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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$\begingroup$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.)
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.
$\endgroup$ 3 $\begingroup$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 uniform 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 $\partial_y f(z) = if'(z) = i\partial_x f(z)$.
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
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).
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$Example of action of a derivative of complex functions. If f(z)=z^2, f'(z)=2z. For example, at z=1+i, the derivative is f'(1+i)=2(1+i). So if you apply this to something like 0.01, a differential along the horizontal axis in the complex plane, you get 2(i+1)(0.01)=0.02(1+i)=0.02+0.02i. The "2" in 2(i+1) stretches 0.01 and the 1+i turns it 45 degrees, since 0.02/0.02=1 and arctan(1)=45 degrees. So you can approximate (1.01+i)^2 as (1+i)^2+2(i+1)(0.01) as a first step. This has the form f(z)+f'(z)dz roughly speaking. Or in words, value of function+amplitwist * differential. (Amplitwist is from Visual Complex Analysis by Needham)
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