What does Cauchy-Riemann mean in polar coordinates? Take z = x+iy = re^{i\theta} so x = r \cos \theta and y = r \sin \theta. By the chain rule, we get \begin{aligned} u_r &= u_x \cos \theta + u_y \sin \theta \\ u_\theta &= -u_x r \sin \theta + u_y r \cos \theta \\ v_r &= v_x \cos \theta + v_y \sin \theta \\ v_\theta &= v_x r \sin \theta + v_y r \cos \theta \end{aligned} We can derive C/R in polar coordinates as r u_r = v_\theta and u_\theta = -r v_r.
Therefore, if f' exists, then f' = u_x + iv_x. By using the polar coordinates expression, we also get f'(z) = e^{-i\theta}(u_r + iv_r).
Formally, we are going to change variables from (x,y) to (z, \bar z), where z = x+iy and \bar z = x-iy. This means that x = (z + \bar z)/2 and y = (z - \bar z)/(2i).
This derivation makes use of the multivariate chain rule. Specifically, if x(t) and y(t) are differentiable functions of t and z = f(x,y) is a differentiable function of x and ythen z = f(x(t),y(t)) is differentiable and \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}.
\begin{aligned} \frac{\partial f}{\partial x} &= \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial f}{\partial \bar z} \frac{\partial \bar z}{\partial x} \\ &= \frac {\partial f}{\partial z} + \frac {\partial f}{\partial \bar z} \\ \frac{\partial f}{\partial y} &= \frac{\partial f}{\partial z}\frac{\partial z}{\partial y} + \frac{\partial f}{\partial \bar z} \frac{\partial \bar z}{\partial y} \\ &= i\frac {\partial f}{\partial z} -i \frac {\partial f}{\partial \bar z} \\ \end{aligned}
Then, \begin{aligned} \frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y} &= 2 \frac{\partial f}{\partial z} \implies \frac{\partial }{\partial z} = \frac 1 2 \left(\frac \partial {\partial x} - i \frac \partial {\partial y}\right) \\ \frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} &= 2 \frac{\partial f}{\partial \bar z} \implies \frac{\partial }{\partial \bar z} = \frac 1 2 \left(\frac \partial {\partial x}+ i \frac \partial {\partial y}\right) \end{aligned} \frac{\partial}{\partial z} and \frac{\partial}{\partial \bar z} are called the Wirtinger operators.
Example: Consider f(z) = z^n = (x+iy)^n. Then, \begin{aligned} \frac{\partial f}{\partial z} &= \frac 1 2 \left(\frac \partial {\partial x} - i \frac \partial {\partial y}\right)(x+iy)^n \\ &= \frac 1 2(n(x+iy)^{n-1} -i^2n(x+iy)^{n-1}) \\ &= \frac 1 2(n(x+iy)^{n-1} +n(x+iy)^{n-1})\\ &= n(x+iy)^{n-1} = nz^{n-1}=f'(z) \\ \frac{\partial f}{\partial \bar z} &= 0\quad \text{(follows from above)} \end{aligned}
For f = u+iv complex differentiable, \begin{aligned} \frac 1 2 \frac{\partial f}{\partial x} &= \frac 1 2 (u_x + iv_x) \overset{\text{CR}}= \frac 1 2 (v_y -iu_y) \\ &= -\frac i 2 (u_y + iv_y) = -\frac i 2\frac{\partial f}{\partial y} \end{aligned} So C/R holds if and only if \frac{\partial f}{\partial \bar z} = 0. This is version II of the Cauchy-Riemann equations.
But why is this partial derivative equal to the full derivative? From f' = u_x + iv_x, \begin{aligned} \frac{df}{dz} &= u_x + iv_x = \frac{\partial f}{\partial x} \\ &= -i \frac{\partial f}{\partial y}\quad \text{(by CR)} \\ &= \frac 1 2 \left(\frac{\partial f}{\partial x} -i \frac{\partial f}{\partial y}\right) \\ &= \frac{\partial f}{\partial z} \end{aligned} Example 1: Find f'(z) for f(z) = e^z. First, we check the sufficient conditions for f' to exist. Writing f(z) = u+iv = e^{x+iy} = e^x(\cos y + i \sin y), it is defined on \mathbb C. Moreover, the components are u = e^x \cos y and v = e^x \sin y which have partials defined and continuous on \mathbb C. Then, we need to check C/R by testing u_x = v_y and u_y = -v_x or just checking \frac{\partial f}{\partial \bar z} = 0.
Example 2: When is g(z) = |z|^2 differentiable? Note that g(z) = z \bar z = x^2 + y^2. Checking C/R II, \frac{\partial g}{\partial \bar z} = 0 \implies z = 0, so g cannot be differentiable for z \ne 0 because C/R is necessary. At z = 0, we check the sufficient conditions. It is easy to show that u, v, u_x, v_x, u_y, v_y are defined and continuous on a neighbourhood of 0. Therefore, g'(0) = 0.
Exercise: Go through the same exercise for z \mapsto 1/z on \mathbb C_*.
Definition. A function f : \Omega \to \mathbb C is analytic at z_0 if f is differentiable on a neighbourhood of z_0.
Definition. A function is singular at z_0 if it is not analytic at z_0 but is analytic at some point in any neighbourhood of z_0. For example, f(z) = 1/z is analytic on \mathbb C_* and singular at 0.
That is, given B_\epsilon(0), f is analytic on B_{\epsilon'}(z_0) for some z_0 \in B_\epsilon(0) and \epsilon' < |z_0|.
Definition. A function is entire if it is analytic on all of \mathbb C. For example, polynomials, sine, cosine, exponential, etc.
Note: If a function is differentiable at precisely one point, it is not analytic there or anywhere (e.g. |z|^2).
Also, note that we are calling once-differentiable functions analytic. In real analysis, analytic functions were smooth and equal to their power series (infinitely differentiable). What’s going on?