Is the complex derivative "speed"?

Solution 1:

The real secret here is that the real derivative is the best linear approximation. If $f:\mathbb{R}\to\mathbb{R}$, here's what the derivative means: if you zoom in really close to a point $x_0$, then $f(x)$ looks very much like the linear function $f(x_0) + f'(x_0)(x-x_0)$. This is why we can think of the real derivative as "speed" --- because that's the interpretation of the slope of a straight line.

Key point: Linear functions on $\mathbb{R}$ are exactly given by multiplication by a constant.

Now let's look at complex derivatives. First, let's think of $\mathbb{C}$ as $\mathbb{R}^2$. Then if $f:\mathbb{C}\to\mathbb{C}$ is real-differentiable, when we zoom in close to a point $z_0\in\mathbb{C}$, $f$ looks very much like $f(z_0) + A(z-z_0)$, where $A$ is a two-by-two matrix, i.e., a linear transformation of $\mathbb{R}^2$.

This is problematic since $\mathbb{C}$ is cool because we can multiply vectors. So let's only pay attention to functions where the linear approximation is multiplication by a complex number: $f$ is complex differentiable if up close to $z_0$, $f$ looks very much like $$f(z_0) + f'(z_0)(z-z_0)$$ where now the multiplication is of complex numbers, not of vector-by-matrix.

This is the similarity between real and complex differentiation: locally given by multiplication.

(By the way, this is a nice way to see where the Cauchy-Riemann equations come from. Work out the matrix for multiplication by a complex number. Apply this condition to the Jacobian of $f$ and the Cauchy-Riemann equations pop right out.)

Solution 2:

Working over $\mathbb{R}$, you can say things like "if the derivative of $f$ is positive, then $f$ is increasing." Note that both the words positive and increasing only make sense because $\mathbb{R}$ is an ordered field. You really need the order to make sense of such things. For this reason, such statements become meaningless when you work over a non-ordered field like $\mathbb{C}$.

Suppose $f$ is a holomorphic function on $\mathbb{C}$. For any point $z\in \mathbb{C}$, you can look at the derivative $f'(z)$, and you can also look at the size of the derivative $|f'(z)|$.

The derivative $f'(z)$ should be thought of as giving a linear approximation to $f$ at $z$, i.e., $$f(z + w)\approx f(z) + f'(z)w$$ for small $w$. This is analogous to the derivative for real functions giving the slope of the tangent line. The complex derivative $f'(z)$ can be thought of as the "slope" of $f$ at $z$ (though I wouldn't actually use this terminology).

The size of the derivative $|f'(z)|$ gives a measurement of how quickly $f$ is changing at $z$. The larger $|f'(z)|$ is, the bigger the difference you expect to see between $f(z)$ and $f(z + w)$ for small $w$. If $f'(z) = 0$, the function $f$ is "instantaneously constant," just like in the real case. One way to interpret this is to consider curves through $z$ of the form $\gamma_\theta\colon \mathbb{R}\to \mathbb{C}$, $\gamma_\theta(t) = z + te^{i\theta}$. Then $f(\gamma_\theta(t))$ is exactly the function $f$ along the curve $\gamma_\theta$, and you can take the real-variable derivatives just like normal. If $f'(z) = 0$, then the derivative of $f(\gamma_\theta(t))$ at $0$ will be $0$ as well, showing that $f$ is instantaneously constant along the curve at $z$.

There is no good analogue of "local maximum" and "local minimum" in the complex setting, since once again maximum and minimum require an ordered field. One thing you can do, however, is look at where the absolute value $|f(z)|$ is largest and smallest. One curious thing about holomorphic functions is that they do not have local maxima unless they are constant (this is the maximum modulus principle), and thus the real-variable theory does not carry over at all. The only place a (nonconstant) holomorphic function can minimize $|f(z)|$ is at its roots, which is also different from the real case.