In Calculus 1 the derivative was the limit of a difference quotient. The complex derivative uses the same quotient, but now the step $h$ is a complex number, so it can shrink to $0$ from every direction. That one change makes complex differentiability far stronger than the real kind.
🔑 Key idea
$f$ is complex differentiable at $z_0$ if
$$f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$$
exists, with $h$ complex and the same limit from every direction. The usual rules hold: sums, products, quotients, the chain rule, and $(z^n)' = nz^{n - 1}$. Near such a point, $f(z_0 + h) \approx f(z_0) + f'(z_0)\,h$: tiny pieces of the plane are turned by $\arg f'(z_0)$ and stretched by $|f'(z_0)|$.
💡 Why it works
For $f(z) = z^2$: $\frac{(z_0 + h)^2 - z_0^2}{h} = \frac{2z_0h + h^2}{h} = 2z_0 + h$, which tends to $2z_0$ no matter how $h \to 0$. For $f(z) = \overline{z}$: $\frac{\overline{z_0 + h} - \overline{z_0}}{h} = \frac{\overline{h}}{h}$. With $h$ real this is $1$; with $h = it$ it is $\frac{-it}{it} = -1$. The limit depends on the direction, so $\overline{z}$ is differentiable nowhere. The algebra for products, quotients and powers is word for word the same as in real calculus, so those rules carry over.
A small square at $z_0$ and its image under $z^2$: turned and scaled, still nearly a square. Drag $z_0$; the readout gives $f'(z_0)$. Try the conjugate below for comparison.Under $\overline{z}$ the square is flipped (look at the red edge's arrow). No turn-and-scale does that, so $\overline{z}$ has no complex derivative.
Divide by $h$: $z_0\frac{\overline{h}}{h} + \overline{z_0} + \frac{|h|^2}{h}$. The last term tends to $0$.
Real $h$: $z_0 + \overline{z_0} = 2$. Imaginary $h$: $-z_0 + \overline{z_0} = -4i$. Different, so $|z|^2$ is not differentiable at $1 + 2i$ (only at $0$, where both are $0$).
Answer: $2$ and $-4i$
Why is "turned and stretched" the right picture? Near $z_0$, the change $f(z_0 + h) - f(z_0)$ is about $f'(z_0)\,h$, and multiplying $h$ by the complex number $f'(z_0) = \rho e^{i\varphi}$ turns it by $\varphi$ and stretches it by $\rho$, the same for every direction of $h$. A real map of the plane can do much more: stretch one direction more than another, shear, or flip. Complex differentiability allows only turning and uniform stretching. That restriction is the source of everything special about analytic functions.
⚠️ Common mistake
Assuming a smooth formula in $x$ and $y$ must be complex differentiable. $\overline{z} = x - iy$ and $|z|^2 = x^2 + y^2$ have partial derivatives of every order, yet neither has a complex derivative (except $|z|^2$ at the single point $0$). Complex differentiability asks the difference quotient to agree from all directions at once. Quick test: compare $h$ real with $h$ imaginary; if they disagree, $f$ is not differentiable there.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.