Café Math : Cauchy-Riemann equation
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Cauchy-Riemann equation

Sat, 22 Jan 2012 06:34:01 GMT


Written with the help of my friend Cyrus Divecha.

1. Foreword

Today I want to expand on a remark one of us made about holomorphic functions. What are holomorphic functions, why are they so stunning, what makes them so useful ?

Definition 1. A holomorphic function $f(z)$ is a complex valued function of the complex variable such that $$ \frac{f(z+h) - f(z)}{h} $$ has a limit when $h$ tends to zero without being equal to zero.

Of course if the limit exists then its value is denoted by $f'(z)$ and is called the derivative of $f$ at $z$. You may think this is simply the definition of differentiability we all learned when we were children, but here the variable $h$ is supposed to be complex and the limit has to exist and be the same independently of the way $h$ tends to zero.


2. Differentiability in the real variable sense

Consider a complex valued function $f(z)$ of the complex variable $z$, defined in an open set $U$. This means that the function $f$ associates to any complex number $z$ in the open set $U$ a complex number $f(z)$. A first question is : what is the possible meaning of $d\,f(z)$ ?

First let's rephrase that in the real variable sense. We recall that complex numbers are expressions like $z = x + i y$ where $x$ and $y$ are both real numbers and $i$ is a square root of $-1$, which means simply that we compute with the extra relation $i^2 = -1$. Such a function can be written $f(x + i \,y) = f_1(x, y) + i\, f_2(x, y)$ where $f_1$ and $f_2$ are two real valued functions of the real variables $x$ and $y$. The function $f_1(x, y)$ is the real part of $f$ and $f_2(x, y)$ is its imaginary part.

Example 1. $$ \begin{aligned} f(z)&= z^3 \\ &= (x + i \,y)^3 \\ &= x^3 + 3\,iyx^2 - 3\,xy^2 - i\,y^3 \\ &= x^3 - 3\,xy^2 + i\,(3\,yx^2 - y^3) \\ &= f_1 + i\,f_2 \end{aligned} $$ where $f_1(x, y) = x^3 - 3\,xy^2$ and $f_2(x, y) = 3\,yx^2 - y^3$ where $f_1(x, y)$ the real part and $f_2(x, y)$ is the imaginary part of $f(z) = z^3$.

Following the book we can write, $$ \begin{aligned} d\,f(x,y) = \left[\, \frac{\partial f_1}{\partial x}(x,y)\,dx + \frac{\partial f_1}{\partial y}(x,y)\,dy\,\right] + i\,\left[\, \frac{\partial f_2}{\partial x}(x,y)\,dx + \frac{\partial f_2}{\partial y}(x,y)\,dy\,\right] \end{aligned} $$

There are still complex number concepts in this formula. The first step in the `reification' of our function $f$ was to consider it a function of the vector with real coordinates $x$ and $y$. The second and last step in this reification is to consider it vector valued. $$ f\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} f_1(x,y) \\ f_2(x,y) \end{bmatrix} $$ So now $d\, f$ can be written in matrix form with only real numbers. $$ d\,f\,\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \dfrac{\partial f_1}{\partial x}(x,y) & \dfrac{\partial f_1}{\partial y}(x,y) \\ \\ \dfrac{\partial f_2}{\partial x}(x,y) & \dfrac{\partial f_2}{\partial y}(x,y) \end{bmatrix} \begin{bmatrix} dx \\ dy \end{bmatrix} $$

I know some of you may ask what the hell are those $d$, $dx$, $dy$ in the formula. A complete answer would be a bit lengthy but genuinely interesting. Let me tell you a story : when I still was at the University, there were barely no one to answer that question with the right answer while almost everyone had its own interpretation. Some were close to be true, some were dangerously wrong. We may have another conversation at the café on that topic. The expressions $dx$ and $dy$ are differential forms (this is some kind of mathematical objects, they may look like functions but it's dangerously wrong to think that way). You can take them as new extra variables and compute with them as if they were some coordinates. The thing that has to be remembered is the pullback formula $$ \begin{aligned} d \, f(x) = f'(x) \,dx \end{aligned} $$ By the way, this is the same $dx$ you see in the integrals.

Let's move on. Here $dx$ and $dy$ are the coordinates of a tangent vector at the point of coordinate $(x,y)$ and $d\, f(x,y)$ is the corresponding tangent vector when transformed through $f$. With formula \ref{matrix:form:differential} you can see one important thing : the coordinates of $d\,f(x,y)$ linearly depend on the coordinates $(dx, dy)$. Hence the matrix in the middle. That matrix is called the jacobian matrix (after the name of Jacobi).

So that's it ? Is it what it means to be holomorphic ? To have partial derivatives with respect to the real coordinates so that the jacobian matrix is defined ? $$ \begin{aligned} d\,f = \begin{bmatrix} \dfrac{\partial f_1}{\partial x}(x,y) & \dfrac{\partial f_1}{\partial y}(x,y) \\ \\ \dfrac{\partial f_2}{\partial x}(x,y) & \dfrac{\partial f_2}{\partial y}(x,y) \end{bmatrix} \end{aligned} $$ Actually no. All this means, if those partial derivatives are continuous, is that the function $f$ is differentiable in the real variable sense.

Example 2. Consider the function $f(z) = \bar{z}$


3. Differentiability in the comlex variable sense

To get a good notion of differentiability in the complex variable sense, one key point is the pullback formula (c.f. previous section). More concretely, one wants that the derivative of a complex valued function of the complex variable is itself a complex function of the complex variable. This way, $f'(z)$ has to be a complex number, which means that the jacobian matrix has to transform the complex plane as it were multiplication by a complex number. Recall that any complex number $z = x+i\,y$ behave as a real matrix. From the product $(x+i\,y)(x'+i\,y') = (xx' - yy') + i\,(xy' + yx')$ one can see the linear relation on real coordinates : $$ \begin{bmatrix} xx' - yy' \\ xy' + yx' \end{bmatrix} = \begin{bmatrix} x & -y\\ y & x \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix} $$

Again you can see that this matrix can't be any matrix, there are constraints on its entries. The two coefficient of the main diagonal are equal and the two coefficients of the other diagonal are opposed. If you want the jacobian matrix of the previous section to have the exact same structure it means that you have a system of two relations. $$ \left\{ \begin{aligned} \dfrac{\partial f_1}{\partial x}(x,y) &= \dfrac{\partial f_2}{\partial y}(x,y) \\ \dfrac{\partial f_2}{\partial x}(x,y) &= - \dfrac{\partial f_1}{\partial y}(x,y) \end{aligned} \right. $$ Those simple relations are called the Cauchy-Riemann equations.

Theorem (Cauchy-Riemann). Let $f$ be a complex function of the complex variable, then the two following conditions are equivalent,

  1. the function $f$ is holomorphic.
  2. the function $f$ is real differentiable and satisfies the Cauchy-Riemann equations.

Exercise. Those who cares are invited to prove the equivalence, personally I don't.

Let's consider now some examples and counter examples.

Example 3. What is the jacobian matrix of the function $f(z) = z^3$ ? With $z = x+i\,y$ as before, $$ \begin{align*} \frac{\partial f_1}{\partial x}(x,y) &= 3\,x^2 - 3 \, y^2, & \frac{\partial f_1}{\partial y}(x,y) &= -6\, xy, \\ \frac{\partial f_2}{\partial x}(x,y) &= 6\, xy, & \frac{\partial f_2}{\partial y}(x,y) &= 3\,x^2 - 3 \, y^2. \end{align*} $$ So the jacobian matrix is, $$ d \,f = \begin{bmatrix} 3\,x^2 - 3 \, y^2 & -6\, xy \\ 6\, xy & 3\,x^2 - 3 \, y^2 \end{bmatrix} $$ Here we are, the Cauchy-Riemann equations are satisfied and the jacobian matrix looks like the multiplication by $f'(x+i\,y)=(3\,x^2 - 3 \, y^2) + i\, 6\, xy$. Just expand the expression $3\,z^2 = 3\, (x+i\,y)^2$ and see this is equal to $f'(z)$.

Example 4. Now consider the function $f(z) = \bar{z}$ that is, $f(x+i\,y) = x - i\,y$. the jacobian matrix is, $$ d \,f = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$ The Cauchy-Riemann equations are not satisfied, so according to the Cauchy-Riemann theorem, the limit of Definition 1 cannot exist. And indeed, if you look at the limit of, $$ \dfrac{f(z+h) - f(z)}{h} $$ when $h$ tends to zero with values only on the real axis (without zero) you find $1$, while the limit of the same expression when $h$ tends to zero with values only on the imaginary axis (without zero) you find $-1$. Those two quantities are different and so the function can't be holomorphic.

3 Comments

Julcan  posted 2012-05-02 07:12:32

I was wondering if you ever thuoght of changing the page layout of your blog? Its very well written; I love what you've got to say. But maybe you could do a little more in the way of content so people could connect with it better. You've got an awful lot of text for only having 1 or two images. Maybe you could space it out better?

Samuel VIDAL  posted 2012-05-02 21:04:37

Hey ! Thanks Julcan for your friendly comment. Your remarks on presentation are really true, and your suggestions are really helpful.

I'm considering improving the layout and and the overall look in various ways.

  1. Change the font, color, and spacing.
  2. More pictures of course as you suggest.
  3. More examples.
  4. Shorter line of reasonning.
  5. Some posts should talk about more elementary mathematics.

There are problems that should be addressed in no time.

  1. Put a protection against spam by robots.
  2. Give the option to be notified by mail when someone responds to a comment.
  3. Put a 'like' button on comments and articles.
  4. The buttons of the site looks ugly as hell when viewed on Internet Explorer.
  5. Give the option to edit one's post, or at least to see a preview of the post.

I'll keep you posted ;-)

julien  posted 2013-01-30 19:41:42

very sweet introduction to a useful pair of equations :) for instance, cauchy riemann equations have very important implications in practice in computer graphics. have a look at that paper for example (one of the major breakthroughs of the early 2000s in geometry processing):

http://www.loria.fr/~petitjea/papers/siggraph02.pdf

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