Cauchy–Riemann equations, harmonic functions

In one sentence

Complex differentiability asks for a single complex number where the real derivative offers four, and the Cauchy–Riemann equations are exactly the two conditions that force the four to collapse into one.

Why the exam asks it

Because the word "differentiable" carries over from real analysis and means something far stronger here. In two real variables, differentiability is a mild condition. In one complex variable it forces infinite differentiability, analyticity, harmonicity of the parts, and conformality — and candidates who have not felt that gap will accept options that treat the two as parallel.

The idea

Write $f(z) = u(x,y) + iv(x,y)$ with $z = x + iy$.

$f$ is complex differentiable at $z_0$ if $\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}{h}$ exists, where $h \to 0$ in the plane — from every direction, with the same limit. Approaching along the real axis gives $u_x + iv_x$; along the imaginary axis gives $v_y - iu_y$. Equating them:

$$u_x = v_y, \qquad u_y = -v_x .$$

These are the Cauchy–Riemann equations.

The correct statement. $f$ is complex differentiable at $z_0$ $\iff$ $u$ and $v$ are real-differentiable at $z_0$ and the Cauchy–Riemann equations hold there.

Both halves matter. The equations alone are not enough, and this is examined.

Geometrically: the real derivative of $(u,v)$ is the matrix $\begin{pmatrix}u_x & u_y\\ v_x & v_y\end{pmatrix}$, four independent numbers. Cauchy–Riemann forces it into the shape $\begin{pmatrix}a & -b\\ b & a\end{pmatrix}$, which is precisely multiplication by the complex number $a + ib$ — a rotation composed with a scaling. That is the whole content, and it is why holomorphic maps preserve angles wherever $f' \ne 0$.

A practical sufficient condition: if the partials exist, are continuous near $z_0$, and satisfy Cauchy–Riemann, then $f$ is holomorphic there. Continuity of the partials gives real differentiability for free.

Holomorphic is enormously stronger than real differentiable

If $f$ is holomorphic on an open set then $f$ is infinitely differentiable there and equals its own Taylor series locally. Nothing like this is true on $\mathbb{R}$: $x^2\sin(1/x)$ is differentiable with discontinuous derivative, and $e^{-1/x^2}$ is smooth without being analytic. Neither has a complex analogue. One derivative in the complex sense buys all of them.

Consequences that come up as options, all for $f$ holomorphic on a connected open set:

• $f' \equiv 0$ $\Rightarrow$ $f$ constant.
• $|f|$ constant $\Rightarrow$ $f$ constant.
• $\operatorname{Re}f$ constant, or $\operatorname{Im}f$ constant $\Rightarrow$ $f$ constant.
• $\overline{f}$ also holomorphic $\Rightarrow$ $f$ constant.
• $f$ real-valued $\Rightarrow$ $f$ constant.

Every one is the same phenomenon: a holomorphic function has no room to vary in only one direction.

Harmonic functions

If $f = u + iv$ is holomorphic then $u$ and $v$ are harmonic: $u_{xx} + u_{yy} = 0$. So $x^3$ is the real part of no holomorphic function, since its Laplacian is $6x \ne 0$.

Given a harmonic $u$, a harmonic conjugate is a $v$ with $u + iv$ holomorphic. It is found by integrating the Cauchy–Riemann equations, and it is unique up to an additive real constant.

Existence is a topological question. On a simply connected domain a harmonic conjugate always exists. On $\mathbb{C}\setminus\{0\}$ it need not: $u = \log|z|$ is harmonic there, and its only candidate conjugate is $\arg z$, which cannot be defined single-valuedly on a punctured plane — going once round the origin adds $2\pi$. So $\log|z|$ is a harmonic function on a domain with no harmonic conjugate, and it is the standard counterexample.

Note also that $u$ and $v$ being harmonic separately is not enough: $u = x$ and $v = x$ are both harmonic, and $x + ix$ is not holomorphic, failing $u_x = v_y$ at once. They must be a Cauchy–Riemann pair, in that order.

Where intuition breaks

"Cauchy–Riemann holds at $z_0$, so $f$ is differentiable at $z_0$." False without real differentiability. The standard witness is $f(z) = \overline{z}^{\,2}/z$ for $z \ne 0$ and $f(0) = 0$: it satisfies Cauchy–Riemann at the origin, and the difference quotient has different limits along different rays, so $f'(0)$ does not exist. The equations compare only two directions; differentiability is about all of them — the same structure as partial derivatives failing to give differentiability in two real variables.

"$u$ and $v$ are harmonic, so $u + iv$ is holomorphic." $u = v = x$. They must satisfy Cauchy–Riemann as an ordered pair.

"Every harmonic function has a harmonic conjugate." $\log|z|$ on the punctured plane. Simple connectedness is the hypothesis, and options drop it.

"$f$ differentiable at a point, so holomorphic near it." Holomorphic means differentiable on an open set. $f(z) = |z|^2$ is complex differentiable at $0$ and nowhere else — differentiability at a single point buys nothing.

"$f$ holomorphic and $\overline{f}$ holomorphic, so $f$ is interesting." $f$ is constant. Worth recognising as a true statement, since it appears among false ones.

"$|f|$ has a maximum inside, so nothing is wrong." On a connected open set, $|f|$ attaining an interior maximum forces $f$ constant — that is the maximum modulus principle, and it follows from the same rigidity.

"$f$ is smooth as a function of $x$ and $y$, so it is holomorphic." $f(z) = \overline{z}$ is as smooth as anything and satisfies $u_x = 1$, $v_y = -1$. Real smoothness is not the issue; the pairing is.

The exam's angle

1. Check Cauchy–Riemann first, then check differentiability of $u$ and $v$. Options are built on the second step being skipped.
2. For "is $u$ the real part of a holomorphic function", compute the Laplacian. If nonzero, done immediately.
3. For harmonic conjugates, look at the domain. Punctured or annular domains are where existence fails, and the question is usually about $\log|z|$ in disguise.
4. Any hypothesis fixing one real quantity forces constancy. $|f|$, $\operatorname{Re}f$, $\operatorname{Im}f$, $\arg f$ — all of them.
5. Remember there is no complex $e^{-1/x^2}$. Any option asserting a smooth-but-not-analytic complex function is false.

The night before

• Complex differentiable $\iff$ $u, v$ real-differentiable and $u_x = v_y$, $u_y = -v_x$.
• Cauchy–Riemann alone is not sufficient: $\overline{z}^2/z$ at the origin.
• Continuous partials $+$ Cauchy–Riemann $\Rightarrow$ holomorphic.
• Holomorphic $\Rightarrow$ infinitely differentiable and analytic. No complex $e^{-1/x^2}$.
• On a connected domain, $f' \equiv 0$, $|f|$ constant, $\operatorname{Re}f$ constant, $f$ real, or $\overline f$ holomorphic — each forces $f$ constant.
• Real and imaginary parts of a holomorphic function are harmonic. $x^3$ is not harmonic.
• Harmonic conjugates exist on simply connected domains. $\log|z|$ on $\mathbb{C}\setminus\{0\}$ has none.
• Two harmonic functions are not a holomorphic pair unless they satisfy Cauchy–Riemann in order.
• Cauchy–Riemann geometrically: the Jacobian is a rotation-scaling, hence conformal where $f' \ne 0$.