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19.2 Complex Functions 679

1. ( f + g) (z) = f (z) + g (z),

2. ( f − g) (z) = f (z) − g (z),

3. (cf ) (z) = cf (z) for any number c,

4. ( fg) (z) = f (z)g (z) + f (z)g(z),

f g(z) f (z) − f (z)g (z)
5. (z) = .
g (g(z)) 2
6. There is also a complex version of the chain rule for differentiating a composition f ◦ g,
which is defined by ( f ◦ g)(z) = f (g(z)). Assuming that the derivatives exist, then

( f ◦ g) (z) = f (g(z))g (z).

In the Leibniz notation,
d df dw
( f (g(z)) =
dz dw dz
where w = g(z).
In form, this looks just like the chain rule for real functions in calculus.
Not every function is differentiable.
EXAMPLE 19.8
Let f (z) = z. Suppose f is differentiable at z. Then
f (z + h) − f (z)

f (z) = lim
h→0 h
z + h − z h
= lim = lim .
h→0 h h→0 h
For f (z 0 ) to exist, we need this limit to be the same no matter how h approaches zero. If h

approaches 0 along the real axis, h = h and h/h = 1, so this limit would have to be 1. But if h
approaches 0 along the imaginary axis, then h = ik for real k, and
h ik −ik
= = =−1.
h ik ik
Since this quotient is always −1 on the imaginary axis (no matter how close k is to zero), the
limit defining f (z) would have to be −1. We conclude that the limit defining this derivative does

not exist. This function has no derivative at any point.
As with real functions, a complex function is continuous wherever it is differentiable.

THEOREM 19.3

If f is differentiable at z 0 then f is continuous at z 0 .
To see why this is true, begin with

f (z 0 + h) − f (z 0 )
f (z 0 + h) − f (z 0 ) = h .
h
As h → 0, the difference quotient in parentheses on the right approaches f (z 0 ), and the factor of

h approaches 0, so the right side approaches 0. Therefore
lim( f (z 0 + h) − f (z 0 )) = 0
h→0
f (z) = f (z 0 ).
and this is equivalent to lim z→z 0

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October 15, 2010 18:5 THM/NEIL Page-679 27410_19_ch19_p667-694