You are familiar with derivatives of functions from to , and with the motivation of the definition of derivative as the slope of the tangent to a curve. For complex functions, the geometrical motivation is missing, but the definition is formally the same as the definition for derivatives of real functions.

exists. In this case, we denote this limit by and call the

By the definition of limit, we can say that is differentiable at
if
, and is a limit point of
and there exists
a function
such
that is continuous at , and
such that

It is sometimes useful to rephrase condition (10.2) as follows:
is
differentiable at if
, is a limit point of
,
and there is a function
such that is continuous at , and

and let . Then for all ,

If we define by

then is continuous at , so is differentiable at and

We could also write this calculation as

Hence is differentiable at and for all .

Let be defined by

Then is continuous at , so is differentiable at , and

for all .

Also it is not good form to say

without specifying the condition `` for ," since someone reading (10.8) would assume is undefined at .

I claim that does not have a limit at , and hence is a nowhere differentiable function.

Let

Then and are sequences in both of which converge to . For all ,

so and , and hence does not have a limit at .

- a)
- are given complex numbers.
- b)
- .
- c)
- , i.e. .

be complex functions, and suppose and are differentiable at . Suppose is a limit point of . Then is differentiable at and .

Proof: Since are differentiable at , there are functions , such that , are continuous at , and

It follows that

and is continuous at .

We can let
and we see is differentiable at
and

Proof: The proof is left to you.

Proof: From our hypotheses, there exist functions

such that is continuous at , is continuous at and

If , then , so we can replace in (10.15) by to get

Using (10.14) to rewrite , we get

Hence we have

and is continuous at . Hence is differentiable at and

Proof: If
, we saw above that is
differentiable and
. Let be a complex function, and
let
. Suppose is differentiable at , and . Then
. By the chain rule is differentiable
at , and

(Use induction.)

for all

. Suppose and are differentiable at and , and is a limit point of . Show that is differentiable at and