# Continuity and differentiability in the complex case

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:

- 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:

*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 .