I'm reading some problem in Stein Shakarchi's complex analysis.
The question says,
Conversely, prove the following: suppose $f:\Omega\to \mathbb{C}$ is a
complex-valued function, that is real-differentiable at
$z_0\in\Omega$, and $J_f (z_0)\neq 0$. If $f$ preserves angles at
$z_0$, then $f$ is holomorphic at $z_0$ with $f'(z_0)\neq 0$.
I have trouble understanding the question. What is $J_f(z_0)$ and the meaning of real-differentiability? And why do they matter? Can you please explain some context of this question? Also, the question doesn't specify that $\Omega$ is a subset of complex plane, or even metric space. Do I assume that it is a subset of complex plane?
Best Answer
Any complex function $\Bbb C\to\Bbb C$ can be viewed as a function $\Bbb R^2\to\Bbb R^2$. You can ask whether this function is differentiable, in the conventional multivariable calculus sense. That's "real differentiability".
Being real differentiable at $z_0$ means that there is a linear map $A:\Bbb R^2\to\Bbb R^2$ such that $$ f(z_0+z)\approx f(z_0)+Az $$ for a particular formalized meaning of $\approx$. The matrix representation of this linear map is $J_f(z_0)$. It's called the Jacobian matrix, and it happens to consist of the four partial derivatives of $f$ at $z_0$ (in short, $\operatorname{Re}f$ differentiated with respect to $x$ and $y$, and $\operatorname{Im} f$ differentiated with respect to $x$ and $y$, where $z=x+iy$).
Being real-differentiable is a much less strict requirement than being holomorphic / analytic / complex differentiable. For instance, the complex conjugation map is real differentiable (its Jacobian matrix is $\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$ everywhere). However, it is not holomorphic, as it doesn't preserve angles (it reverses them).
Essentially, a function is complex differentiable iff the linear map $A$ corresponds to multiplication by a complex number (that's what a linear map $\Bbb C \to \Bbb C$ is, so it meshes nicely with the conventional definition of "differentiable"), and this complex number is "the derivative". There are many, many linear maps $\Bbb R^2 \to \Bbb R^2$ that do not do this. You are asked to show that a non-zero linear map which preserves angles actually does correspond to multiplication by a (non-zero) complex number. That's a different way to view your problem.
And yes, I think it is safe to assume $\Omega \subseteq \Bbb C$. They really should've specified that, but unfortunately they didn't. Maybe they said so a paragraph or two earlier? This excerpt is clearly written as a follow-up to something else (with the "Conversely" at the start).