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Difference between revisions of "Holomorphic function"

m (Holomorphic moved to Holomorphic function: We're talking about holomorphic functions, right? We could have a "holomorphic function" page that redirects to this, but I think it's more logical this way.)
(added stuff about Cauchy-Riemann equations)
Line 1: Line 1:
−
A '''holomorphic''' function <math>f: \mathbb{C} \to \mathbb{C}</math> is a differentiable [[complex number|complex]] [[function]]. That is, just as in the [[real number|real]] case, <math>f</math> is holomorphic at <math>z</math> if <math>\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}</math> exists. This is much stronger than in the real case since we must allow <math>h</math> to approach zero from any direction in the [[complex plane]].
+
A '''holomorphic function''' <math>f: \mathbb{C} \to \mathbb{C}</math> is a
 +
differentiable [[complex number|complex]] [[function]]. That is, just
 +
as in the [[real number|real]] case, <math>f</math> is holomorphic at <math>z</math> if
 +
<math>\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}</math> exists. This is much stronger
 +
than in the real case since we must allow <math>h</math> to approach zero from
 +
any direction in the [[complex plane]].
 +
 
 +
Usually, we speak of functions as holomorphic on (open) sets,
 +
rather than at points, for when we consider the behavior of a function
 +
at a point, we prefer to consider it in the context of the points
 +
nearby.
  
 
== Cauchy-Riemann Equations ==
 
== Cauchy-Riemann Equations ==
  
−
Let us break <math>f</math> into its real and imaginary components by writing <math>f(z)=u(x,y)+iv(x,y)</math>, where <math>u</math> and <math>v</math> are real functions. Then it turns out that <math>f</math> is holomorphic at <math>z</math> [[iff]] <math>u</math> and <math>v</math> have continuous partial derivatives and the following equations hold:
+
We can obtain an equivalent definition if we break <math>f</math> and <math>z</math>
−
* <math>\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}</math>
+
into real and imaginary components.
−
* <math>\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}</math>
+
 
 +
Specifically, let <math>u, v : \mathbb{R \times R \to R}</math> be definted
 +
by
 +
<cmath> u(x,y) = \text{Re}\,f(x+iy), \qquad v(x,y) = \text{Im}\,f(x+iy) .  </cmath>
 +
If <math>z = x+iy</math>, then
 +
<cmath> f(z) = u(x,y) + i v(x,y). </cmath>
 +
 
 +
It turns out that we can express the idea "<math>f</math> is holomorphic"
 +
entirely in terms of partial derivatives of <math>u</math> and <math>v</math>.
 +
 
 +
'''Theorem.'''  Let <math>D</math> be an open, connected subset of <math>\mathbb{C}</math>.
 +
Let us abbreviate <math>x = \text{Re}\, z</math> and <math>y = \text{Im}\, z</math>.
 +
Then the function <math>f</math> is holomorphic on <math>D</math>
 +
if and only if all the partial derivatives of <math>u</math> and <math>v</math> with respect
 +
to <math>x</math> and <math>y</math> are continuous on <math>D</math>, and the following system holds
 +
for every point <math>z \in D</math>:
 +
<cmath> \begin{align*}
 +
\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} ,\
 +
\frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x}.
 +
\end{align*} </cmath>
 +
These equations are called the '''[[Cauchy-Riemann Equations]]'''.
 +
 
 +
For convenience, we may abbreviate
 +
<cmath> \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x}
 +
+ i \frac{\partial v}{\partial x}, \qquad
 +
\frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} +
 +
i \frac{\partial v}{\partial y} . </cmath>
 +
With this abuse of notation, we may rewrite the Cauchy-Riemann
 +
equations thus:
 +
<cmath> \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x} . </cmath>
 +
 
 +
''Proof of theorem.''  First, suppose that <math>f</math> is
 +
complex-differentiable at <math>z</math>.  Then at <math>z</math>,
 +
<cmath>\begin{align*}
 +
\frac{\partial f}{\partial y} = \lim_{h\to 0} \frac{f(z+ih)-f(z)}{h}
 +
&= i \cdot \lim_{h\to 0} \frac{f(z+ih) - f(z)}{ih} \
 +
&= i \cdot f'(z) \
 +
&= i \cdot \lim_{h\to 0} \frac{f(z+h)-f(z)}{h}
 +
= i \cdot \frac{\partial f}{\partial x} . </cmath>
 +
Breaking <math>f</math> into real and imaginary components, we see
 +
<cmath> \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}
 +
= \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x}
 +
= -\frac{\partial v}{\partial x} + i \frac{\partial u}{\partial y}. </cmath>
 +
Setting real and imaginary components equal, we obtain the
 +
Cauchy-Riemann equations.  It follows from the
 +
[[Cauchy Integral Formula]] that the second derivative of <math>f</math> exists
 +
at <math>z</math>; thus the derivative of <math>f</math> is continuous at <math>z</math>, and so
 +
are the partial derivatives of <math>u</math> and <math>v</math>.
  
−
These equations are known as the [[Cauchy-Riemann Equations]].
+
Now, suppose the Cauchy-Riemann equations hold a point <math>z</math>, and
 +
that the partial derivatives of <math>u</math> and <math>v</math> exist and are continuous
 +
in a [[neighborhood]] of <math>z</math>.  Let <math>h = h_1 + i h_2</math> be an arbitrarily
 +
small complex number, with <math>h_1, h_2 \in \mathbb{R}</math>.  Then
 +
<cmath> \begin{align*}
 +
\frac{f(z + h) - f(z)}{h} &= \frac{f(z+h_1+ih_2)-f(z+h_1)}{h_1+ih_2}
 +
+ \frac{f(z+h_1)-f(z)}{h_1+ih_2} \
 +
&\approx \frac{ih_2}{h_1+ih_2} \frac{\partial f}{\partial y}(z+h_1)
 +
+ \frac{h_1}{h_1 + ih_2} \frac{\partial f}{\partial x}(z) \
 +
&\approx \frac{ih_2}{h_1+ih_2} \frac{\partial f}{\partial y}(z)
 +
+ \frac{h_1}{h_1 + ih_2} \frac{\partial f}{\partial x}(z) ,
 +
\end{align*} </cmath>
 +
with the first approximation from the definition of the partial derivatives
 +
and the second from the continuity of the partial derivatives.
 +
We may force <math>h</math> to be small enough that both approximations
 +
are arbitrarily accurate.  Now, by the Cauchy-Riemann equations,
 +
<cmath> \frac{i h_2}{h_1+ih_2} \frac{\partial f}{\partial y}(z)
 +
+ \frac{h_1}{h_1 + ih_2} \frac{\partial f}{\partial x}(z)
 +
= \frac{\partial f}{\partial x} (z) .</cmath>
 +
Therefore
 +
<cmath> \lim_{h\to 0} \frac{f(z+h)-f(z)}{h} = \frac{\partial f}{\partial x}
 +
(z) . </cmath>
 +
In particular, the limit exists, so <math>f</math> is differentiable at <math>z</math>.
 +
Since <math>z</math> was arbitrary, it follows that <math>f</math> is differentiable
 +
everywhere in <math>D</math>. <math>\blacksquare</math>
  
 
== Analytic Functions ==
 
== Analytic Functions ==
  
−
A related notion to that of homolorphicity is that of analyticity. A function <math>f:\mathbb{C}\to\mathbb{C}</math> is said to be '''analytic''' at <math>z</math> if <math>f</math> has a convergent [[power series]] expansion on some [[neighborhood]] of <math>z</math>. Amazingly, it turns out that a function is holomorphic at <math>z</math> if and only if it is analytic at <math>z</math>.
+
A related notion to that of homolorphicity is that of analyticity. A
 +
function <math>f:\mathbb{C}\to\mathbb{C}</math> is said to be '''analytic''' at
 +
<math>z</math> if <math>f</math> has a convergent [[power series]] expansion on some
 +
[[neighborhood]] of <math>z</math>. Amazingly, it turns out that a function is
 +
holomorphic at <math>z</math> if and only if it is analytic at <math>z</math>.
  
 
[[Category:Complex analysis]]
 
[[Category:Complex analysis]]

Revision as of 00:16, 7 April 2009

A holomorphic function $f: \mathbb{C} \to \mathbb{C}$ is a differentiable complex function. That is, just as in the real case, $f$ is holomorphic at $z$ if $\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$ exists. This is much stronger than in the real case since we must allow $h$ to approach zero from any direction in the complex plane.

Usually, we speak of functions as holomorphic on (open) sets, rather than at points, for when we consider the behavior of a function at a point, we prefer to consider it in the context of the points nearby.

Cauchy-Riemann Equations

We can obtain an equivalent definition if we break $f$ and $z$ into real and imaginary components.

Specifically, let $u, v : \mathbb{R \times R \to R}$ be definted by \[u(x,y) = \text{Re}\,f(x+iy), \qquad v(x,y) = \text{Im}\,f(x+iy) .\] If $z = x+iy$, then \[f(z) = u(x,y) + i v(x,y).\]

It turns out that we can express the idea "$f$ is holomorphic" entirely in terms of partial derivatives of $u$ and $v$.

Theorem. Let $D$ be an open, connected subset of $\mathbb{C}$. Let us abbreviate $x = \text{Re}\, z$ and $y = \text{Im}\, z$. Then the function $f$ is holomorphic on $D$ if and only if all the partial derivatives of $u$ and $v$ with respect to $x$ and $y$ are continuous on $D$, and the following system holds for every point $z \in D$: \begin{align*} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} ,\\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x}. \end{align*} These equations are called the Cauchy-Riemann Equations.

For convenience, we may abbreviate \[\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}, \qquad \frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y} .\] With this abuse of notation, we may rewrite the Cauchy-Riemann equations thus: \[\frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x} .\]

Proof of theorem. First, suppose that $f$ is complex-differentiable at $z$. Then at $z$, 

\begin{align*}
\frac{\partial f}{\partial y} = \lim_{h\to 0} \frac{f(z+ih)-f(z)}{h}
&= i \cdot \lim_{h\to 0} \frac{f(z+ih) - f(z)}{ih} \\
&= i \cdot f'(z) \\
&= i \cdot \lim_{h\to 0} \frac{f(z+h)-f(z)}{h} 
= i \cdot \frac{\partial f}{\partial x} . (Error compiling LaTeX. Unknown error_msg)

Breaking $f$ into real and imaginary components, we see \[\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y} = \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x} = -\frac{\partial v}{\partial x} + i \frac{\partial u}{\partial y}.\] Setting real and imaginary components equal, we obtain the Cauchy-Riemann equations. It follows from the Cauchy Integral Formula that the second derivative of $f$ exists at $z$; thus the derivative of $f$ is continuous at $z$, and so are the partial derivatives of $u$ and $v$.

Now, suppose the Cauchy-Riemann equations hold a point $z$, and that the partial derivatives of $u$ and $v$ exist and are continuous in a neighborhood of $z$. Let $h = h_1 + i h_2$ be an arbitrarily small complex number, with $h_1, h_2 \in \mathbb{R}$. Then \begin{align*} \frac{f(z + h) - f(z)}{h} &= \frac{f(z+h_1+ih_2)-f(z+h_1)}{h_1+ih_2} + \frac{f(z+h_1)-f(z)}{h_1+ih_2} \\ &\approx \frac{ih_2}{h_1+ih_2} \frac{\partial f}{\partial y}(z+h_1) + \frac{h_1}{h_1 + ih_2} \frac{\partial f}{\partial x}(z) \\ &\approx \frac{ih_2}{h_1+ih_2} \frac{\partial f}{\partial y}(z) + \frac{h_1}{h_1 + ih_2} \frac{\partial f}{\partial x}(z) , \end{align*} with the first approximation from the definition of the partial derivatives and the second from the continuity of the partial derivatives. We may force $h$ to be small enough that both approximations are arbitrarily accurate. Now, by the Cauchy-Riemann equations, \[\frac{i h_2}{h_1+ih_2} \frac{\partial f}{\partial y}(z) + \frac{h_1}{h_1 + ih_2} \frac{\partial f}{\partial x}(z) = \frac{\partial f}{\partial x} (z) .\] Therefore \[\lim_{h\to 0} \frac{f(z+h)-f(z)}{h} = \frac{\partial f}{\partial x} (z) .\] In particular, the limit exists, so $f$ is differentiable at $z$. Since $z$ was arbitrary, it follows that $f$ is differentiable everywhere in $D$. $\blacksquare$

Analytic Functions

A related notion to that of homolorphicity is that of analyticity. A function $f:\mathbb{C}\to\mathbb{C}$ is said to be analytic at $z$ if $f$ has a convergent power series expansion on some neighborhood of $z$. Amazingly, it turns out that a function is holomorphic at $z$ if and only if it is analytic at $z$.

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