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Apologies for a novice question. I'm wondering about the "rigidity" of a function being complex differentiable. A function $f : D \subset \Bbb C \to \Bbb C$ is said to be complex differentiable if $z$ is an interior point of $D$ and $$f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h}$$ exists.

Now every complex-valued function $f :\Bbb C \to \Bbb C$ can be identified with a function $f :\Bbb R^2 \to \Bbb R^2$ so I naively expected that every function $$f:\Bbb R^2 \to \Bbb R^2$$ that is differentiable in the real sense would also be differentiable in the complex case, but this seems not to be true. What might the problem be here?

Wondera
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    see Cauchy-Riemann equations: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations – Randall Jul 14 '24 at 16:13
  • I know the equations and I know that one can check analyticity by seeing if $f$ satisfies these, but I cannot figure out from here why there is a difference? @Randall – Wondera Jul 14 '24 at 16:14
  • Because not all real and imaginary parts obey the C-R equations, but they might still be differentiable "enough" to give a differentiable function on $\mathbb{R}^2$. – Randall Jul 14 '24 at 16:16
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    If you want people to find a mistake in your reasoning, you have to provide more of an attempted argument than just "I naively expected." – Moishe Kohan Jul 14 '24 at 16:25
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    The derivative at a point is a linear map. But be careful now: linear in what sense? Complex linear or real-linear? This is the distinction. The extra requirement of complex-differentiability requires the derivative to be $\Bbb{C}$-linear and this is where the rigidity comes from (this manifests itself as the Cauchy-Riemann equations once you write it out in terms of the standard basis). Anyway, see the beginning of my answer to Generalizing the Analyticity of holomorphic functions for more details (and generality). – peek-a-boo Jul 14 '24 at 18:20
  • And same thing but at a not-so-general level: here. – peek-a-boo Jul 14 '24 at 18:29

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The derivative of the function on $\mathbb R^2$ is given by the 2x2 matrix of partial derivatives. If you take the operation "multiplication by the complex number $a$", and see how it acts on the real and complex parts of a number, this also correspond to a 2x2 matrix. A function is complex differentiable iff the matrix of partials is of the right form to be "multiplication by some complex number". And that is exactly the content of the Cauchy-Riemann equations.

JonathanZ
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Consider $$A = \left(\matrix{1&0\\0 &-1}\right).$$ This matrix is a linear mapping from $\mathbb{R}^2$ into itself and is linear, therefore in particular differentiable in $\mathbb{R}^2$. The analog in $\mathbb{C}$ would be the complex conjugate $a + bi \mapsto a - bi$, which is known to be non-holomorphic. The problem is posed by the Chauchy Riemann equations, which must apply IN ADDITION to the reel differentiability. The problem is that $\mathbb{C}$ as a body extension of $\mathbb{R}$ has much more structure (multiplication) than $\mathbb{R}^2$, where you only have the addition as a vector space.

Noctis
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There is a huge difference in the qualitative behavior of real and complex differentiable functions.

Real differentiable functions, even smooth (infinitely differentiable) functions, have very little that can be said about their global behavior: for any open $U \subset \mathbb{R}^n$ there is a smooth function $\mathbb{R}^n \to \mathbb{R}^m$ that is zero outside $U$, but nonzero inside. (A very useful tool in analysis, as it lets one construct partitions of unity.)

However, a holomorphic (complex differentiable) function is necessarily analytic, i.e. locally determined by its own Taylor series. This has big implications: it is uniquely determined by its behavior near a single point (assuming the domain is connected). We also get results such as:

  • Liouville's theorem states that there are no bounded holomorphic functions $\mathbb{C} \to \mathbb{C}$ except for constant functions.
  • The Schwarz lemma says that any holomorphic function $f \colon D \to \overline{D}$ on the unit disc $D = \{ z : |z| < 1 \}$ with $f(0) = 0$ must have $|f'(0)| \le 1$ and $|f(z)| \le |z|$ for all $z$, and if $|f'(0)| = 1$ or $|f(z)| = |z|$ for some $z$ then $f$ is linear.
  • The Riemann mapping theorem tells us that for any simply connected domains $U,U' \subset \mathbb{C}$ which are not all of $\mathbb{C}$, there is a biholomorphic map $f \colon U \to U'$ (i.e. a holomorphic bijection whose inverse is holomorphic). Moreover, by the Schwarz lemma it is unique if $f(z_0)$ and $\arg f'(z_0)$ are prescribed for some $z_0 \in U$.

Much of this can be seen as a consequence of the fact that because of the Cauchy-Riemann equations, the real and imaginary parts of a holomorphic function are harmonic (i.e. satisfy Laplace's equation $\nabla^2 u = 0$); harmonic functions similarly enjoy a lot of rigidity results.

arkeet
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