Derivatives of Complex Functions

Question

For single variable function, it is considered to be differentiable at a point when left derivative equal to right derivative. But in the case of complex function we need to have derivative that approach our point of interest in all direction to be equal in order to be differentiable. My question here is that a complex function have two real variables as multi variable function but why multi variable function don’t have that “approaching in all direction” problem when we define its derivative?

Answer 1

Actually multivariable functions do have the "approaching from all directions" problem. An example is $f(x,y) = |xy|$ which has $f'_x(0,0)=0=f'_y(0,0)$ since $f(x,0)=0=f(0,y),$ but isn't differentiable at $(0,0).$

Answer 2

It may be true that a function $f:\mathbb R\to\mathbb R$ is differentiable at $x\in\mathbb R$ if and only if the left- and right derivative at that point exists and are equal. But in my opinion, that's not a good way to think about differentiability, or limits in general. Informally, a limit of the form $\lim\limits_{x\to a}f(x)$ is $L$, if $f(x)$ gets close to $L$ when $x$ gets close to $a$. No matter how $x$ gets close to $a$. It could be from the left. Could be from the right. Could be in a zig zag, too. In 1d, it just so happens that this is already true if it's true for just two directions: from the left and from the right. The same applies to limits of the form $\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$, and thus to derivatives.

This doesn't work for higher dimensions. In two dimensions already there is a disproportionate multitude of new paths along which $a$ can be approached. Parabolic arches, spirals, wiggly waves, or other completely wild paths. There is no more nice set of directions which are enough to determine convergence, or the existence of a derivative.

So in a way, the problem doesn't just vanish in higher dimensions. It is amplified. However, people stop thinking in terms of a few "nice" directions. In higher dimensions, convergence always implies convergence along all possible approaches. Well, it already implied that in one dimension. It just wasn't that interesting, because convergence along two paths alone already lead to convergence along all paths. Which is why those two paths sometimes get undue attention (imho).

Answer 3

We can adopt your idea of having same derivative for all direction in multivariable calculus, for example, for function $f\colon \mathbb{R}^2\to \mathbb{R}$. The notion of differentiability can be defined as follows; $f \colon \mathbb{R}^2\to\mathbb{R}$ is said to be differentiable at $x$ if there exist a function $\epsilon\colon U \to \mathbb{R}$, where $U$ is an open neighborhood of $x$ i.e. $U =\{z \in \mathbb{R}^2 | \|z-x\|<r\text{ for some (possibly very small) }r>0\}$, that satisfies $\left|\epsilon(h)\right| \to 0 $ as $\|h\| \to0$, and $$f(x+h) = f(x) + D h + h \epsilon(h).$$ Here think of $x$ and $h$ as 2 by 1 matrices and $D$ to be 1 by 2 matrix, and $Dh$ as a matrix multiplication (in this case, yes it can be thought as an inner product.) Here $D$ is something like $f'(x)$ in 1-variable case. (Make an 1-variable case notion of this definition and think about why is it a generalization.)

Here $|\epsilon(h)|\to0$ as $\|h\|\to0$, or $ \lim_{h\to(0,0)} |\epsilon(h)|=0$, encodes the idea for "in all direction" notion. However, it is a little bit complicated than some simple ideas; for example, "in all linear direction" is not enough, and we have to examine all of the curves whose one endpoint is $x$. There are some concrete counterexamples. (for example, Refer to this link.) In this sense, So we have the approaching in all direction problem.

If function $f \colon \mathbb{R}^2\to\mathbb{R}$ is differentiable at x, then for any (smooth) curve to $x$, we can think of the tangent vector of the curve at $x$, and find the derivative of function along the curve with a matrix $D$ above and the tangent vector.

In complex analysis, the notion of differentiability for complex function $f(z)$ is same with the differentiability of $\mathbb{R}^2\to\mathbb{R}^2$ function $g(u, v) = (Re(f(u+iv)), Im(f(u+iv))$. However, since there are additional structure in complex numbers, we can see that more strong condition for differentiability is needed. For example, As a two-variable function, $F \colon \mathbb{R}^2 \to\mathbb{R}^2;$ $F(x,y) = (x, -y)$, is differentiable; but we don't say that the complex conjugation, $f(x+iy) = x-iy$ is complex-differentiable (or holomorphic) in complex analysis. (here $x, y \in \mathbb{R}$) Google Cauchy-Riemann condition for more information.

Since the differentiability condition is very restrictive, the differentiable functions in complex analysis are super well-behaving. For example,

Theorem. If a complex function $f$ is holomorphic at $x$, it has $n$th derivative for all $n\ge1$ at $x$, and the taylor series at $x$ always converges to $f$ itself for some open neighborhood of $x$. (In this sense, we often call such $f$ analytic.)

Theorem. (Liouville) If $f$ is holomorphic on $\mathbb{C}$ and bounded, then $f$ is constant.