1.
Functions of a Complex Variable
Let E be a set of complex numbers. A function f defined on E is a rule that assigns to each z in E a complex number w . The number w is called the value of f at z and is denoted by
( )
; that is, w = ( )
. The set E is called the domain of definition of f , written
=
{ z ∈ » : ( ) defined
}
.
( )
,
Example 1
Determine the domain definition of = z + 1 z
2 + + 1
and
( ) = z
2 z 1
Solution:
= { z ∈ » :
=
{ z ∈ » :
( ) defined
}
2 z z 1 0
}
=


 z ∈ » : z
1 1
≠ − ±
2 2 i 3



( ) = { z ∈ » :
= »
}
2.
Limits
Let a function f be defined at all points z in some deleted neighborhood of z
0
. The statement that the limit of
( )
as z approaches z is a number
0 w , or
0 lim z → z
0
( ) = w
0
(2) means that the point w = ( )
can be made arbitrarily close to w
0
if we choose the point z close enough to z
0 but distinct from it. Equation (2) means that, for each positive number
, there is a positive number δ such that
( ) − w
0
< ε whenever 0 < z − z
0
<
. (3)

Figure 4. Limit function
Example 1
Show that lim z
2 − = 2 z → i
Solution:
From the properties of modulli, we have
( z
2 − 1
) ( ) = z
2
1 z i z i .
Observe that when z » is in the region z i 1 , z i z i 2 i ≤ z i 2 i 1 2 3 .
Hence, for z » such that z i 1 ,
( z
2 − 1
) ( ) = z
2
1 z i z i 3 z i .
For any positive number
, get δ = min
 ε
,1

 
 3 
such that 0 < z i
,
( z
2 − 1
) ( ) < 3 z i 3
ε
 
 
= ε . ■
Note that when a limit function
( )
exist at a point z
0
, it is unique.
Example 2
If z
= , then z z lim
→ 0 does not exist.
When z = ( x , 0
) is a nonzero point on the real axis,
=
− 0
+ 0
= 1 and when z = (
0, y
)
is a nonzero point on the imaginary axis,

=
0 − iy
0 + iy
= − 1 .
Thus, by letting z approach the origin along real axis, we would find that the desired limit is 1.
An approach along imaginary axis would, on the other hand, yield the limit -1. Since the limit is unique, we must conclude that z lim
→ 0 z z does not exist.
Since limits of the latter type are studied in calculus, we use their definition and properties freely.
Suppose that
( ) = (
,
) + (
,
)
, z
0
= x
0
+ iy
0
, and
Then if and only if lim
( ) → ( x
0
, y
0
)
(
,
) = u
0 lim z → z
0
Example
Find z lim i

 z
2 +
1 z


.
( ) = w
0 and
Observe that z
2
1
( x iy
) 2
+ = + + z
1 w
0
= u
0
+ iv
0
. lim
( ) → ( x
0
, y
0
)
(
,
) = v
0
=
( x
2 − y
2 )
+ x x
2 + y
2
+ i



2 xy − y x
2 + y
2



.
We have
( ) =
( x
2 − y
2
)
+ x x
2 + y
2
and
( ) = 2 xy − y x
2 + y
2
. From Theorem 1, lim
( ) ( )



( x
2 − y
2
)
+ x x
2 + y
2



=
1
2 and lim 2
( ) ( )


 xy − y x
2 + y
2



=
3
2
, thus z lim i

 z
2
1  1 3
+ = + z 2 2 i .
(4)
(5)
If lim z → z
0
, lim z → z
0 exist and c ∈ » , then

(1) lim z → z
0
(
(3) lim z → z
0
(
( ) + ( ) ) exist and lim z → z
0
(
(2) lim z → z
0
( ) exist and lim z → z
0
( ( ) ) = c lim z → z
0
( )
) exist and lim z → z
0
(
( ) + ( ) ) = lim z → z
0
)
= lim z → z
0
+ lim z → z
0
( ) lim z → z
0
( )
(4) lim z → z
0



 exist and lim z → z
0







 = lim z → z
0 lim z → z
0
, whenever lim z → z
0
Limits Involving the Point at Infinity
We have three point about limits that is involving the point at infinity:
1 lim z → z
0
= ∞ if and only if lim z → z
0
= 0 .
≠ 0 z lim
→∞
( ) = w
0 if and only if z lim
→ 0 f
 
 
= w
0
. z lim
→∞
= ∞ if and only if z lim
→ 0 f
1
( ) z
= 0 .
Example 1
Observe that z lim
→− i i + 3 z + i z + i
= ∞ since z lim
→− i i + 3
= 0 .
Example 2
Observe that z lim
→∞
3 − z + 2
= 3 since z lim
→ 0
3
( ) z
− i
+ 2 z
= lim z → 0
3 − iz z
= 3 .
Example 3
Observe that z lim
→∞
3 z
4 − i z
3 + 2
= ∞ since z lim
→ 0
3 z z
+ 2
− i
= z lim
→ 0 z + 2 z
4
3 − iz
4
= 0 .
Exercises
1.
For each of the functions below, describe the domain of definition that is understood
(a) = z 2
1
+ 4
(c) = cos
( z
2 − i
)
(6)
(7)
(8)

(b) = z + 2 i z + z
2.
Write the function
( ) = z
3 + 2 z i in the form
( ) = (
,
) + (
,
)
.
3.
Let z c denote complex constant. Use definition (3) to prove that
0
,
(a) lim z → z
0 c = c (b) z lim i
( x + 2 ) = − i
4.
Let
2 z
= z
2 a.
Find z lim
→ 0 b.
Find z lim
→ 0 c.
Find z lim
→ 0 along the line y = x
along the line y = 2 x
along the parabola y = x
2 d.
What can you conclude about the limit of
5.
Using (6), (7) and (8) of limits, show that
(a) z lim
→∞ z
4 − z
3 + 2 z
( z + 1
) 4
= 1
( ) along
(c) z lim
→∞ z
2 − 1
= ∞ z + 1 z → 0
(b) z lim
→ 2 i z
( z − 2 i
) 2
= ∞
6.
Find the value of limits below
(a) z lim
→ + i z 2 + 2 z − 1
(b) z lim
→− 2 i z
2 + 2 z − 1 z
2 − 2 z + 4 z
3 + 8
(c) z
( lim
→ + i 3
) z
4 + 4 z
2 + 16
(d) lim z → i z
4 − 1
(e) lim z → i z
2 + 4 z + 2 z + 1
(f) z lim i z
2 1 3 i z
2 − 2 z + 2