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.