Differentiability and Cauchy-Riemann equations

Complex differentiability is a one-limit, all-directions condition. $f(z)$ is differentiable at $z_0$ if

exists as a single value, no matter how $\Delta z \to 0$ in the complex plane — along the real axis, the imaginary axis, or any spiral in between. where: $\Delta z$ is a complex increment, and the limit must be the same finite number $f'(z_0)$ for every path of approach. Real differentiability only ever asks about approach along a line in $\mathbb{R}^2$ restricted to one variable at a time; complex differentiability asks about every direction in the plane simultaneously, which is why it is so much harder to satisfy.

Analyticity upgrades a point condition to a neighbourhood condition. $f$ is analytic (equivalently holomorphic) at $z_0$ if it is differentiable not just at $z_0$ but at every point of some open disc around $z_0$. Differentiability at an isolated point does not make a function analytic there — you need a whole neighbourhood of good behaviour before the deep machinery of complex analysis (Cauchy's integral theorem, power-series expansions, residue calculus) becomes available.

The Cauchy-Riemann equations translate the limit condition into partial derivatives. Writing $f(z) = u(x,y) + iv(x,y)$, if $f'(z_0)$ exists then $u$ and $v$ have first partials at $(x_0,y_0)$ satisfying

where: $u$ is the real part and $v$ the imaginary part of $f$, both functions of the two real variables $x,y$; the four quantities $u_x,u_y,v_x,v_y$ are ordinary partial derivatives. These two identities are necessary for differentiability at a point — every differentiable $f$ obeys them — but on their own they are not enough to guarantee differentiability, which is the single most tested subtlety in this topic.

Continuity of the partials upgrades necessity to sufficiency. If $u_x,u_y,v_x,v_y$ exist and are continuous in a neighbourhood of $(x_0,y_0)$ and satisfy the C-R equations there, then $f$ is differentiable — in fact analytic — at $z_0$, with $f'(z_0)=u_x+iv_x$. Drop the continuity requirement and C-R holding at a point no longer guarantees anything, because the underlying real-variable theorem (differentiability of a function of two variables) itself needs continuous partials. The figure below visualises why direction-independence is the crux: two different paths into $z_0$ must produce the same difference-quotient limit, and equating what each path forces on $u$ and $v$ is exactly how the C-R equations are born.

Harmonic conjugates fall out for free. If $f=u+iv$ is analytic, differentiate the C-R equations once more: $u_{xx}=v_{yx}$ and $u_{yy}=-v_{xy}$. Adding, and using equality of mixed partials $v_{xy}=v_{yx}$, gives $u_{xx}+u_{yy}=0$ — so $u$ is harmonic, and by the same argument so is $v$. They are called harmonic conjugates: given one, C-R determines the other up to an additive real constant.

Problem. Show that $f(z) = z^2$ is differentiable everywhere and find $f'(z)$ using the Cauchy-Riemann equations; then show $f(z)=\bar z$ (complex conjugation) is nowhere differentiable.

Step 1. Split $z^2$ into real and imaginary parts. With $z=x+iy$, $z^2 = (x+iy)^2 = x^2-y^2+2ixy$, so $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$.

Step 2. Compute the four partials. $u_x=2x$, $u_y=-2y$, $v_x=2y$, $v_y=2x$.

Step 3. Check C-R and continuity. $u_x=2x=v_y$ holds, and $u_y=-2y=-v_x=-(2y)$ holds — for all $(x,y)$. All four partials are polynomials, so they are continuous everywhere. By the sufficiency direction, $z^2$ is analytic on all of $\mathbb{C}$ (an entire function), with $f'(z) = u_x+iv_x = 2x+i(2y) = 2(x+iy) = 2z$, matching the familiar formula $\frac{d}{dz}(z^2)=2z$.

Step 4. Now test $f(z)=\bar z = x-iy$, so $u=x$, $v=-y$, giving $u_x=1,\,u_y=0,\,v_x=0,\,v_y=-1$. The first C-R equation requires $u_x=v_y$, i.e. $1=-1$, which is false at every point. Since the necessary condition fails everywhere, $\bar z$ is nowhere differentiable — even though $u$ and $v$ are perfectly smooth real functions on their own, differentiability of the components separately is not enough.

Answer: $f(z)=z^2$ is entire with $f'(z)=2z$; $f(z)=\bar z$ is differentiable at no point of $\mathbb{C}$.

Competitive stretch

For $f(z)=|z|^2 = x^2+y^2$: $u=x^2+y^2,\,v=0$, so $u_x=2x,\,u_y=2y,\,v_x=0,\,v_y=0$. C-R needs $u_x=v_y \Rightarrow 2x=0$ and $u_y=-v_x \Rightarrow 2y=0$, both true only at the single point $(0,0)$. So $f$ is differentiable only at $z=0$ and, since no neighbourhood of $0$ is entirely made of points where it's differentiable, $f$ is nowhere analytic — a sharp illustration that differentiability at one point does not give analyticity.

• Confusing C-R sufficiency with necessity. C-R holding at a point is necessary but not sufficient by itself — you also need the four partials continuous near that point before concluding differentiability.
• Checking C-R at one point and calling the function analytic. Analyticity needs differentiability throughout a neighbourhood; $f(z)=|z|^2$ satisfies C-R only at the origin and is nowhere analytic.
• Sign error in the second equation. It is $u_y=-v_x$, not $u_y=v_x$; this is the single most common slip.
• Forgetting the harmonic conjugate is only unique up to a constant. Given $u$ harmonic, $v$ is determined by integrating the C-R relations but always carries a free additive real constant.

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