TITLE: Critique my geometrical understanding of the complex derivative
QUESTION [1 upvotes]: I wrote the following bit on the geometry of the complex derivative based on my understanding from reading several books and from some side comments my lecturer has made. I wanted to put it all in one place. Could you critique it according to correctness, completeness, pedantism (if that's the correct word to use here), conciseness and how didactic it is?

As with real functions, a geometrical understanding of the derivative is possible with complex differentiation. Given $X \subset \mathbb{C}$ and $f : X \rightarrow \mathbb{C}$ differentiable at $a \in X$, we have
$$f'(a) = \lim_{z \to a} \frac{f(z) - f(a)}{z - a}.$$
We now introduce an error term $\epsilon \in \mathbb{C}$ such that
$$\frac{f(z) - f(a)}{z - a} = f'(a) + \epsilon.$$
Clearly, $\epsilon$ is a function of $z$. Hence, we will write $\epsilon = \epsilon(z)$. Moreover, we must have
$$\lim_{z \to a} \epsilon(z) = 0.$$
Isolating $f(z)$ on the left-hand side with
$$f(z) = f(a) + f'(a)(z-a) + \epsilon(z)(z - a),$$
and then factoring,
$$f(z) = f(a) + (f'(a) + \epsilon(z))z - (f'(a) + \epsilon(z))a.$$
Thus, we have obtained an alternate expression for the values of the function $f$ in terms of its derivative at a single point and an error function which we know becomes arbitrarily small near that point. This expression becomes interesting and carries geometrical significance when $f'(a) + \epsilon(z) \approx f'(a)$. This is easily accomplished by letting $z$ approach $a$. Suppose $z$ is close enough to $a$ that we can make the above approximation. Then,
$$f(z) \approx f(a) + f'(a)z - f'(a)a.$$
Note that this expression is linear in $z$. Thus, we refer to it as the first order approximation, or linearization, of $f$ at $a$. The fact that the approximation is linear provides us with an intuitive geometrical understanding of the derivative of a function when it exists. To see this, consider the linear expression $w = az + b$, where $w, a, b, z \in \mathbb{C}$. The multiplication of $a$ and $z$ can be viewed geometrically as rotating and dilating the vector $(\operatorname{Re}(z), \operatorname{Im}(z)) \in \mathbb{R}^2$. The term $b$ simply translates a vector in the plane. Thus, the linear function $w$ will map every point on the unit circle to a new circle of radius $|a|$ centered at $b$. Going back to our linearization of $f$, we see that, if $z$ is sufficiently near $a$, the existence of the derivative of $f$ at $a$ signifies, geometrically, that $f$ maps circles onto circles. This can be understood to mean that $f$, in a sense, preserves shapes near $a$. The error term $\epsilon(z)$ has the effect of distorting these shapes, such that a circle may not map to another circle in the plane. Let $f : X \rightarrow \mathbb{C}$ be a function such that
$$f(z) = z^2 + 2z.$$
Since $f$ is a polynomial, it is differentiable on it domain, with derivative
$$f'(z) = 2z + 2.$$
At $z = 0$, $f'(z) = 2$. Then, the linearization of $f$ at $0$ is
$$f(z) \approx f(0) + f'(0)z - f'(0) \cdot 0= 2z.$$
Thus, if the domain is a small circle centered at $0$, the image should be a concentric circle of twice the radius. In the first image of the figure below, the inner circle is the domain $X = \{z \ ; \ |z| = 0.01\}$ and the outer circle is the image $f(X)$. Visually, the linearization seems to accurately predict the image. In the second image below, with the domain $X = \{z \ ; \ |z| = 0.99\}$, the distortion is, visually speaking, significant, and our linear approximation seems to break down. We can evaluate the error term directly to see whether our assessment above is valid or not.
$$\epsilon(z) = \frac{f(z) - f(0)}{z - 0} - f'(0) = z \quad (z \neq 0 \text{ in either case})$$
Indeed, in our first case, $|f'(0)| = 2 \gg |\epsilon(z)| = |z| = 0.01$. In our second case however, $|\epsilon(z)| = |z| = 0.99$ becomes comparable to $|f'(0)|$.

REPLY [1 votes]: There's one very important thing I think you should have mentioned. While real and complex derivatives are both limits, the former exists when two one-sided limits exist and are equal, but the latter requires existent and equal directional limits from (uncountably) infinitely many different directions. Differentiability is therefore a very stringent constraint on complex functions (with consequences that take a lot of work to flesh out, but let's limit our scope here.) This is why, for example, $z^\ast$ isn't differentiable:$$\lim_{r\to0^+}\frac{(z+re^{i\theta})^\ast-z^\ast}{re^{i\theta}}=e^{-2i\theta}$$is direction-dependent. But it's not too stringent; you need only verify two equations to prove a derivative exists, as it does for e.g. all polynomials in $z$. It's common to write functions in terms of $z,\,z^\ast$: the rule of thumb is if $z^\ast$ shows up, there's no derivative.
Here's another perspective on this, albeit one you probably needn't include in your notes. The univariate chain rule $df=\frac{\partial f}{\partial x}dx$ for real-to-real functions has a more complicated variant in number systems of some dimension $n>1$ over $\Bbb R$. If $f$ is a function from one such number to another - in your case, $n=2$ - the $i$th part of $f(z)$ satisfies$$df_i=\frac{\partial f_i}{\partial z_j}dz_j,$$where $z_j$ is the $j$th part of $z$ and I've summed over $j$. The display-line equation above looks like $n$ multivariate chain rules for functions from $\Bbb R^n$ to $\Bbb R$, doesn't it?
But that's not the whole story. For $df=g(z)dz$ to be true in this number system, with $g=f^\prime$, we need $g$ to be a number, not a matrix. There are only $n$ degrees of freedom in that number, but they have to do the work of $n^2$ DOFs. If we write numbers as $a=a_ie_i$ and multiplication obeys $a_ie_ib_je_j=c_{ijk}a_ib_je_k$ (and feel free to compute the eight $c_{ijk}$ for multipication on $\Bbb C$), a differentiable $f$ satisfies$$df_k=c_{ijk}g_idz_j\implies\frac{\partial f_k}{\partial z_j}=c_{ijk}g_i.$$Needless to say, not all matrices of derivatives will be of this form. (If you did the exercise I suggested a moment ago, you can verify the equations in my first hyperlink emerge from applying this logic to complex analysis.) More generally, the $n^2$ values $\frac{\partial f_k}{\partial z_j}$ have to be linear combinations of $n$ values $g_i$, which impose $n(n-1)$ constraints which, if satisfies, allow $n$ DOFs in the value of $g(z)$ to do the work of $n+n(n-1)=n^2$ DOFs.
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