Holomorphic function

Set

context $\mathcal O\subset \mathbb C$
definiendum $f\in \mathrm{it}$
inclusion $f:\mathcal O\to\mathbb C$
for all $z_0\in\mathcal O$
postulate $\left(\lim_{z \to z_0} {f(z) - f(z_0) \over z - z_0 }\right)\in\mathbb C $

Discussion

The following discussion is an elaboration/derivation on holomorphic functions from the viewpoint of analysis on $\mathbb R^n$. The article Linear approximation is prerequisite for the following.

The derivative of a real function again gives us a real function because there is only one direction to go. For $\mathbb R^n, n>1$ you must choose a direction and so this clearly can't be true - the derivative is given by the Jacobi matrix. Let's identify $a+i\,b\in\mathbb C$ with $\langle a,b\rangle\in\mathbb R^2$. The complex numbers are an algebraic field and are hence equipped with the structure of complex multiplication map $\mathbb C\times \mathbb C\to\mathbb C$. If $j=j_1+i\,j_2$ and $d=d_1+i\,d_2$, then the multiplication is given by

$j\cdot d = (j_1\cdot d_1-j_2\cdot d_2)+i (j_2\cdot d_1+j_1\cdot d_2) \overset{\sim}{=} \langle j_1\cdot d_1+(-j_2)\cdot d_2,j_2\cdot d_1+j_1\cdot d_2\rangle$

Let's compare this with the $2\times 2$ Jacobi matrix resulting when computing the derivative of a function taking values in $\mathbb R^2$, to see if the linear approximation at a point $z_0=x_0+i y_0$ of a complex function along $d=\langle d^1,d^2\rangle$ can be formulated as a multiplication within $\mathbb C$, i.e. by multiplying a $d$-independent function $f'$ with the number $d^1+i d^2$. Given the similarity with the real function derivative, it's not too surprising that this condition is equivalent to the existence of the limit in the definition above. So we could also say

$f:\mathcal O\to \mathbb C$ … holomorphic on $\mathcal{O}\ \equiv\ \forall(z_0\in\mathcal O).\ \exists (f':\mathcal O\to \mathbb C).\ J_{z_0}^f(d)=f'(z_0)\cdot(d^1+i d^2) $

This viewpoints leads us directly to a very important property of holomorphic functions. Comparing the first components tells us that

$j_1=\frac{\partial u}{\partial x},\hspace{.5cm} j_2=-\frac{\partial u}{\partial y}$

and comparing the second components then gives a strong necessary condition on $f$, which can also shown to be sufficient.

Cauchy–Riemann equations
$\dfrac{ \partial u }{ \partial x } = \dfrac{ \partial v }{ \partial y }$
$\dfrac{ \partial u }{ \partial y } = -\dfrac{ \partial v }{ \partial x }$

Hence

$f$ … holomorphic $\implies f'(x+i y)=\left(\dfrac{\partial}{\partial x}-i \dfrac{\partial}{\partial y}\right)u$

So for a holomorphic function $f$, the total change (a complex value) is determined by the real part (or alternatively complex part) of $f$ alone.

Non-example

The limit definition for a function requires that variation of function values stops being large once you get sufficiently close to the fixed point $0$.

For functions like the complex conjugation, $f(z):=\overline{z}$, the finite difference

$\dfrac{f(z)-f(0)}{z-0}=\dfrac{\overline{z}}{z}={\mathrm e}^{-2i\,\mathrm{arg}(z)}$

varies strongly with the $z$'s argument, even if $z$ is varied in only an arbitrarily small area around $0$, and so there is no limit. The function $f$ doesn't have a derivative.

Since $\mathrm{Re}(z)=\frac{1}{2}(z+\overline{z}), \mathrm{Im}(z)=\frac{1}{2i}(z-\overline{z})$, this translated to the projections to the real and imaginary axis, and in turn to all functions which can only be defined in terms of those (as oppsed to be defined in terms of $z$ itself, e.g. $z^3$).

Theorems

The Looman–Menchoff theorem gives a converse to the statement above: If $f$ is continuous, $u$ and $v$ have first partial derivatives (but not necessarily continuous), and they satisfy the Cauchy–Riemann equations, then $f$ is holomorphic.

If a function is holomorphic, it's continuous. (That's not in general true for the derivative of functions $\mathbb R^n\to \mathbb R^n$ if $n>1$, e.g. consider the function $f(x,y):=xy/(x^2+y^2)$ on $x,y\neq 0$ and else $f(0,0):=0$.)

A holomorphic function, interpreted as a map from the complex plane to the complex plane, is (at each point with at a point where $f'(z)\ne 0$) conformal, i.e.preserves the angles of crossing curves. A holomorphism isomorphism is conformal.

If $z\in\mathbb C$ and $b\in \mathbb R$, then $f(z):=z^b$ is holomorphic with $f'(z)=b\ z^{b-1}$, except from maybe at $z=0$. By linearity, this extends to polynomials and infinite sums $\sum_k a_k\ z^{b_k}$.

$ \mathrm{det}(J_z^f)=|f'(z)|^2 $

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