A clean case: f(z) = z²
First, a function that is differentiable, so the definition feels familiar.
Step 1 — form the quotient. Expand
(z + h)^2 = z^2 + 2zh + h^2:
\frac{(z + h)^2 - z^2}{h} = \frac{2zh + h^2}{h}.
Step 2 — cancel the h. Divide top and bottom:
\frac{2zh + h^2}{h} = 2z + h.
Step 3 — let h \to 0. The leftover
h vanishes regardless of direction, so the limit is the
same from every side:
f'(z) = 2z.
Just like the real power rule. The direction of h never mattered —
which is exactly why the derivative exists.
The headline: f(z) = z̄ is NOT differentiable
Now take conjugation, f(z) = \bar{z} — a function that looks tame
but secretly depends on direction.
Step 1 — form the quotient. Using
\overline{z + h} = \bar{z} + \bar{h}, the
\bar{z} terms cancel:
\frac{\overline{z + h} - \bar{z}}{h} = \frac{\bar{z} + \bar{h} - \bar{z}}{h} = \frac{\bar{h}}{h}.
Everything now rests on \bar{h}/h as
h \to 0 — and that ratio depends entirely on the
direction of h.
Step 2 — approach along the real axis (h = t, real
and going to 0). Then \bar{h} = t, so
\frac{\bar{h}}{h} = \frac{t}{t} = +1.
Step 3 — approach along the imaginary axis (h = it).
Then \bar{h} = -it, so
\frac{\bar{h}}{h} = \frac{-it}{it} = -1.
Step 4 — compare. One direction gives +1, the
other gives -1. The limit has two different values, so it
does not exist:
f(z) = \bar{z} \ \text{is nowhere complex-differentiable.}
Conjugation is perfectly smooth as a map of the real plane — yet it fails to have a complex
derivative anywhere. The all-directions rule has teeth.
The complex derivative
f'(z) = \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} with complex
h:
-
must give the same limit for every direction of
h \to 0 — far stronger than the real case;
-
for f(z) = z^2 it exists and equals
2z (the direction never appears);
-
for f(z) = \bar{z} the quotient is
\bar{h}/h, which is +1 along the
real axis but -1 along the imaginary axis — so it is
not differentiable anywhere.
Write h = \rho\, e^{i\alpha} in polar form. Then
\bar{h} = \rho\, e^{-i\alpha}, and the troublesome ratio is
\frac{\bar{h}}{h} = \frac{\rho\, e^{-i\alpha}}{\rho\, e^{i\alpha}} = e^{-2i\alpha}.
Its modulus is always 1, but its angle is
-2\alpha — it spins twice around the unit circle as the approach
direction \alpha sweeps once. There is no single limiting value
to settle on, which is the whole problem in one line.
On the real line, "differentiable" is a mild request — most functions you meet have a
derivative almost everywhere. It is tempting to carry that intuition into
\mathbb{C} and assume any smooth-looking map is differentiable.
It is not. Because the limit
\lim_{h \to 0}(f(z+h)-f(z))/h must return the
same value no matter which of the infinitely many
directions
in the plane h shrinks along, complex
differentiability at a point is a severe constraint.
The proof is right above: f(z) = \bar{z} is continuous
everywhere — a perfectly tame map of the plane — yet it is complex-differentiable
nowhere. The same fate befalls f(z) = \operatorname{Re}(z) = x:
along the real axis its quotient is 1, but along the imaginary axis
it is 0, so no limit exists at any point. There is simply
no analogue of this in real calculus — a continuous real function that is
differentiable at not a single point is exotic, but for complex derivatives it is the
everyday behaviour of ordinary-looking maps. Never read "complex differentiable" as merely
"smooth".