have
q  1.
If q > 1, then the series diverges.
Example: Suppose q = ½.
zn  
.
1 n
2
n
sn   
n 0

1 k
2
 1  12  14  ...
1

 2.
1
1 2
Example: Suppose q = 9/10.
zn  
.
9 n
10
n
sn   
n 0

9 k
10
81
 1  109  100
 ...
1

 10.
9
1  10
Example: Suppose q = -1/2.
z n  
n
.
1 n
2
sn   
n 0

1 k
2
 1  12  14  ...
1
2


3.
1
1 2
(3) Ratio Test: Take the limit of the ratio of one term
to the previous term. If the limit is less than one, the
series converges
If
n
sn   z k ,
k 1
then we take
zk 1
wk 
.
zk
If
lim wn  1.
n
then the series converges.
As a justification (hardly a proof) for this test,
consider definining zk in terms of wk:
zk 1
wk 
.
zk
zk 1  wk zk .
So,
z2  w1 z1.
z3  w2 z2  w2 w1 z1.
n
n
k
sn   zk  z1  wm .
k 1
k 1 m 1
The term
k
w
m
m 1
 (max wm ) .
k
1 m k
If
max wm  1,
1 m  k
then the series behaves like a convergent
geometric series:
n
n
k
n
sn   zk  z1   wm  z1  (max wm ) .
k
k 1
k 1 m 1
k 1
1 m k
Example: Determine if
n
k
2
sn  
k  0 k!
converges.
Solution: Taking the ratio test
k 1
zk 1
2
k!
2
wk 


 0.
k
k  1! 2 k  1
zk
We see that the series converges.
(4) Root Test: Take the k-th root of the k-th term. If
the limit is less than one, the series converges.
If
n
sn   z k ,
k 1
then we take
wk  k zk .
If
lim wn  1.
n
then the series converges.
As a justification (better than that of the ratio test but
still not a proof) for this test, consider
wk  k zk .
wk 
k
 zk .
n
n
k 1
k 1
sn   zk   wk  .
k
If wk < 1, then the series is a convergent geometric
series.
Example: Determine if
n
k
e
sn   2
n 0 k
converges.
Solution: Taking the root test
wk  k z k  k
ek
e
e
e
e
 2/ k 
 2 ln k / k  0  e.
2/ k
2
ln
k
k
k
e
e
e
 
We see that the series diverges. (ek dominates k2.)
(5) Integral Test: compare the sequence to an
integrand of a known integral.
n
sn   z k .
k 1
The term zk is really a function of k. We can
represent that function as z(k) or z(x). The
convergence of the integral


1
z ( x)dx
can be used to determine whether or not the series
converges.
Example: Consider the series
n
1
sn   .
k 1 k
This series is greater than the integral

n
1
1
dx.
x
Since the integral diverges [ln n], the series
diverges.

Integral 
dx/x
Versus
Series

1/n
1
n=1
2
1/x
1.5
1
0.5
1
2
3
4
5
x
6
7
8
9
10
Example: Consider the series
n
n
1
1
sn   2  1   2 .
k 1 k
k 2 k
This series from k=2 to k=n is less than the integral

n
1
1
dx.
2
x
Since the integral converges [1/n], the series
converges.
2

2
Integral 
dx/x
Versus
Series

1
/
n
1
n=2
2
1/x
1.5
1
0.5
1
2
3
4
5
x
6
7
8
9
10
Taylor Series
A Taylor series is a power series representation for a
function.

f ( z )   ck z  a  .
k
k 1
A Taylor series is much like a Fourier series (which
is a harmonic series).
To find a Taylor series, all we need to do is find the
coefficients (much like Fourier series).
To find these coefficients let us start with Cauchy’s
integral formula:
f (z )
f ( z) 
d
z
.
2p j C z  z
1
We will attempt to express this formula in a power
series about z = a.
We start with the fraction in the integral:
1
z z
The object is to express this in terms of powers of (z
– a). So, we try to get this fraction in the form of an
infinite geometric series:
1
z aaz
1
1
1


z  a  z  a  z  a  1  z  a
z a
k
 za 
1

 .


z  a  k 0  z  a 

The last is true if z is on a curve C at a distance r
from a and z is within the close curve.
r
Im{z}
z is on C
a
z
Re{z}
z  a  z  a.
k
 za 
f (z )

 dz
f ( z) 


2p j C z  a  k 0  z  a 

1
f (z )
  z  a 
d
z
.
k

1

C 
2
p
j
z  a
k 0

k
1
The coefficient of (z – a)k inside the summation
f (z )
d
z
n

1
2p j C z  a 
1
is similar to the k-th derivative of f(a) from the
corollary to Cauchy’s integral theorem:
f
(k )
k!
f (z )
(a) 
d
z
.
k

1
2p j C z  a 
So,
f (z )
f ( z )   z  a 
dz
k

1

C 
2
p
j
z  a
k 0


  z  a 
k 0
1
k
k
f
(k )
(a)
.
k!
Hence, we have our Taylor series coefficients.
Example: Find the Taylor series for
f1 ( z )  e .
z
f 2 ( z )  cos z.
f 3 ( z )  sin z.
f1 ( z )  e .
(n)
z
n

 1 2  cos z
n even

f2 ( z)  
.

n
 1  2  sin z n odd 
n

2

  1 sin z n even
(n)
f3 ( z)  
.

n
 1  2  cos z n odd 
(n)
We will evaluate the Taylor series about z = 0 :
f1 ( 0 )  1 .
(n)
f2
(n)
f3
(n)
n

 1 2 

( 0)  
 0
n even
.
n odd 
n even
0
( z)  
.
n



2



1
n odd 

So,
2
3
4
z
z
z
f1 ( z )  1  z     .
2! 3! 4!
2
4
z
z
f 2 ( z )  1    .
2! 4!
3
z
f 3 ( z )  z   .
3!
You can use these Taylor series to prove
e  cos z  j sin z.
jz
Example: Find the Taylor series for
1
f ( z)  .
z
Here, we will evaluate the series about a = 1.
f
(n)
n!
( z )   1 n 1 .
z
n




1
z  1
z  1
z  1
z  1
 1  1!
 2!
 3!
 4!

z
1!
2!
3!
4!
2
3
4
 1  z  1  z  1  z  1  z  1  .
2
3
4
Example: Find the Taylor series for
f ( z )  ln z.
Here, we will evaluate the series about a = 1.
f
(n)

n  1!
n  1.
( z )   1
n
n 1
z


z  1
z  1
ln z  0 
 1!
2
1!
2!

z  1 z  1


2
1
2

z  1
 2!
3
3!

z  1

3
3

z  1
 3!
4
4!

z  1

4
4
 .
If z is not close to one, this series is very slow to
converge.

Exercise: Find the Taylor series for
f ( z )  ln z.
Evaluate the series about an arbitrary a.
Conformal Mapping
How do we “graph” complex functions? The
difficulty lies in the dimensionality: we have two
independent variables (x,y) and two dependent
variables (u,v).
w  f ( z)  u( x, y)  jv( x, y).
To “graph” this function, we start with a family of
curves corresponding to constant values of x and
constant values of y. These curves are represented
by dashed green lines on the following slide.
y = Im{z}
y=4
y=3
y=2
y=1
x = Re{z}
x=1 x=2 x=3 x=4
To what do these curves correspond to in the u-v
plane?
Let us start with a simple example
w  f ( z )  2 z.
w  u  jv  2 x  j 2 y.
u  2 x.
v  2 y.
v = Im{w}
y=2
y=1
u = Re{w}
x=1
x=2
Example: Find the conformal map for
w  f ( z)  e .
z
We expand ez from the real and imaginary parts of z.
w  f ( z)  e  e
z
x  jy
e e .
x
jy
This expansion is best handled using polar
coordinates.
j
w  e e  re ,
x
jy
where
r e ,
  y.
x
The resultant curves will be a set of circles of radii
ex. Constant values of y correspond to rays at angle
y.
v = Im{w}
y=2
y=1
u = Re{w}
x=1
x=2
Negative values of x correspond to circles of radius
e-|x|. Negative values of y correspond to rays at
angle -|y|.
v = Im{w}
x = -2
x = -1
u = Re{w}
x=2
y = -2
y = -1
Example: Find the conformal map for
z 1
w  f ( z) 
.
z 1
We represent z and w in terms of their real and
imaginary components:
( x  jy )  1
w  u  jv 
.
( x  jy )  1
We then try to make the denominator real:
( x  1)  jy ( x  1)  jy
u  jv 

( x  1)  jy ( x  1)  jy
( x  1)(x  1)  y  jy ( x  1  [ x  1])

( x  1) 2  y 2
2
x2  y 2 1 j2 y

.
2
2
( x  1)  y
x  y 1
u
.
2
2
( x  1)  y
2y
v
.
2
2
( x  1)  y
2
2
A plot of the constant x and the constant y curves (in
the u-v plane) is shown on the following slide.
w = (z-1)/(z+1)
1
y=1
x=0
y=2
0.5
x=1
x=2
jv
y=0
0
-0.5
y = -2
y = -1
-1
-1
-0.5
0
u
0.5
1
The resultant graph is that of a Smith Chart. This
chart is used in radio-frequency electronics. The
graph is a conformal map of line impedance onto
complex reflection coefficient.
The Argument Principle
Let
w  f (z )
be a function of the complex variable z.
As z follows a path in the z-plane, what path does w
follow in the w-plane?
Let z follow a closed path in the z-plane.
z  re
jq
(r  1, 0  q  2p ).
Im{z}
C
Re{z}
As z follows this closed path, what path will w=f(z)
follow?
As an example, let
f ( z)  z .
2
w  f ( z)  z  r e
2
2
j 2q
e
j 2q
.
As z goes around the circle once, w goes around
the same circle twice.
w z .
2
Im{z}
Im{w}
C’
C
Re{z}
Re{w}
As another example, let
1
f ( z)  .
z
1 1  jq
 jq
w  f ( z)   e  e .
z r
As z goes around the circle counter-clockwise, w
goes around the same circle clockwise.
1
w .
z
Im{z}
Im{w}
C
Re{z}
Re{w}
C’
Let us choose two points on the z-path: z1 and z2
Im{z}
C
z1
Re{z}
z2
The two points z1 and z2 are very close together;
their radii are the same, but their angles are
different.
The two points z1 and z2 are very close together;
their radii are the same, but their angles are
different.
z1  r1e jq1 ,
z 2  r2 e jq 2 .
r1  1, q1  0,
r1  1, q 2  2p .
The corresponding points in the w-plane f(z1 ) and
f(z2 ) are very close together. Like z, the radii of f(z1 )
and f(z2 ) are the same, but their angles are
different.
j1
w1  r1e ,
w2  r 2 e j2 .
r1  r 2 .
1  2 .
As z follows the circular closed path, what path will
f(z) follow?
To get an idea of the path f(z) will follow, let us look
at log f(z).
j
log f ( z)  log re  log r  j.
log w1  log f ( z1 )  log r1e j1  log r1  j1.
log w2  log f ( z2 )  log r2e j2  log r2  j2 .
The difference between log w1 and log w2 is
log w2  log w1  log r 2  log r1  j 2  1 
 j 2  1 .
Now log w can be written as an indefinite integral:
dw
log w  
w
The difference between log w1 and log w2 can be
written as a definite integral:
log w2  log w1  
2
1
dw
w
The points 1 and 2 in the integral

2
1
dw
w
correspond to z1 and z2. So,
log w2  log w1  
z2
z1
'
z 2 f ( z ) dz
df ( z )

.
z1
f ( z)
f ( z)
Combining this expression with
log w2  log w1  j 2  1 ,
we have
log w2  log w1  
z2
z1
f ' ( z )dz
 j 2  1 .
f ( z)
So, the integral from z1 to z2 is the same as the
integral around the closed curve in the z-plane.

C
'
f ( z )dz
 j 2  1 .
f ( z)
To summarize, as z follows a closed path in the zplane, w=f(z) follows a closed path in the w-plane.
The angular rotation that w takes is equal to 2 – 1.
Since w=f(z) follows a closed path in the w-plane,
the angular rotation, 2 – 1, must be an integral
multiple of 2p.

C
f ' ( z )dz
 j 2p N .
f ( z)
Suppose G(s) is a polynomial fraction:

s  z1 s  z2 
G( s) 
.
s  p1 s  p2 
Let s take on an infinite circular path in the righthalf complex plane.
Im{s}
s  +j
r
s  +
Re{s}
C
s  -j
What kind of path will G(s) take?
To answer this question, let us try to evaluate the
integral
'
G ( s)ds
C G(s) ,
where C is the curve described on the previous
slide.
We will try to evaluate this integral by removing
poles and zeroes in the right-half plane. Suppose
p1 is a right-half plane pole. We can define a new
function H(s) with this pole removed:

s  z1 s  z2 
s  p1 G(s) 
 H ( s).
s  p2 
So,
H ( s)
G( s) 
.
s  p1 
If
H (s)
G( s) 
,
s  p1 
we have
'
H
(
s
)
H
( s)
G ' ( s)  

.
2
s  p1  s  p1 
If
H (s)
G( s) 
,
s  p1 
and
H ( s)
H ' ( s)
G ( s)  

.
2
s  p1  s  p1 
'
we have
G ' ( s)
1
H ' ( s)


.
G( s )
s  p1  H (s)
The integral of
G ' ( s)
1
H ' ( s)


.
G( s )
s  p1  H (s)
is equal to
G ' (s) ds
ds
H ' (s) ds
C G(s)  C s  p1   C H (s) .
By the Cauchy Integral Formula, the integral
ds

 2p j.
C 
s  p1 
So
G ' (s) ds
H ' (s) ds
C G(s)  2p j  C H (s) .
Now suppose z1 is a right-half plane zero. We can
define a new function F(s) with this zero removed:
s  z2   F (s).
G( s)

s  z1  s  p1 s  p2 
So,
G(s)  s  p1 F (s).
If
G(s)  s  z1 F (s).
we have
G' (s)  F (s)  s  z1 F ' (s).
If
G(s)  s  z1 F (s).
and
G' (s)  F (s)  s  z1 F ' (s).
we have
G ' ( s)
1
F ' ( s)


.
G(s) s  z1 
F ( s)
The integral of
G ' ( s)
1
F ' ( s)


.
G(s) s  z1 
F ( s)
is equal to
G ' (s) ds
ds
F ' (s) ds
C G(s)  C s  z1   C F (s) .
By the Cauchy Integral Formula, the integral
ds
C s  p1   2p j.
So
G ' ( s) ds
f ' ( s) ds
C G(s)  2p j  C f (s) .
If we continue eliminating poles and zeroes, we get
a term of -2pj for every pole and a term of +2pj for
every zero.
So
G ' ( s) ds
C G(s)  2p j(Z  P),
where Z is the number of zeroes, and P is the
number of poles.
Now from our previous result,

C
'
'
f ( z )dz
G ( s)ds
 j 2p N  
 j 2p N ,
C G( s)
f ( z)
we see that
N  Z  P,
where N is the number of rotations of G(s), Z is the
number of right-half plane zeroes, and P is the
number of right-half plane poles.
By looking at the number of clockwise rotations of
G(s), we can find the number of right-half plane
zeroes minus the number of right-half plane poles.
We have already done two examples:
f ( z)  z 2 .
1
f ( z)  .
z
We have already done two examples:
f ( z)  z 2 .
1
f ( z)  .
z
The first function did two counter-clockwise rotations
for each counter-clockwise rotation of z.
The second function did one clockwise rotation for
each counter-clockwise rotation of z.
Equivalently, for
G( s )  s 2 .
1
G (s)  .
s
The first function will do two clockwise rotations for
each clockwise rotation of s. These clockwise
rotations correspond to two zeroes (at zero).
The second function will do one counter-clockwise
rotation for each clockwise rotation of s. This
counter-clockwise rotation corresponds to one pole
(at zero).
Example: How many times does the following
transfer function circle the origin as s goes from -j
to +j ?
s2
G(s) 
.
s 1
Solution: There is one RHP pole and no RHP zeros
(The term s+2 corresponds to a LHP zero: s = -2.)
The plot of G(s) will circle the origin in the counterclockwise direction once.
In MATLAB, we can plot G(s) using the control
function nyquist().
>> EDU» s = tf('s');
>> H = (s+2)/(s-1);
>> nyquist(H)
The Nyquist plot is on the following slide.
Nyquist Diagrams
From: U(1)
2
1.5
1
To: Y(1)
Imaginary Axis
0.5
0
-0.5
-1
-1.5
-2
-3
-2.5
-2
-1.5
-1
-0.5
Real Axis
0
0.5
1
1.5
2
In control systems, we are often concerned about
the poles of a closed-loop transfer function
G ( s)
T (s) 
.
1  G ( s)
The poles of T(s) are the zeros of 1+G(s).
The right-half plane poles of T(s) are the right-half
plane zeros of 1+G(s).
If we did a plot of
1  G( s)
and 1+G(s) had no right-half plane poles, then the
number of clockwise rotations around the origin is
equal to the number of right-half plane zeros of
1+G(s) or the number of right-half plane poles of
T(s) .
The number of clockwise rotations around the
origin of
1  G( s)
is equal to the number of clockwise rotations of
G (s )
around s = -1.
Therefore, if 1+G(s) has no right-half plane poles,
the number of clockwise rotations around s = -1 of
G (s )
is equal to the number right-half plane poles of
G ( s)
T (s) 
.
1  G ( s)