Complex Analysis/Complex differentiability

In the case $n=1$, this condition is equivalent to the existence of the limit

$\underset{w\to z}{\lim }\frac{f(z)-f(w)}{z-w}$.

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

The Cauchy–Riemann equations

Suppose that $f:U\to ℂ$ is a function which is complex differentiable in a ball $B_r(z_0)$, where $z_0\in U$ is an element of $U$ and $r>0$ is a small constant (to which we shall refer as a radius).

A complex function can be uniquely written as $f(z)=u(z)+iv(z)$, where $u$ and $v$ are functions $ℂ\to ℝ$. The function $u$ corresponds to the real part of $f$, whereas the function $v$ corresponds to the imaginary part of $f$, so that for all $z\in U$

$u(z):=\mathrm{Re}(f(z))$, $v(z):=\mathrm{Im}(f(z))$.

Now since $f$ was supposed to be complex differentiable, it was supposed not to matter from which direction $h$ approaches $0$. In particular, $h$ may approach $0$ along the $x$-axis of the complex plane

$\{(x,y)\in ℂ|y=0\}$

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

Note that this means that $u$ is a harmonic function.

Computation rules

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.

Proof:

First note that the maps of addition and multiplication

$+:ℂ\times ℂ\to ℂ$

and

$\cdot :ℂ\times ℂ\to ℂ$

are continuous; indeed, let for instance $B_r(z_0)\subset ℂ$ be an open ball. Take $0<ϵ\le 1$ such that $B_{ϵ}(ab)\subset B_r(z_0)$. Now suppose that we have

$(c,d)\in B_{\delta }(a)\times B_{\delta }(b)$,

where $\delta$ is to be determined later. Then we have

$|cd-ab|=|(a+z)(b+w)-ab|=|aw+bz+wz|$,

where $|w|,|z|<\delta$. Upon choosing

$\delta =\frac{ϵ}{3\max \{1,|a|,|b|\}}$,

we obtain by the triangle inequality

$|aw+bz+wz|\le |aw|+|bz|+|wz|<ϵ$,

whence ${\cdot }^{-1}(B_r(z_0))$ is open. If then $U\subseteq ℂ$ is an open set, then ${\cdot }^{-1}(U)$ will also be open, since $U$ 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 $f,g$ are functions such that

$\underset{z\to z_0}{\lim }f(z)=a$ and $\underset{z\to z_0}{\lim }g(z)=b$,

then

$\underset{z\to z_0}{\lim }(f(z)+g(z))=a+b$

and

$\underset{z\to z_0}{\lim }f(z)g(z)=ab$;

indeed, this follows from the continuity of $+$ and $\cdot$ at the respective points. In particular, if $f$ is constant (say $f\equiv \alpha$ where $\alpha \in ℂ$ is a fixed complex number), we get things like

$\underset{z\to z_0}{\lim }\alpha g(z)=\alpha \underset{z\to z_0}{\lim }g(z)$.

1. Now suppose indeed that $f,g:U\to ℂ$ ($U$ open, so that we have a neighbourhood around $z_0$ and the derivative is defined in the sense that the direction in which $h$ goes to zero doesn't matter) are differentiable at $z_0\in U$. We will have

$\underset{h\to 0}{\lim }\frac{\alpha f(z_0+h)+\beta g(z_0+h)-(\alpha f(z_0)+\beta g(z_0))}{h}=\alpha \underset{h\to 0}{\lim }\frac{f(z_0+h)-f(z_0)}{h}+\beta \underset{h\to 0}{\lim }\frac{g(z_0+h)-g(z_0)}{h}=\alpha f^{\prime }(z_0)+\beta g^{\prime }(z_0)$.

4. Let indeed $f:U\to V$ be a bijection between $U$ and $V\subseteq ℂ$ which is differentiable in a neighbourhood of $z_0\in U$. By the inverse function theorem, $f^{-1}$ is real-differentiable at $f(z_0)$, and we have, by the chain rule for real numbers,

$I_2=(f^{-1}{)}^{\prime }(f(z_0))\cdot f^{\prime }(z_0)$ ($I_2$ denoting the identity matrix in ${ℝ}^{2\times 2}$ and the primes (e.g. $f^{\prime }(z_0)$) denoting the Jacobian matrices of the functions $ℂ\to ℂ$ seen as functions ${ℝ}^2\to {ℝ}^2$, "$\cdot$" denoting matrix multiplication),

since we may just differentiate the function $f^{-1}\circ f$. However, regarding $ℂ$ and ${ℝ}^{2\times 2}$ as $ℝ$-algebras (or as rings; it doesn't matter for our purposes), we have a morphism of algebras (or rings)

$\mathrm{\Phi }:ℂ\to {ℝ}^{2\times 2},x+iy\mapsto \left(\right)$.

Moreover, due to the Cauchy

Holomorphic (and meromorphic) functions

Let $U\subseteq ℂ$ be an open subset of the complex plane, and let $f:U\to ℂ$ be a function which is complex differentiable in $U$ (that means, in every point of $U$). Then we call $f$ holomorphic in $U$.

If $U$ happens to be, in fact, equal to $ℂ$, so that $f$ is complex differentiable at every complex number, $f$ 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

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

Now all elements of $ℂ[z]$ are entire functions. This is seen as follows:

Analogous to real analysis (with exactly the same proof), the function $z\mapsto z^n$ is complex differentiable. Thus, any polynomial

$p(z)=a_n{z}^n+\cdots +a_{1}z+a_0$ ($a_0,a_1,\dots ,a_n\in ℂ$ complex coefficients, ie. constants)

is complex differentiable by linearity.

We may also define $\mathbb{Q}(i)$, an extension of $\mathbb{Q}$ in $ℂ$. This extension turns out to be equal to

$\{x+iy|x,y\in \mathbb{Q}\}$.

From this, there arises a polynomial ring $\mathbb{Q}(i)[z]$. Let now $K\subseteq ℂ$ 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, $\mathbb{Q}(i)[z]$ is dense in $ℂ[z]$. Alternatively, one finds that

Exercises

1. Prove that whenever $f,g:{ℂ}^n\to {ℂ}^n$ are holomorphic, then $\overline{\partial }(f_k\circ g)=\sum _{l=1}^n\frac{\partial f_k}{\partial z_l}\frac{\partial g_l}{\partial {\overline{z}}_j}+\sum _{l=1}^n\frac{\partial f_k}{\partial {\overline{z}}_l}\frac{\partial {\overline{g}}_l}{\partial {\overline{z}}_j}$