1.
Derivatives
Let f be a function whose domain of definition contains a neighborhood of a point z0 . The
derivative of f at z0 , written f ′ ( z0 ) , is defined by the equation
f ′ ( z0 ) = lim
z → z0
f ( z ) − f ( z0 )
z − z0
,
(13)
provided this limit exist.
By expressing the variable z in definition (13) in terms of the new complex variable
∆z = z − z0 , we can write that definition as
f ′ ( z0 ) = lim
∆z → 0
f ( z0 + ∆z ) − f ( z0 )
.
∆z
(14)
Note that, because f is defined throughout a neighborhood of z0 , the number f ( z0 + ∆z ) is
always defined for ∆z sufficiently small. Let ∆w = f ( z0 + ∆z ) − f ( z0 ) , then if we write
dw
for
dz
f ′ ( z ) , equation (14) becomes
dw
∆w
= lim
.
dz ∆z →0 ∆z
(15)
Example
Suppose that f ( z ) = z 3 . At any point z,
3
z + ∆z ) − z 3
(
∆w
dw
2
2
lim
= lim
= lim 3z 2 + 3z ( ∆z ) + ( ∆z ) = 3z 2 . Hence
= 3z 2 or f ′ ( z ) = 3 z 2 .
∆z →0 ∆z
∆z →0
∆
z
→
0
∆z
dz
2.
Differentiation Formulas
Let c be a complex constant, and let f be a function whose derivative exist at a point z. It is
easy to show that
d
c=0,
dz
d
z = 1,
dz
d
cf ( z ) = cf ′ ( z ) .
dz
(16)
Also, if n is a positive integer
d n
z = nz n−1
dz
(17)
Theorem
If the derivation of two function f and g exist at a point z , then
(1)
d
 f ( z ) + g ( z )  = f ′ ( z ) + g ′ ( z )
dz 
(2)
d
 f ( z ) g ( z )  = f ′ ( z ) g ′ ( z )
dz 
(3)
d  f ( z )  f ′ ( z ) g ( z ) − g′ ( z ) f ( z )
, when g ( z ) ≠ 0 .

=
2
dz  g ( z ) 
( g ( z ))
Example
To
find
d (2z2 + i)
dz
the
derivative
of
( 2z
2
5
+ i) .
According
to
the
theorem,
5
4
4
= 5 ( 2 z 2 + i ) ( 4 z ) = 20 z ( 2 z 2 + i ) .
Exercises
1. Apply definition (15) of derivative to find f ′ ( z ) when
a.
1
f ( z) = , z ≠ 0
z
b.
b. f ( z ) = 3z 2 − 2 z
c.
c. f ( z ) =  z +


1

2
4
2. Apply definition (15) of derivative to find f ′ ( z ) when f ( z ) = z 3 − 4 z at point
a.
z = z0
b. z = i
3. Use result in Sec. 2 to find f ′ ( z ) when
a.
f ( z ) = 3z 2 − 4 z + 1
we
have
b.
f ( z ) = (1 − 3 z 3 )
c.
f ( z) =
2
z +i
3z − 2
4. If the derivation of two function f and g exist at a point z, proof that
d
 f ( z ) + g ( z )  = f ′ ( z ) + g ′ ( z ) .
dz 
5. Show that f ′ ( z ) does not exist at any point z when f ( z ) = z .