# Complex Analysis/Complex differentiability

**Definition (complex differentiability)**:

Let , and let be a function. Let . We say that is **complex differentiable** in if and only if there exists a -linear function such that

- .

In the case , this condition is equivalent to the existence of the limit

- .

Indeed, if this limit is , then the -linear map in the above definition is just multiplication by , and vice-versa, any linear map is simply multiplicaton by an element of , which is then the limit.

**Proposition (Osgood's lemma)**:

Let be a continuous function.

## The Cauchy–Riemann equations Edit

Suppose that is a function which is complex differentiable in a ball , where is an element of and is a small constant (to which we shall refer as a *radius*).

A complex function can be uniquely written as , where and are functions . The function corresponds to the *real part* of , whereas the function corresponds to the *imaginary part* of , so that for all

- , .

Now since was supposed to be complex differentiable, it was supposed not to matter *from which direction* approaches . In particular, may approach along the -axis of the complex plane

or the -axis (which is defined in a similar way).

**Definition (Cauchy–Riemann equations)**:

**Theorem (Cauchy–Riemann equations)**:

Let be a continuously differentiable function and . Then is holomorphic if and only if it satisfies the Cauchy–Riemann equations.

**Proposition (differential equations satisfied by components of holomorphic functions)**:

Let be holomorphic, and write , where are real-valued. Then we have

and

- .

**Proof:** We have, by the Cauchy–Riemann equations and Clairaut's theorem,

and

- .

Note that this means that is a harmonic function.

## Computation rules Edit

In the case of real differentiable functions, we have *computation rules* such as the *chain rule*, the *product rule* or even the *inverse rule*. In the case of complex functions, we have, in fact, precisely the same rules.

**Theorem 2.2**:

Let be complex functions.

- If and are complex differentiable in and , the function is complex differentiable in and (
**linearity of the derivative**) - If and are complex differentiable in , then so is (pointwise product with respect to complex multiplication!) and we have the
**product rule** - If is complex differentiable at and is complex differentiable at , then is complex differentiable at and we have the
**chain rule** - If is bijective, complex differentiable in a neighbourhood of and , then is differentiable at (
**inverse rule**) - If are differentiable at and , then (
**quotient rule**)

**Proof**:

First note that the maps of addition and multiplication

and

are continuous; indeed, let for instance be an open ball. Take such that . Now suppose that we have

- ,

where is to be determined later. Then we have

- ,

where . Upon choosing

- ,

we obtain by the triangle inequality

- ,

whence is open. If then is an open set, then will also be open, since is the union of open balls and inverse images under a function commute with unions.

The proof for addition is quite similar.

But from these two it follows that if are functions such that

- and ,

then

and

- ;

indeed, this follows from the continuity of and at the respective *points*. In particular, if is constant (say where is a fixed complex number), we get things like

- .

1. Now suppose indeed that ( open, so that we have a neighbourhood around and the derivative is defined in the sense that the direction in which goes to zero doesn't matter) are differentiable at . We will have

- .

4. Let indeed be a bijection between and which is differentiable in a neighbourhood of . By the inverse function theorem, is real-differentiable at , and we have, by the chain rule for real numbers,

- ( denoting the identity matrix in and the primes (e.g. ) denoting the Jacobian matrices of the functions
*seen as*functions , " " denoting matrix multiplication),

since we may just differentiate the function . However, regarding and as -algebras (or as rings; it doesn't matter for our purposes), we have a *morphism of algebras* (or rings)

- .

Moreover, due to the Cauchy

## Holomorphic (and meromorphic) functions Edit

Let be an open subset of the complex plane, and let be a function which is complex differentiable in (that means, in every point of ). Then we call *holomorphic* in .

If happens to be, in fact, *equal* to , so that is complex differentiable at every complex number, is called an **entire function**. We will see examples of entire functions in the chapter on trigonometry, where the exponential, sine and cosine function play central roles. Another important class of entire functions are *polynomials*.

### Polynomials are entire functions Edit

In algebra, one studies polynomial rings such as , or, more generally, , where is a ring (one then has theorems that "lift" properties of to , eg. if is an integral domain, a UFD or noetherian, then so is ).

Now all elements of are entire functions. This is seen as follows:

Analogous to real analysis (with exactly the same proof), the function is complex differentiable. Thus, any polynomial

- ( complex coefficients, ie. constants)

is complex differentiable by linearity.

We may also define , an extension of in . This extension turns out to be equal to

- .

From this, there arises a polynomial ring . Let now be any compact subset of the complex plane, or even a bounded subset. Then it is easy to see from direct arguments that with respect to the topology of uniform convergence, is dense in . Alternatively, one finds that

## Exercises Edit

- Prove that whenever are holomorphic, then