We provide the defintions of continuous and differentiable function in
.
We will see that it is not possible to simply extend the properties of real functions.
Definition
We consider
and
. We say that
is continuous at the point
if:

By recalling that
we have that
and so that
, we state and prove the following
Theorem
We consider
Also
. The function
is continuous at the point
if and only if the functions
as defined in (1.1) are continuous, in the notion of the continuity of functions defined on
made during analysis II, in point
.
Proof
Let
be continuous at the point
.
We have that:




We used the definition of
and the elementar inequality:
In an analogous way:

We know, by hypothesis, that
are continuous at
, that is:


Therefore we have that, called
:

Vice versa let
be continuous at
. From the definition of continuity for complex function we have that:

We see that:


So, given
and
, we obtain:


That is,
are continuous at
Now, we introduce the definition of differentiable functions in
; it will be clear that, while the definition of continuity is simply an extension of the known one for the functions in
, the notion of differentiability is more complicated that the pure extension of a known fact.
Definition
Consider
and
. We say that
is differentiable at the point
if the following limit exists and is unique:

The value of this limit is the derivative of the function
at the point
.
Remark 1: Stating that the limit is unique is equivalent to state that it must exists and be the same for every path followed to reach
.
Now, we provide some necessary conditions for differentiability of a complex function.
We will see that, in some cases, these conditions can become sufficient.
We repeat that the limit
must exist for every path used to reach
; we study two simple paths: those for which we arrive at
along parallel axis to those of
in the complex plane:
- Along the axis parallel to
: A generic
who belongs to this axis has the form
. Therefore we have that:

By taking the limit for
we have that:

![{\displaystyle =\lim _{x\rightarrow x_{0}}\left[\frac{u(x,y_{0})-u(x_{0},y_{0})}{x-x_{0}}+i\frac{v(x,y_{0})-v(x_{0},y_{0})}{x-x_{0}}\right]= }](//restbase.wikitolearn.org/en.wikitolearn.org/v1/media/math/render/svg/481a3cb07582202835c896da7f287b9052454dc7)

- Along the axis parallel to
: A generic
who belongs to this axis has the form
. Therefore, as before, we have that:

We remind that if
is differentiable at
the limit at
must exist and be unique. The two found limits, if
is differentiable at
, must be the same. We give the following
Definition
The condition:

are called the
Cauchy-Riemann equations.
For what said about the differentiability of a function at
, these equations are necessary conditions for the differentiability of a function at
. But, they are not sufficient: if the function is differentiable the two limits computed by getting closer along the real and the imaginary axis must be the same because the uniqueness of the limit along each path is valid.
If the function is not differentiable, the two limits can coincide even if the limit is not unique along each path. It can exist a path that denies the existence of a limit and so making the function not differentiable, even if it satisfies the Cauchy-Riemann equations.
Theorem
We consider
and
. If
and it is derivable at the point
, then it satisfies the Cauchy-Riemann condition
Remark 2:
If Cauchy-Riemann equations are satisfied, then:

Proof
We notice that:


So, the Cauchy-Riemann equations are equivalent to:

We compute
as derivative of a composed function, since we can write:

so we have:

We have three necessary and equivalent conditions for the differentiability of a composed function in
:



Definition
We consider
and
. If
is differentiable at the point
, we say that
is holomorphic in
.
If
wesay that
is holomorphic in
if it has this property at every points
.
As already said, in some cases the necessary conditions for the differentiability of
can also be sufficient. In fact, more rigorously:
Theorem
Given
and
, we consider
. If partial derivatives of
exist and are continuous in
, so Cauchy-Riemann equations are necessary and sufficient conditions for the differentiability of
in
Equivalently, we can say that
is holomorphic in
if and only if it satisfies the Cauchy-Riemann equations
we do not provide a proof of this fact, but we consider the following examples:
, that is the complex function that
.
We see that
. Cauchy-Riemann equations are not satisfied, so we conclude that
is not holomorphic.
. By using Cauchy-Riemann equations:

By seeing that it is true
we conclude that
is holomorphic.
,
. First of all we see that
, while
We want to apply the Cauchy-Riemann equations, from which we want to obtain that
.


First of all, we see that both the partial derivatives, if computed respectively at
are zero.
By computing the incremental limit of
along a line passing through the origin of the axis in the plane
we have:

By taking the limit for
it is evident that it depends on
that is the chosen path to reach
.
So,
cannot exist.
Remark 3:
We will see a relationship between the solutions of the Laplace equation
and the class of the holomorphic function in
.
We consider Laplace equation in cartesian coordinates:

Let be
a generic holomorphic function on
and be
. We can state that:

If
is partially differentiable up to the second order, we can derive with respect to
the previous equation, and we have:


The class of the holomorphic function in
coincides with the class of the solutions of the Laplace equation in two dimensions.
We do not go beyond in the theory of differentaibility in
, beacuse the possibility to derive to orders greater than the second (in particular, to derive a function infinite times) will be strictly related to the theory of the integration in
.