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( )
( )
( )
0
0 0
0 lim
z z
f z f z f z
z z →
−
′ =
− , (13)
provided this limit exist.
By expressing the variable z in definition (13) in terms of the new complex variable 0
z z z
∆ = − , we can write that definition as
( )
(
0)
( )
00 limz 0
f z z f z f z
z ∆ →
+ ∆ −
′ =
∆ . (14)
Note that, because f is defined throughout a neighborhood of z0, the number f z
(
0+ ∆z)
is always defined for ∆z sufficiently small. Let ∆ =w f z(
0+ ∆z)
− f z( )
0 , then if we write dwdz for
( )
f′ z , equation (14) becomes
0 lim
z
dw w
dz ∆ → z ∆ =
∆ . (15)
Suppose that f z
( )
=z3. At any point z,(
)
( )
( )
3 3
2 2
2 2
0 0 0
lim lim lim 3 3 3
z z z
z z z
w
z z z z z
z z
∆ → ∆ → ∆ →
+ ∆ −
∆
= = + ∆ + ∆ =
∆ ∆ . Hence
2 3 dw
z
dz = or
( )
2 3 f′ z = z .2. Differentiation Formulas
0 d
c
dz = , 1
d z
dz = ,
( )
( )
d
cf z cf z dz
′
= . (16)
Also, if n is a positive integer
1
n n
d
z nz dz
−
= (17)
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 zdz = ′ ′
(3)
( )
( )
( ) ( )
( ) ( )
( )
(
)
2f z f z g z g z f z d
dz g z g z
′ − ′
= , when g z
( )
≠0.To find the derivative of
(
2)
52z +i . According to the theorem, we have
(
)
(
)
( )
(
)
5 2
4 4
2 2
2
5 2 4 20 2
d z i
z i z z z i
dz +
= + = + .
Exercises
1. Apply definition (15) of derivative to find f′
( )
z when a. f z( )
1, z 0z
= ≠
b. b. f z
( )
=3z2−2zc. c.
( )
4 1 2 f z = z+
2. Apply definition (15) of derivative to find f′
( )
z when f z( )
=z3−4z at pointa. z= z0 b. z=i
b.
( )
(
3)
21 3
f z = − z
c.
( )
3 2
z i f z
z + =
−
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 + = ′ + ′ .
