Complex derivative

The complex derivative of a complex function $f$ at $z_0$ is

$f^{\prime }(z_0)={\lim }_{\Delta z\to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z},$

provided the limit exists. The definition looks identical to the real-variable case. The difference is that $\Delta z$ is a complex number, and the limit must give the same value along every direction of approach in the plane.

That’s a much stronger condition than two-dimensional real differentiability of the components. Functions that pass turn out to be infinitely differentiable, equal to their Taylor series, and rigid in ways real functions never are.

A computation that succeeds

$f(z)=z^2$. Difference quotient:

$\frac{(z+\Delta z{)}^2-z^2}{\Delta z}=\frac{2z\Delta z+(\Delta z{)}^2}{\Delta z}=2z+\Delta z\to 2z.$

The limit is $2z$ regardless of how $\Delta z$ approaches $0$. So $f^{\prime }(z)=2z$ everywhere, same as the real-variable formula.

A function with no complex derivative anywhere

$f(z)=\bar{z}$. Difference quotient:

$\frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}=\frac{\overline{\Delta z}}{\Delta z}.$

Along the real axis ($\Delta z=\Delta x$, real): the quotient equals $1$.

Along the imaginary axis ($\Delta z=i\Delta y$): the quotient equals $\overline{i\Delta y}/(i\Delta y)=-i\Delta y/(i\Delta y)=-1$.

Different directional limits, so $\bar{z}$ is complex differentiable nowhere.

What’s odd is that $\bar{z}$ is smooth in every real-variable sense: $u(x,y)=x$ and $v(x,y)=-y$ are linear and infinitely differentiable. It fails complex differentiability not because it’s badly behaved as a function of $(x,y)$, but because it’s orientation-reversing (a reflection), while complex differentiability requires the function to behave locally like rotation-and-scaling.

More diagnostic failures

$f(z)=\mathrm{Re}(z)=x$. Difference quotient $\mathrm{Re}(\Delta z)/\Delta z$. Along reals: $1$. Along imaginaries: $0$. Differentiable nowhere.

$f(z)=|z{|}^2+iz$. Write $f=u+iv$ with $u=x^2+y^2-y$ and $v=x$. Computing along the real and imaginary axes:

• Along $\Delta z=h$ real: the difference quotient is $z+\bar{z}+h+i\to 2x+i$.
• Along $\Delta z=ih$ imaginary: the quotient is $-z+\bar{z}+i+\cdots \to -2iy+i=i(1-2y)$.

These agree only when $2x+i=i(1-2y)$, i.e., when $x=0$ and $y=0$. So $f$ is differentiable only at the origin, and not analytic anywhere: analyticity requires differentiability on a whole open set, but a single point isn’t open.

Differentiation rules

Despite the strong condition, the standard rules carry over for differentiable $f,g$ and complex constant $c$:

Rule	Statement
Sum	$(f+g{)}^{\prime }=f^{\prime }+g^{\prime }$
Constant multiple	$(cf{)}^{\prime }=c{f}^{\prime }$
Product	$(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$
Quotient	$(f/g{)}^{\prime }=(f^{\prime }g-f{g}^{\prime })/g^2$
Chain	$(f(g(z)){)}^{\prime }=f^{\prime }(g(z))g^{\prime }(z)$
Power	$(z^n{)}^{\prime }=n{z}^{n-1}$

Proofs don’t use that the variable is real, only that $\mathbb{C}$ has field structure. So polynomials, rational functions, $e^z$, $\sin z$, $\cos z$ all differentiate as in real calculus, with the same formulas.

The genuinely complex content is in the prior question of whether the function is differentiable at all. Polynomials and rationals are; products, quotients, compositions of differentiables are; but plenty of smooth-looking real-variable functions ($\bar{z}$, $|z|$, $\mathrm{Re}(z)$, $\mathrm{Im}(z)$) are not, anywhere.

The test: Cauchy–Riemann

The clean test for complex differentiability is the Cauchy-Riemann equations. Writing $f=u+iv$, $f$ is complex differentiable at $z_0$ iff $u_x=v_y$ and $u_y=-v_x$ at $z_0$, plus continuity of the partials in a neighborhood. The derivative is then $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$.

Differentiable on an open set: analyticity

A function differentiable on an open neighborhood of $z_0$ is analytic at $z_0$. The main results of complex analysis (Cauchy’s theorem, the Cauchy integral formula, the residue theorem) are statements about analytic functions, not merely differentiable ones.