8. Functions of a Complex Variable
8.1. Revision of Complex Numbers
A complex number is a number of the form z
 a
 ib where a , b

R and i
2  
1 vector space of complex numbers. a is the real part of z , written a = Re z , b is the imaginary part of z , written b =
Im z . Sometimes z
 a
 ib is written as the ordered pair ( a , b ). a

0 i

( a , 0 ) is a real number
0
 bi

( 0 , b ) is an imaginary number
0
 i 0

( 0 , 0 ) is the complex number zero a
 ib is the complex conjugate of z
 a
 ib denoted by z
If
(i)
(ii)
(iii)
(iv) z
1 z z z
1

1 z
 z a
1
2
 z
2

 a
2
( ib z
1
 z
2
iff
 a a
1

1
1
, and
 b a

2
1 ib a

1
- a
2
)( a z
 a
2
2
2 i (
 b

1 a
2
and
 ib
2
 b
1 ib b
2
)

)
real number

2 a z
1
then, b a
2

2 z
 b
1 b
2
 i ( a
1 b
2

2 a

2 Re z
 a
2 b
1
)
- a real number
(v) z
 z

2 ib

2 i Im z - an imaginary number
(vi) z
1 z
2
 z
1 z
2 z
2 z
2
 a
1 a
2 a
2
2

 b
1 b
2 b
2
2
 i b
1 a
2 a
2
2

 a
1 b
2 b
2
2
A complex number z
 a
 ib

( a , b ) can be represented geometrically by a point in the z -plane or the Argand diagram
Im z b
 r
 a
 jb

 z
Re z
A complex number may also be written in polar form using polar coordinates ( r ,

) , in the complex plane z
 a
 jb

 r r cos
  jr
 cos
  j sin sin


 where r
 a
2  b
2  z z is called the modulus of z and denoted by z and

, satisfying cos
  a / r , sin
  b / r is called the argument of z and denoted by arg z .
1

Note that z lying in
 
2 k

for k

0 ,

1 ,

2 , ....
all represent the same complex number. The value of arg
     
is known as the principal value of arg z .
The following properties hold:
(i)
(ii)
(iii)
(iv) z
 z , z
1 z
2 z
1
 z
1

 z
2 z
2 z
1

 arg z
2
, z
 arg

 z
1 arg z
2
 z
 z
1
 z
2 z
1
 z
2 arg z
1
 arg z
2
Euler’s formula is
(i)
(ii) e i
 e
 i




1 cos
  e i
 j sin
 cos
  i sin
 giving
 and which allows the polar form of z to be written as z
 re i

or z
 re i (
 
2 k )
, k

0 ,

1 ,

2 , ...
and gives De Moivre’s theorem z n 
 r
 cos
  j sin
   n  r n
 cos n
  j sin n
  which holds for all rational numbers n . This theorem allows us to find the nth roots of a complex number z . z
1 n

 r
 r
1 n
 cos
 

 cos
 j sin

2 k

 
1 n
 n j sin
 n
2 k
 
 for k = 0, 1, 2, … , n1 giving n different values for z
1 / n
.
Remark:
By Taylor’s Theorem, we have e i
 
1
 i

1 !

( i

2 !
)
2

( i

3 !
)
3

( i

)
4
4 !

1
 i

1 !


2 !
2
 i

3
3 !


4
4 !
  
  

4 k
( 4 k )!
 i

( 4 k
( i

)
4 k
( 4 k )!

4 k

1

1 )!

( i

( 4 k
)
4 k

1

1 )!


( 4 k
4 k

2

2 )!

( i

( 4 k
)
4 k

2

2 )!
 i

( 4 k
4 k

3

3 )!
 
( i

( 4 k
)
4 k

3

3 )!
 

1


2 !
2
 cos



4
4 !
 i sin

  

4 k
( 4 k )!


4 k

2
( 4 k

2 )!
   i



1 !


3 !
3
  

( 4 k
4 k

1

1 )!


( 4 k
4 k

3

3 )!
 


As we learned that by Taylor’s Theorem, we have cos
 sin


1


2 !
2


1 !


3
3 !


4 !
4
 



( 4
 k


4 k
( 4 k )!

4 k

1

1 )!

( 4

4 k

2 k

2 )!
 
( 4

4 k

3 k

3 )!
 
8.2. Curves and Regions in the Complex Plane
2

If x and y are real variables, then z
 x
 iy is complex variable. Curves and regions in the complex plane may be represented by equations and inequalities involving z .
Example
If z
0
 x
0
 iy
0
is a fixed complex number, z
 z
0
is the distance between z and z
O
, and z
 z
0
= r is the equation of a circle centered at z with radius r .
0
Im z r z o
Re
Re z z
 z
0
 r describes any point inside the circle and specifies an open circular disc. z
 z
0
 r includes points on the circle and is a closed circular disc. z
 z
0
 r is the region exterior to the circle. Similarly, r
1
 z
 z
0
 r
2
describes an open annulus
Im z r
1 z o r
2
Re
Re z
Example
(i) arg z
  
1 half line
3

Im z
Im arg z
 

Re z
(ii) y > 0 upper half plane y < 0 lower half plane x > 0 right half plane x < 0 let half plane
(iii)
2

Im z

5 a strip of width 3
Im z
5
2
Re z
0
Example
Determine the locus of z

2
 z

3
Sol:
Geometrically: z must be equi-distant from the points –2 and –3, i.e. the locus is the line x = 5/2.
“Mathematically” : Let z=x+jy , then

 x

2

 x
2

 jy y
2



1 2
2




 x x


3
 jy
2


 y
3
2

1 2
Im z x
2 
4 x

4
 y
2 x

 x
2

5

6 x
2

9
 y
2
Re z

3 
5
2
2 0
8.3. Functions of a Complex Variable
4

If, to each value of a complex number z in some region of the complex plane there corresponds a unique value of another complex variable w , then w is a function of z , i.e. a function of complex variable, written w
 f ( z ) . Let u and v be the real and imaginary parts of w , then since z
 x
 iy , u and v will in general be functions of x and y and we write w = u ( x , y ) + jv ( x , y )
Example w
 z
2  z is defined for all z

C .
Since w

( x
 j y )
2 
( x
 j y )

( x
2  y
2  x )
 j(2 xy
 y ) we have u ( x , y )
 v ( x , y )
 x
2 
2 xy
 y y
2  x
For example, f (3+2j) = 2+10j
Example and f ( 4
 w

3 z


3 ( x
 z jy )

( x


4 x

 u ( x ,
2 jy y )
 j )

16

2 j jy )
4 x , v ( x , y )
 
2 y
In order to plot a “graph” of the function w
 f ( z ) we use two separate Argand diagrams, one for the z-plane and one for the w-plane.
Example w
 z
2

( x
 j y )
2
 x
2  y
2 
2 j xy
 u
 x
2  y
2 and v

2 xy or in terms of polar coordinates, if z
 re j

, w
 r
2 e
2 j

 w
 r
2 and arg w

2

5

v y w z z -plane
 r w -plane
2
 r
2 x u
A region R in the z -plane will be mapped into a region R
’ in the w -plane. For example if R is the first quadrant of the z -plane R : 0
 r
 
, 0
  

2
then its image R ’ in the w -plane will be the upper half plane. y v x zplane w -plane
R : O
 r
R

: O
 r
 
,
 
,


2

 


 
2
 maps to
A curve C in the z -plane will map to a curve C
’ in the w -plane.
For example:
(i) C : y = x maps to C
’: u = 0
(ii) C : y = x + 1 maps to
C
 u v

 x
2
2 x

 x

1

2
 x
1





2 x
2
2 x


1
2 x
Eliminating x gives u
2 
2 v

1 -a parabola
(iii) C : y = a maps to
C


: u
 x
2 v

2 xa
 v
2 
4 a
2 a
2
 u
 a
2

- a parabola
Hence: u
6

B y
P a
A v
2
2 a A
’
 a
O a x

P ’ a
2 u z -plane w -plane

2 a
2
B
’ x -axis

positive u -axis
0 P

0’ P
’
(
 a , a )

( 0 ,

2 a
2
)
8.4. Analytic Functions
We need to define the concepts of limit, continuity and differentiability for functions of a complex variable. A function f ( z ) has the limit  as z
 z
0
, if for any real
 
0 ,

real
 
0 such that for all z
 z
0
such that z
 z
0
 
, f ( z )
   
, we write lim z
 z
0 f ( z )
  or f ( z )
  as z
 z
0
Geometrically, this means that f ( z ) must lie within an open disc with centre  and arbitrary small radius

whenever z lies within an open disc of radius

and with centre z
0 y
 z z
O v
  f ( z ) z -plane x wplane u
Note that z may approach definition for a function z from any direction and is hence more restrictive than the equivalent
0 f ( x ) of a real variable. Indeed it is closely related to the definition for a function of two real variables.
Example y
Find z lim

3
 j
( z
2  z ) along the 3 paths shown.
( c )
( b )
1
Proof:
(a) z = x +j 0
3
( a ) x
7

z
2  z

( x
 j )
2 
( x
 j)
 x
2  x

1
 j ( 2 x

1 )

5

5 j as z

3
 j i.e.
x

3
(b) z =3+j y z 2  z

( 3
 j y ) 2 
( 3
 j y )

6
 y
2 
5 y j

5

5 j as y

1
(c) z = k (3 + j), k -real z
2  z
 k
2
( 8

6 j )


5

5 j as k k ( 3


1 j )
In each case the limit is the same.
Example
Find lim z

0 z z
along (a) the real axis, (b) the imaginary axis.
Sol:
(a) z
 x , z
 x
 z z

1 and the limit is 1
(b) z
 j y , z
  j y
 z z
 
1 and the limit is –1
More generally if z
 e j

, z
 re
 j

and z / z
 e

2 j

which will give different limits for every different straight line to the origin. Hence lim z

0 z z
does not exist.
If f ( z ) and g ( z ) each possess a limit at z
 z
0
, the usual rules for the limit of the sum, difference, product and quotient function apply.
A function f(z) is continuous at z
 z
0
if (i) f ( z
0
) exists (ii) z lim
 z
0 f ( z ) exists and equals f ( z
0
) .
8

Example z lim

1
 j 
( z
2 z 2

1 )( z
 z


1 j)




[( 1
 j )
2 
1 ][ 1
 j


 z lim

1
 z j
( z
2 lim

1

 j
(
1 z 2
)

 z lim

1
 j
( z z

1 )
 j]/[(1
 j)
2 
( 1
 j)

1] j)



8
 j
 f
13
( 1
 j)
 continuous at z

1
 j
Example f ( z )
 z
2 z

4
is continuous everywhere except at z

2 j where the function is not defined.
If fg f ( z ) and g ( z ) are continuous in so e region (i.e. at all points of the region) then so are
and f / g (if g

0 ). f
 g ,
A function limit f ( z ) is differentiable at some point f ' ( z
0
)
 lim
 z

0

 f ( z
0
  z )
 z
 z
 z
0
if the f ( z
0
)

 exists. This limit is then the derivative of f ( z ) at z
0
.
Example f ( z )
 z
2
is differentiable for all z and f ' ( z )

2 z since f ' ( z )


( lim
 z

0 lim
 z

0

2 z z


 z )
 z
 z

2


2 z z
2
All the usual rules of differentiation for functions for a real variable (sum, difference, product, quotient, chain rule etc) hold for functions of a complex variable.
If f ( z ) is differentiable at z , it is also continuous at
0 z . However there are many simple functions
0 which do not possess a derivative.
Example f ( z )
 z
 x
 j y
Consider
9

 f ( z

 x x
  z )
 z


 j
 y j
 y f ( z )

 x
  x
 j ( y
 x


 y )

 j
 y
( x
 j y )
Along the path parallel to the x -axis,
Along the path parallel to the y -axis,
 y

0 and the limit is 1.
 x

0 and the limit is –1.
Hence z is not differentiable at any point z . y

1 z 
1 x
O
A function f ( z ) is said to be analytic in a region R if it is differentiable at all points of R . A function is analytic at a point z
 z
0
. If it is analytic in a small region surrounding z
0
. The terms regular and holomorphic are sometimes used for analytic.
If a function f ( z ) is analytic except at a finite number of points, these points are called singularities.
Example
(i) 1 , z , z
2
, z
3
etc and more generally the polynomial f
 a
O
 a
1 z
 a
2 z
2    a n z n
are analytic everywhere.
(ii)
(iii) f ( z )

1
1
 z is analytic everywhere except at z =1 which is a singularity. f ( z )
 z is analytic nowhere.
Example
Show that f ( z )
 z
2 is differentiable at 0 but not analytic at 0.
Proof:
We have f ( 0 )

0 . And
10

 z lim

0 lim z

0 f ( z )

( 0 )
 x z
2


0 y
2
(
 x f
2  y x

2
 z lim

0 yi ) z z
2
 z lim

0
 z lim

0
( x
 x
2 x

 y yi
2 yi )

 z lim

0 lim x

0
( x y

0

 x
2  y
2
(

( x
 yi )( x x

 yi ) yi ) yi )

0
For z
0
 x
0 f ( z )
 z
 f z
0
( z
0
)
 y
0 i

0 , consider
 x
2  x
 y x
0
2 

 x
0
2

 y
 y y
 i
0
0
2

 x
2 x

 x
0 x
0
2

 y
2

 y
 y y
 i
0
0
2
Along the line y
 y
0
 m ( x
 x
0
) y
 y
0 z lim
 z
0
 m ( x
 x
0
)
 x lim
 x
0 x
2 f ( z
 x
0
2
)

 f ( z
0
) z
 z
0
2 y
0 m ( x
( x
 x
0
 lim x
 x
0 x
2 x

 x
0 x
0
2
 x
0
)( 1

)
 mi ) m
2
( x




 y y
0
0 x
0
)
2

 m ( x m ( x

 x
0 x
0
)
)

2

 lim x
 x
0 x
 x
0
 y
0 y
 i
0
2

2 y
0 m
1

 mi m
2
( x
 x
0
)

2 x
0

1

2 y
0 m mi
As we can see, the limit of the above depends on the slope of the line given so different slopes will give different limits. So f ( z )
 z
2 is not differentiable at any z
0
 x
0
 y
0 i

0 .
Thus, although f ( z ) is differentiable at z = 0, f ( z ) is not analytic at z = 0.
8.5. The Cauchy-Riemann Equations
We shall now obtain conditions on u and v where f (z) = u +j v for f ( z ) to be analytic.
Assume f ( z ) is analytic at z , then
 f ' ( z )
 lim
 z

0 f ( z
  z )
 z
 lim
 z

0

 u ( x
  x , y
  y )
 f ( z ) j v ( x

 x
 x ,
 y j
 y
  y )
 u ( x , y )
 j v ( x , y )

 and this limit is independent of the path.
(i)
Suppose
 y

0 i.e. a path parallel to the xaxis f ' ( z )


 lim
 x

0

 u ( x
  x , y lim
 x

0

 u (
 u
 x
 j
 v
 x x


 x ,

 f x y )
 x
)

 u j v ( x
( x , y
  x ,
 x
)
 y )
 u ( x , y )
 j v ( x
  x , y
 x
) j v ( x , y )


 v ( x , y )


(ii)
Now suppose
 x

0 i.e. a path parallel to the y -axis
11

f ' ( z )
 lim j
 y

0

 u ( x , y
  y )
 j v ( x , y
  j
 y y )
 u ( x , y )
 j v ( x , y )


 lim
 j

0

 u ( x , y
  y )
 u ( x , y ) j
 y
 v ( x , y
  y )
 y
 v ( x , y )



1
 j
 u
 y

 v
 y


 u j
 y
 v
 y

1 j

 f
 y
These two expressions must be the same if imaginary parts to give
 u
 x

 v
 y
and
 v
 x f ( z ) is analytic and so we may equate real and
 
 u
 y
. These conditions are known as the
Cauchy-Riemann equations.
We have proved that these conditions are necessary for f ( z ) to be analytic. Although we shall not give a proof here, sufficient conditions for f ( z ) to be analytic are that u , v , u x
, u y
, v x and v y
are continuous and that u and v satisfy the C-R equations.
Example
Show that f ( z )
 z Re z is differentiable at z = 0 and nowhere analytic.
Proof: f ( z )
 u ( x , y ) z

Re z

( x
 x
2
, v ( x , y )
 yi ) x
 xy x
2  xyi then u ( x , y )
 x
2
, v ( x , y )

are differentiable everywhere. xy . It is obvious that
In addition, u x

2 x , u
Y

0 , v x
 y , v
Y
 x . So at (0, 0) we have


 u u y x
(
( 0 , 0 )
0 , 0 )


 v v y x
( 0 , 0 )
( 0 , 0 )
.
Therefore f ( z )
 z Re z is differentiable at z = 0. However, for z
0
 x
0
 y
0 i

0 , u x
 v y or u y
  v x
, thus, f ( z )
 z Re z is not differentiable at z

0 . It follows that f ( z )
 z Re z is analytic nowhere.
Example (optional)
Given f ( z )
 u
 vi where u ( x , y )


 x
2 xy

2 y
2 if
0 if
( x
(
, x , y ) y )


( 0 , 0 )
(0,0)
and u ( x , y ), v ( x , y ) satisfy the Cauchy-Riemann condition at z = 0 but v ( x , y )

0 , show that f ( z )
 u
 vi is not differentiable at z
= 0.
Proof:
12

 u
 x
( 0 , 0 )
 lim x

0 u ( x , 0 ) x
 u (

0
0 , 0 )
 lim x

0
0 x

0

0

0 ;
 u
 y
( 0 , 0 )
 lim y

0 u ( 0 , y ) y
 u (

0
0 , 0 )
 lim y

0
0 y

0

0

0
Therefore,


 u u y x

 v y
 v x


0
0
In addition, we know that u ( x , y ) is not differentiable at ( 0 , 0 ) .
Along the line y=x , z lim

0 y
 x f ( z z
)


0 f ( 0 )
 lim z

0 y
 x xy
2 x
2  x
 y yi
2 x
3
 x lim

0 x
2 x

2 xi

1
2 ( 1
 i )
.
However, along y= 0, z lim

0 y

0 f ( z ) z
 f

0
( 0 )
 z lim

0 y

0 xy
2 x
2 x

 y yi
2
 x lim

0
0
 x
0
Example
(i) f ( z )

 u z
2

 x
2 x
2

 y
2 y
2

2 j xy v

2 xy
 u
 x
 u
 y


2 x ,

2 y ,
 v
 y
 v

 y
2
 x
2 y
Both equations are satisfied
 x , y and hence f ( z )
 z
2
is analytic everywhere
(ii) f ( z )
 z
 x
 j y u
 u


 x
 u
 y
 x
1
0 v
  y
 v
 y
 v
 x



1
0
13

The first C-R equation is not satisfied anywhere and hence f ( z )
 z is not analytic anywhere.
(iii) f ( z )
 z z
 z
2  z
2  y
2 u
 u


 x
 u
 y
 x
2 
2 x
2 y y
2 v

0
 v
 y
 v
 x

0

0
The C-R equations are only satisfied at z = 0 and hence f ' ( z ) exists only at z = 0 (when f ' ( z )

0 ). However f ( z ) is not analytic anywhere – why? Note that f ( z )
 z
2
is continuous everywhere.
In polar form, if f ( z )
 u ( r ,

)
 jv ( r ,

) the C-R equations become
 u
 r

1 r
 v
  and
 v
 r


1
 u r
 
Example
Suppose that u , v are the real part and imaginary part of the function that u , v are expressed in polar coordinates r ,
 w
 f ( z )
 u
 iv . Suppose
. Find the Cauchy-Riemann condition in polar coordinates.
Sol:
Suppose z

As we know
0 , i.e.
r x y


 r sin
0 r cos


.
, then
 u
 r
 v
 

 u
 x
 x
 r

 u
 y
 y
 r

 u
 x cos
 
 u
 y sin

 ( 1 )

 v
 x
 x
 

 u
 y
 y
 

 v
 x
(
 r sin

)

 v
 y
( r cos

)

1 r
 v
 
 
 v
 x sin
 
 v
 y cos

 ( 2 )
From (1) & (2), we have
 u
 r

1 r
 v
 



 u
 x

 v
 y

 cos
 


 u
 y

 v
 x

 sin

Also
14

 u
 

 
1 r
 u
 x
 u
 
 x
 


 u
 x
 u
 y sin
 y
 
 

 u
 y
 u
 x
(
 r sin cos


 ( 3 )
)

 u
 y
( r cos

)
 v
 r

 v
 x
 x
 r

 v
 y
 y
 r

 v
 x cos
 
 v
 y sin
  ( 4 )
From (3) & (4), we have

1 r
 u
 

 v
 r



 u
 x

 v
 y

 sin
 


 u
 y

 v
 x

 cos












 u x u x




 v
 y v y



 cos sin











 u y u y




 v
 x v x



 sin cos





 u
 r
1 r

1
 u r
 

 v
 
 v
 r


 cos
 sin
  sin
 cos









 u

 x u
 y


 v

 y v
 x







 u

 r
1 r

1
 u r
 


 v


 v r

 




 u

 x u
 y


 v

 y v
 x






 cos
 sin
 sin

 cos




1


 u

 r
1 r

1
 u r
 


 v


 v r


We see that if u , v satisfy the Cauchy-Riemann condition in Cartisian coordinates, that is,




 u x
 y
 u




 v y

 v x
then we have
 u
 r

1 r
 v
 

0 and

1 r
 u
 

 v
 r

0 , that is,






 u
 r
1 r
 u



1
 v r

 
 v
 r
Similarly, if u , v satisfy the Cauchy-Riemann condition in polar coordinates, that is,






 u
 r
1 r
 u



1
 v r

 
 v
 r then we have




 u x
 y
 u




 v y

 v x
We note that it can be shown that if f ( z ) is analytic in a region then it possesses derivatives of all orders in that region.
If w
 u
 iv may be written as a function of z only then w
 f ( z ) will be analytic. If however z appears in the expression, w will not be analytic (Exercise).
15

8.6 Harmonic Functions
If we have an analytic function f ( z ) = u ( x , y ) + jv ( x , y ), then the C-R equations are
 u
 x

 v
 y
 u
 y
 
 v
 x
Differentiating again gives
 2 u
 x
2

 2 v
 x
 y and
 2 u
 y
2
 
 2 v
 y
 x and so
 2 u
 x
2

 2 u
 y
2

0 . Similarly
 2 v
 x
2

 2 v
 y
2

0
Hence the real and imaginary parts of a analytic complex function satisfy Laplace’s equation
 2
 
 2

 x
2

 2

 y
2

0
A solution of this equation having continuous second order partial derivatives is called a harmonic function. Hence u and v above are harmonic functions.
If two harmonic functions u and v satisfy the C-R equations, i.e. they are the real and imaginary parts of some analytic complex function f ( z ) , they are said to be conjugate harmonic functions.
Example
Show that u ( x , y )
 x ( 1
 y ) is harmonic and find its conjugate harmonic function v ( x , y ). Write u ( x , y )+ iv ( x , y ) as f ( z ) .
Proof: u
 x
 l
 y


 2 u
 x
2

 2 u
 y
2

0
 u x u xx
 l


0 , y , u yy u y


0
 x and u is harmonic. To find v we use the C-R equations
(1)

 v y

 u
 x

1
 y and (2)

 v x
 

 u y
 x
Integrating (1) gives v
 y

1
2 y
2  g and equating with (2) gives
16

 v
 x
 g
  
 x
 g

1
2 x
2  c c -c o n s t a n t
 v
 y

1
2 y
2 
1
2 x
2  c
 f ( z )
 u
 j v
 x
 xy
 j ( y

1
2 y
2 
1
2 x
2
)
 j c f ( z )
 z

1
2 j z
2  j c
(either by inspection or x

1
( z
 z ), y

1
( z
 z ) ).
2 2 j
If u and v are conjugate harmonic functions, the level curves of u and v are orthogonal at points where f ' ( z )

0 .
Consider two level curves u
 
 u o
, v
Using the chain rule
 v o
0

 u
 x

 dy dx dy dx u u

 u

 y u y dy dx
 u x
 dy dx v
 u
0

 u x
 u y dy
 v y
 dx
 v v x
 v
 x
 v x

 v y

 v
 y
1 dy dx v and the curves are orthogonal.
8.7. Standard Functions
(a) Polynomials
The functions f ( z )
 z n
, n
= 0,1,2,….are analytic everywhere in the complex plane (such functions are sometimes called entire functions) and hence so is the polynomial of degree n , f
 a
O
 a
1 z
 a
2 z
2    a n z n where a
0
, a
1
,....., a n
are complex constants with a n

0 . Such a function had n roots (not necessarily distinct). If all the coefficients are real and z = a +j b is root then the complex conjugate z
 a
 j b is also a root.
(b) The exponential function
We define e z  e x
 jy  e x
(cos y
 j sin y ) which is analytic everywhere (check the C-R equations are satisfied). It is easy to show that,
17

( i) e z  e x
( ii) arg e z
(iii) e
(z

2k
 j)
 y
 e z k
 
1,

2,

3,....
(iv) d dz e z  e z
(c) Trigonometric Functions
We define cos z
 e j z 
2 e
-j z
, sin z
 e j z  e
-j z
, which we both analytic everywhere.
2 j
It is easy to show that,
(i)
(ii)
(iii) d cos dz sin cos
2
 z z
1
 z
 
 cos z
2
2
 sin


1 z , cos d dz z
1 sin z cos z
2
 cos z
 sin z
1 sin z
2
Indeed all the real trigonmetric identities carry over in the complex case.
(iv) cos( x
 j y )
 cos x cos j y
 sin x sin j y , sin( x
 j y )
 sin x cos j y
 cos x sin
(v)
(vi) cos j y cos( x

 cosh j y )
 y , sin j y cos cosh y

 j sinh j sin y x sinh y , sin( x
 j y )
 sin x cosh y
 j y j cos x sinh y
Similarly other trigonometric functions may be defined e.g. tan z
 sin z cos z
, sec z

1 cos z
with d dz tan z
 sec
2 z etc.
(d) Hyperbolic Functions
We define cosh z
 e z  e
 z
, sinh z
2
 e z  e
 z
2 which are analytic everywhere. All the real variable results are true for complex variables also.
(e) Logarithm
The natural logarithm of z , log z , is defined as the inverse of the exponential function, i.e. w =log z , for z

0 , satisfies
Let w
 u
 iv , z
 e w  re i
 z .
, then giving e u  r and e i v  e i

.
Thus u
 log r
 log z and v
   arg z so that
 j arg z
However, although we usually have
   arg z
 
, all values
 
2 k

, k

0 ,

1 ,

2 , .....
give the same complex number z , but different values for the logarithm i.e. log z is a multi-valued function.
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log z
 log z
 j (
 
2 k

) . Taking k = 0 gives the principal value of log z denoted by Log z . Then log z

Log z

2 k
 j
Example
(i) log( 1
 j)
 log 2
 j (

/ 4

2 k

)
(ii) Log(1
 j)
 log 2
 j

/ 4
Log z is analytic for z

0 and arg z
 
(where Log z is not continuous) and d dz
Log z

1
(z z

0, arg z
 
)
The usual rules of logarithms
(i)
(ii) log log
 z
1 z
1 z
2 z

2

  log
 log z
1 z
1

 log log z
2 z
2 do not hold for principal values but do hold if the arguments are suitably adjusted by multiples of
Example
Take z
1

3 z
2

2 j then z
1 z
2
 
6 j
Log

But, z
Log
1
 z
1
Log log

3

Log z
 j, Log z
2
 log 6

3
 j
2
2
 z
1 z
2
 log 6


2 j log 2
 j

/ 2
However these values only differ by 2 k
 j with k = 1.
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