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Differentiation has an obvious geometric interpretation, and the Cauchy Riemann equations are closely linked with differentiation.

Do the Cauchy Riemann equations have a geometric interpretation?

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  • $\begingroup$ @YvesDaoust What's an "iso curve" and what's an "orthogonal pencil"? $\endgroup$ – dfdfg Nov 17 '14 at 15:50
  • $\begingroup$ @YvesDaoust Sorry, I have absolutely no idea what you're talking about. Could you please expand on your comment? This is my first complex analysis course and my third course in calculus so I don't know a lot. $\endgroup$ – dfdfg Nov 17 '14 at 15:56
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    $\begingroup$ You should read Visual Complex Analysis by Tristan Needham. $\endgroup$ – hjhjhj57 Nov 23 '14 at 21:52
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Suppose $f: \mathbf{C} \rightarrow \mathbf{C}$ is analytic near $a \in \mathbf{C}$. Geometrically, this can be interpreted as saying that the effect of $f$ on a small vector $h$ emanating from $a$ is just multiplication by the complex number $f'(a)$: $$ f(a+h) \approx f(a) +f'(a)h,$$ where the error tends to zero sublinearly as $|h| \rightarrow 0$. That is, the local effect of an analytic function is a rotation plus a scaling. (There's an equivalent interpretation of the real derivative.)

enter image description here

In general, if we multiply $z = x + iy$ by a fixed complex number $a + ib$, we have

$$(x + iy) \mapsto (a +ib)(x+iy) =(ax-by) + i(bx+ay).$$

Thinking of $\mathbf{C}$ as $\mathbf{R}^2$, this transformation corresponds to the matrix multiplication $$ \begin{bmatrix}a&-b\\b&a\end{bmatrix} \begin{bmatrix} x\\y\end{bmatrix} = \begin{bmatrix} ax - by\\bx+ay\end{bmatrix}.$$

Notice the form of this matrix - it's the composition of a dilation and a rotation about the origin.

Back in $\mathbf{C}$, if $u$ and $v$ are continuously differentiable mappings from $\mathbf{R}^2$ into $\mathbf{R}$, then the Jacobian $J$ of $f = u + iv$ is

$$J = \begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix} .$$

$J$ describes the local effect of $f$. Now, for $f$ to be analytic, the Cauchy-Riemann equations require that $u$ and $v$ satisfy $ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} $. But if that's true, then $J$ must be a dilative rotation, like the matrix above. So the Cauchy-Riemann equations express the fact that the local effect of an analytic mapping is a dilative rotation, that is, multiplication by a complex number.

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  • $\begingroup$ Nice answer! +1 $\endgroup$ – Markus Scheuer Nov 23 '14 at 22:44
  • $\begingroup$ @Owen Like it +1. Maybe add something akin to the figure 1. on page 2 of this document. Just so people have a visual indication (in this case for a conformal mapping whose prerequisite is the fulfillment of the Cauchy-Riemann equations). $\endgroup$ – WalyKu Nov 27 '14 at 12:03
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    $\begingroup$ @Kurtovic Good idea. (Picture taken from Tristan Needham, Visual Complex Analysis, pg 196.) $\endgroup$ – Owen Colman Nov 28 '14 at 21:35
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The version of the Cauchy-Riemann equations that I prefer is (assuming we are using $x,y$ as coordinates for the real and imaginary part on the domain) $$ i\partial_x f = \partial_y f $$

This has a direct geometrical interpretation: the partial derivative with respect to the real part rotated by $\pi/2$ (which is what multiplying by $i$ does to a complex number) is the partial derivative with respect to the imaginary part.

The geometric interpretation of the complex derivative itself $$ f(z+h) = f(z) + f'(z)h + o(h) $$ is that $f$ can be approximated at the first order near $z$ by an affine transformation with a linear part represented by complex multiplication. If $f'(z)=re^{i\theta}$, then $h\mapsto f'(z)h$ is a rotation by $\theta$ followed (or preceded) by a uniform scaling of factor $r$.

This is much more restrictive than real differentiability for functions $\mathbb R^2\to\mathbb R^2$, which allows approximation by arbitrary affine transformations. For example the mapping $c: (x, y) \mapsto (x, -y)$ has itself as a real differential everywhere (because it's a linear transformation). Geometrically $c$ is a reflection, which changes the orientation of the plane (it sends the base $(1, 0),\ (0, 1)$ which is clockwise or anticlockwise oriented depending on how you draw the axes, to $(1, 0),\ (0, -1)$ which has the opposite orientation). When thought of as a transformation of the complex plane, $c$ is just the complex conjugation $z \mapsto \overline z$. From a geometric viewpoint $c$ has no complex derivative anywhere because if such a derivative $c'(z)$ existed the linear transformation $h\mapsto c'(z)h$ would have to be a reflection, which is impossible to obtain by composing scalings with rotations.

The Cauchy-Riemann equations can be seen as a consequence of the relation between the complex derivative and the real differential. Because of the approximation argument given above and the unicity of the real differential, the complex linear transformation $h\mapsto f'(z)\cdot h$ in $\mathbb C$ corresponds in $\mathbb R^2$ to the real differential $$ df(z_x, z_y): \begin{bmatrix}h_x\\h_y\end{bmatrix}\mapsto \begin{bmatrix}\partial_x f(z_x, z_y) & \partial_y f(z_x, z_y)\end{bmatrix} \begin{bmatrix}h_x\\h_y\end{bmatrix} $$ where the real partial derivatives $\partial_x f(z_x, z_y)$ and $\partial_y f(z_x, z_y)$ are column vectors and correspond to the complex partial derivatives $\partial_x f(z), \partial_y f(z)$. Taking $h = 1$, i.e. $h_x = 1, h_y = 0$, in $\mathbb R^2$ we have $$ \begin{bmatrix}\partial_x f(z_x, z_y) & \partial_y f(z_x, z_y)\end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}\partial_x f(z_x, z_y)\end{bmatrix} $$

which translated back to $\mathbb C$ is $f'(z)\cdot 1 = \partial_x f(z)$. Taking $h = i$ (i.e. $h_x = 0, h_y = 1$) yields $f'(z)\cdot i = \partial_y f(z)$. Together these two identities give the Cauchy-Riemann equations $\partial_y f(z) = if'(z) = i\partial_x f(z)$.

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  • $\begingroup$ Great post, what do you mean by the linear part can change the orientation of the plane? And where is $f'(z) \cdot 1 = \partial_x f(z)$ coming from? $\endgroup$ – csss Apr 24 '17 at 6:42
  • $\begingroup$ Thanks @csss. I should probably edit the post to clarify those two points. $\endgroup$ – Fra Apr 24 '17 at 14:17
  • $\begingroup$ Added a couple of paragraphs, hope it helps $\endgroup$ – Fra Apr 24 '17 at 16:11
  • $\begingroup$ ...does a rotation by 180 degress not equal a reflection? $\endgroup$ – eurocoder Apr 25 '17 at 6:49
  • $\begingroup$ @eurocoder, no. For example think of a clockwise circular arrow, you can rotate it in any way you like (180 deg included), it will still be clockwise. After a reflection it will be anticlockwise $\endgroup$ – Fra Apr 25 '17 at 15:53
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If functions $u$ and $v$ satisfy the Cauchy-Riemann equations then theirs gradients are orthogonal and equal in length: $\nabla u \perp \nabla v$, $|\nabla u|=|\nabla v|$. And this means that level curves of $u$ and $v$ are orthogonal where the gradients do not vanish.

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    $\begingroup$ Thank you. And why do these conditions need to be fulfilled for a function to be differentiable? $\endgroup$ – dfg Nov 23 '14 at 1:58
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As pointed out in the comments, there is indeed a nice geometric interpretation in Tristan Needham's "Visual Complex Analysis". See, especially, the material on the "Polya Vector field". In case you cannot access that book, I've thrown in some point form notes below.

  • Let $f = u + iv$ be a complex-valued function, defined on some open domain $D \subset \mathbb{C}$.
  • If we identify $\mathbb{R}^2$ with $\mathbb{C}$ via $(x,y) \mapsto x+iy$, then we can think of $f = (u,v)$ as a vector field on $D$.
  • The Polya vector field $V$ of $f$ is the the complex conjugate $\overline f$, viewed as a vector field. To be precise, $V = (u, -v)$. Obviously, $V$ is completely equivalent to $f$ i.e. $f$ determines $V$ and vice versa. For convenience, let us make notation for the component functions of the $V$. I'll put $P=u$ and $Q = -v$, so as to have $$V=(P,Q).$$
  • The Cauchy-Riemann equations are: \begin{align*} \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} && && && \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}. \end{align*} In terms of $P = u$ and $Q = -v$, that is \begin{align*} \frac{\partial P}{\partial P} + \frac{\partial Q}{\partial y} =0 && && && \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} =0 \end{align*} which says precisely \begin{align*} \mathrm{Div}(V) = 0 && && && \mathrm{Curl}(V) = 0, \end{align*} in terms of the divergence and curl operators.

Summary:

  • The first Cauchy Riemann equation says exactly that $\overline f$, when viewed as a vector field, has zero divergence. Some people would say "$\overline f$ is solenoidel".
  • The second Cauchy-Riemann equation says exactly that $\overline f$, when viewed as a vector field, has zero curl. Some people would say "$\overline f$ is irrotational".

Thus, understanding the geometry of the Cauchy-Riemann equations really comes down to understanding the geometry of the divergence and curl operators, whose physical interpretations are well documented.

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  • $\begingroup$ As a note, the Polya vector field can be derived formally. Consider the clifford algebra $\mathbb G^2$ that is built upon $\mathbb R^2$. The even subalgebra of $\mathbb G^2$ is isomorphic to $\mathbb C$. Multiplying elements of this subalgebra with any single vector under the clifford product produces a vector result. For the right choice vector to multiply with, you get the Polya vector field and derive the vector calculus analogues of the CR condition. $\endgroup$ – Muphrid Nov 24 '14 at 5:08
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I personally think of Cauchy Riemann equations in terms of differential geometry.

A function $f: \mathbb C \to \mathbb C $ can be thought of as a function from $\mathbb{R} ^2 \to \mathbb{R}^2$, say $f= (u(x,y) , v(x,y))$. Now conformality, (for local diffeomorphisms) is ensured by the fact that First Fundamental form of any surface patch $\sigma$ of $\mathbb{R} ^2$ (say identity function) and First Fundamental form of the patch $f \circ \sigma$ is a scalar multiple of each other.

Lets take $\sigma = id$. First Fundamental form of $\sigma$ is $(1,0,1)$.

Now First Fundamental form of $f \circ \sigma$ is $\left( (u_x^2 + v_x^2), u_x u_y + v_x v_y, (u_y ^2 + v_y^2) \right)$.

So for conformality what we need is $u_x ^2 + v_x ^2 = u_y^2 + v_y ^2$ and $u_xu_y + v_xv_y = 0$.

Note that Cauchy-Riemann equation guarantees that and in fact those equations offer the cauchy-riemann equation(with extra assumption that orientation of angles must also be preserved).

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One might think that being differentiable on $\mathbb{R}^2$ is sufficient for differentiability on $\mathbb{C}$. But the Jacobian of an arbitrary such function doesn't have a natural complex number representation.

$$ \left[ {\begin{array}{*{20}{c}} {\partial u/\partial x} & {\partial u/\partial y} \\ {\partial v/\partial x} & {\partial v/\partial y} \end{array}} \right] $$

Another way of putting this is that no complex-valued derivative (see below for an example) you can define for an arbitrary function fully captures the local behaviour of the function that is represented by the Jacobian.

$$ \frac{df}{dz} = \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right) + i\left(\frac{\partial v}{\partial x}-\frac{\partial v}{\partial y}\right) $$

The idea is that we should be able to define a complex-valued derivative "purely" for the value $z$, without considering directions, i.e. we want to consider $\mathbb{C}$ one-dimensional in some sense (the sense being "as a vector space"). More precisely, the derivative in some direction in $\mathbb{C}$ should determine the derivative in all other directions in a natural manner -- whereas on $\mathbb{R}^2$, the derivatives in two directions (i.e. the gradient) determines the directional derivatives in all directions.

If you think about it, this is quite a reasonable idea -- it's analogous to how not every linear transformation on $\mathbb{R}^2$ is a linear transformation on $\mathbb{C}$ -- only spiral transformations are.

$$ \left[ {\begin{array}{*{20}{c}} {a} & {-b} \\ {b} & {a} \end{array}} \right] $$

How would we generalise differentiability to an arbitrary manifold? Here's an idea: a function is differentiable if it is locally a linear transformation. So on $\mathbb{R}^2$, any Jacobian matrix is a linear transformation. But on $\mathbb{C}$, only Jacobians of the above form are linear transformations -- i.e. the only linear transformation on $\mathbb{C}$ is multiplication by a complex number, i.e. a spiral/amplitwist. So a complex differentiable function is one that is locally an amplitwist (geometrically), which can be stated in terms of the components of the Jacobian as:

$$ \begin{align} \frac{\partial u}{\partial x} & = \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} & = - \frac{\partial v}{\partial x} \\ \end{align} $$

This is precisely why you shouldn't (and can't) view complex differentiability as some basic first-degree smoothness -- there is a much richer structure to these functions, and it's better to think of them via the transformations they have on grids.

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