Difference between the properties of differentiation in $\mathbb{C}$ and $\mathbb{R}^2$

Question

I'm taking a course on Complex Calculus and I've been provided with the following definition for the derivative of a function:

Definition: Let $f$ be a function whose domain contains a neighborhood of a point $z_0$. The derivative of $f$ at $z_0$ is the limit $$f'(z_0) = \lim_{z\to z_0} \frac{f(z) - f(z_0)}{z - z_0}$$

But for a course on Multivariable Calculus, I was provided with the following definition (from Munkres' Analysis on Manifolds):

Definition: Let $A \subset \mathbb{R}^m$, let $f: A\to \mathbb{R}^n$. Suppose $A$ contains a neighborhood of a. We say $f$ is differentiable at a if there is an n by m matrix B such that $$\frac{f(\mathbf{a + h})-f(\mathbf{a}) - B\cdot \mathbf{h}}{|\mathbf{h}|}\ \text{as}\ \mathbf{h}\to 0.$$

My problem here is that my instructor stated that any function $f:D\to \mathbb{C},\ D\subset \mathbb{C}$ that has continuous partial derivatives which satisfy the Cauchy-Riemann equations at some $z\in D$ is also differentiable at $z$. But this was not the case for a function $g:D\to \mathbb{R}^2,\ D\in \mathbb{R}^2$ using the second definition for a derivative. To my knowledge, $\mathbb{C} \cong \mathbb{R}^2$ when viewed as $\mathbb{R}$-modules, so I would expect the differential operator to behave the same as well.

• Why is it the case that these two definitions do not agree?
• Are these two definitions describing different things?
• Is any definition encapsulating the other? (If that makes any sense)

Edit: Initially, I stated that my instructor specified that every complex function continuous at a point is also differentiable at that point. This was an error on my part, as I probably mixed up the statement. I rephrased my question, though the answers are still satisfactory and explanatory of the differences between these two definitions.

Answer 1

Your teacher is wrong about continuity implying differentiability.

It is true that differentiability for a function $\Bbb C\to \Bbb C$ is different from differentiability $\Bbb R^2\to\Bbb R^2$. It is easiest to compare the two notions of differentiability if we change our definition of derivative slightly. The end result will be the same, but it's a different way of thinking about it.

Given a function $f:V\to W$ for vector spaces $V,W$, the total derivative of $f$ at $v\in V$ (if it exists) is the linear map $D_vf:V\to W$ such that $$ o(h)=f(v+h)-f(v)-D_vf(h) $$ gives a function $o:V\to W$ which satisfies $\lim_{h\to0}\frac{o(h)}{|h|}=0$.

With this interpretation of a derivative, note that the (real) total derivative of a function $f:\Bbb R^2\to\Bbb R^2$ at a point is given by a $2\times2$ real matrix, and any such matrix can appear as the total derivative. The (complex) total derivative of a function $g:\Bbb C\to\Bbb C$ at a point is a $1\times 1$ complex matrix, which is to say, a single complex number.

There are simply more degrees of freedom available for a real derivative $\Bbb R^2\to\Bbb R^2$. If we take a complex differentiable function $\Bbb C\to\Bbb C$ with derivative $a+bi$ for real $a,b$, and reinterpret it as a function $\Bbb R^2\to\Bbb R^2$, then its derivative will be $\left[\begin{smallmatrix}a&b\\-b&a\end{smallmatrix}\right]$. A real differentiable function $\Bbb R^2\to\Bbb R^2$ whose derivative isn't of that form at each point won't be complex differentiable if it is reinterpreted as $\Bbb C\to\Bbb C$.

Answer 2

Differentiability of a complex function $f: \mathbb C \rightarrow \mathbb C$ is a stricter condition that differentiability of the corresponding multivariate real function $g: \mathbb R^2 \rightarrow \mathbb R^2$.

Complex differentiability of $f(z)$ at $z_0=x_0 + iy_0$ requires that the limit of $\frac{f(z) - f(z_0)}{z - z_0}$ as $z \rightarrow z_0$ is independent of how $z$ approaches $z_0$. This can be expressed informally by saying that a complex differentiable function maps small disks around $z_0$ to small disks around $f(z_0)$. If $f'(z_0) = ke^{i \theta}$ these small disks (in the limit) are rotated by an angle $\theta$ and expanded/contracted by a factor $k$.

However, the derivative of a real differentiable function $g(x,y)$ at $\mathbf a = (x_0, y_0)$ can take different values as $(x,y)$ approaches $\mathbf a$ from different directions. This is why the derivative of $g$ at $\mathbf a$ is a matrix $B(\mathbf a)$ rather than a single value. Informally, $g$ will map small disks around $\mathbf a$ to small ellipses around $g(\mathbf a)$.

If $f: \mathbb C \rightarrow \mathbb C$ is complex differentiable then the corresponding function $g: \mathbb R^2 \rightarrow \mathbb R^2$ will be real differentiable - but not vice versa.

Answer 3

As has been mentioned above the notion of complex differentiability is a much stronger one than the notion of differentiability in $\mathbb{R}^2$. You can actually compare the notion of complex differentiability to the notion of a divergence and curl free vector field $\mathbf{F}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ in $\mathbb{R}^2$. By use of the Cauchy-Riemann equations one can show that the vector field $\mathbf{F}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ is divergence and curl free if and only if the corresponding complex function $f(z=x+iy) = P(x,y)-iQ(x,y)$ is analytic. Be aware of the minus sign!

This has applications in 2D fluid dynamics. For each analtyic complex function $f(x)$, there is a divergence and curl free vector field corresponding to the function $\overline{f'(x)}$. This will describe a stationary (time independent), incompressible and irrotational fluid flow.