I am trying to get some intuition for the meaning of a complex derivative. When talking about real numbers, the function $f(x)=|x|$ is not differentiable at $x=0$, since there is a "sharp corner" there, i.e., the limits from right and left are not the same, hence the function is not smooth. The math is similar when talking about complex numbers: for $f(z)=\overline{z}$, for any $z_{0}\in\mathbb{C}$, denoting $z-z_0=re^{i\theta}$, we see that $\frac{\overline{z-z_{0}}}{z-z_0}=e^{-i2\theta}$ can have any value in $[-1,1]$ no matter how close to $z_0$ we approach. However, it is somewhat unintuitive for me that $\overline{z}$ is nowhere differentiabe in $\mathbb{C}$. Specifically:
- The complex conjugate is just the number $z$ reflected across the $x$-axis. What makes this kind of reflection impossible to differentiate, while the similar reflection $f(z)=-z$ is differentiable everywhere in the complex plane?
- Is there a "sharp corner" in some essence in the function $f(z)=\overline{z}$, similar to the one in $f(x)=|z|$ in $\mathbb{R}$?
- Separating the function to its action on the real and imaginary parts, we see that $f(x+yi)=u(x,y)+iv(x,y)$ where $u(x,y)=x$ and $v(x,y)=-y$. Both $u$ and $v$ feel "smooth", how is it that $f$ isn't?
Some graphical or other intuitive explanations will be helpful. Thank you.
If $f$ is differentiable at $z_0$, then $f(z) \approx f(z_0)+f'(z_0)(z-z_0)$. When $f(0)=0$, which is the case for $f(z)=\bar z$, we get $f(z) \approx f'(0)z$ near $0$. This means that, to first order, $f$ is a rotation followed by a scaling because that's what multiplication by the complex number $f'(0)$ does. This cannot happen for $f(z)=\bar z$ because on the real axis we get a scaling of $1$ but on the imaginary axis we get an scaling of $-1$.
All this is much better explained in Needham's wonderful book Visual Complex Analysis. See chapter 4: Differentiation: The Amplitwist Concept.