Example: Take the derivative of f(z) = 4z^2 from first principles. Put w = f(z) and take z_0 \in \mathbb C. \begin{aligned} \lim_{\Delta z \to 0} \frac{\Delta w}{\Delta z} &= \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z} \\ &= \lim_{\Delta z \to 0} \frac{4(z_0 + \Delta z)^2 - 4z_0^2}{\Delta z} \\ &= \lim_{\Delta z \to 0} \frac{4z_0^2 + 8z_0\Delta z + 4(\Delta z)^2 - 4z_0^2}{\Delta z}\\ &= 8z_0\\ \implies f'(z) &= 8z \end{aligned}
Example: For f(z) = |z|^2, f' doesn’t exist except at z=0. This is a very different situation from the case in \mathbb R, where the function is differentiable everywhere.
B.C. 23 Ex 2 (8 Ed 22 Ex 2)
Note. Differentiability implies continuity, but the converse does not hold. An example of the converse failing is |z|^2 or |z|.
\begin{aligned} \frac{d}{dz}(c) &= 0\qquad c \in \mathbb C \\ \frac{d}{dz}\,z^n &= n z^{n-1} \quad n \in \mathbb Z\\ \frac{d}{dz} \,e^z &= e^z \\ \frac{d}{dz} \,\sin z &= \cos z \\ \frac{d}{dz} \,\cos z &= -\sin z \end{aligned}
The usual rules apply. For f, g differentiable, \begin{aligned} (f \pm g)' &= f' \pm g' \\ (fg)' &= fg' + f'g \\ (f/g)' &= \frac{gf' - fg'}{f^2} \quad g \ne 0 \end{aligned} We also have the chain rule: if f is differentiable at z_0 and g is differentiable at f(z_0), then the composition g \circ f is differentiable at z_0 and the derivative is (g\circ f)'(z_0) = g'(f(z_0))f'(z_0) and this can be written as \frac{dg}{dz} = \frac{dg}{dw} \frac{dw}{dz}\quad \text{where } w = f(z).
Let z = x+iy and suppose f : z \mapsto w = u(x,y) + iv(x,y) is differentiable at z_0 = x_0 + iy_0. Set \Delta z = \Delta x + i \Delta y, then f'(z_0) = \lim_{\Delta z\to 0} \frac{\Delta w}{\Delta z}.
Key point: If the derivative exists, its value is independent of how \Delta z \to 0.
Note that \begin{aligned} \Delta w = f(z_0 + \Delta z) - f(z_0) &= u(x_0 + \Delta x, y_0 + \Delta y) + iv (x_0 + \Delta x, y_0 + \Delta y) - u(x_0,y_0) - iv(x_0, y_0) \end{aligned} We can decompose the limit into real and imaginary, \begin{aligned} f'(z_0) &= \lim_{(\Delta x, \Delta y) \to (0,0)} \operatorname{Re}\left(\frac{\Delta w} {\Delta z}\right) + i \lim_{(\Delta x, \Delta y) \to (0,0)}\operatorname{Im}\left(\frac{\Delta w} {\Delta z}\right). \end{aligned} These limits must still be independent of the path (\Delta x, \Delta y) \to (0,0). To start, let (\Delta x, \Delta y) \to (0,0) along the x-axis, i.e. along (\Delta x, 0) for \Delta x \ne 0. So, \begin{aligned} \frac{\Delta w}{\Delta z} &= \frac{u(x_0 + \Delta x, y_0) - u(x_0, y_0)}{\Delta x} + i\frac{v(x_0 + \Delta x_0, y_0) - v(x_0, y_0)}{\Delta x} \end{aligned} which implies (below, u_x is the partial derivative of u w.r.t. x) \begin{aligned} \lim_{(\Delta x, \Delta y) \to (0,0)} \operatorname{Re}\left(\frac{\Delta w} {\Delta z}\right) &= u_x(x_0, y_0) = \frac{\partial u}{\partial x}(x_0, y_0) \\ \lim_{(\Delta x, \Delta y) \to (0,0)} \operatorname{Im}\left(\frac{\Delta w} {\Delta z}\right) &= v_x(x_0, y_0) = \frac{\partial v}{\partial x}(x_0, y_0) \end{aligned} We can derive similar expressions for \Delta z \to 0 along the y-axis. For this, we get \begin{aligned} \frac{\Delta w}{\Delta z} &= \frac{u(x_0, y_0 + \Delta y) - u(x_0, y_0)}{i\Delta y} + i\frac{v(x_0, y_0+ \Delta y) - v(x_0, y_0)}{i\Delta y} \end{aligned} Being careful with the i, we get \begin{aligned} \lim_{(\Delta x, \Delta y) \to (0,0)} \operatorname{Re}\left(\frac{\Delta w}{\Delta z}\right) &= v_y(x_0, y_0) \\ \lim_{(\Delta x, \Delta y) \to (0,0)} \operatorname{Re}\left(\frac{\Delta w}{\Delta z}\right) &= -u_y(x_0, y_0) \\ \end{aligned} Together, because the \Delta z \to 0 must be path independent and we’ve found the value along two paths, these must coincide. This gives is the Cauchy-Riemann equations.
Theorem. (Cauchy-Riemann equations) If f = u+iv is differentiable at z_0 = x_0 + iy_0, then u_x = v_y and -v_x= u_y at (x_0, y_0).
Note. We have shown that C/R are necessary for complex differentiability, but they are not sufficient. There are sufficient conditions.
If we know
then f'(z_0) exists.
Note that there are no is in this board; it is a statement on functions of \mathbb R^2.
Remark. There are no necessary and sufficient conditions for complex differentiability. Otherwise, we would have reduced complex analysis to \mathbb R^2 analysis (how boring!).