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




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




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.

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


First note that the maps of addition and multiplication




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  ,





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

  1. Prove that whenever   are holomorphic, then