Lesson 2: Holomorphic Functions & the Cauchy-Riemann Equations

Lesson 0 left a promissory note: writing $f(z)=z^2$ as $u(x,y)+iv(x,y)=x^2-y^2+2ixy$, the map turned out to preserve angles everywhere except at the origin. Why does it preserve angles at all? This lesson answers that.

The limit definition, and why it's restrictive

Real differentiability only has to work along a line — $h\in\mathbb{R}$ approaching $0$ from two directions. The complex analogue looks identical on paper, $$f'(z_0) = \lim_{h \to 0} \frac{f(z_0+h)-f(z_0)}{h}, \qquad h \in \mathbb{C},$$ but $h$ now ranges over an entire plane of approach directions, and the limit has to agree along every one of them. That's a dramatically stronger requirement than the real case, where continuity plus a two-sided limit is enough. Most functions fail this test; the ones that pass, where the complex derivative exists on an open set, are called holomorphic, and they turn out to be extraordinarily special.

Key takeaway: $f'(z_0)$ requires the same limit along every direction $h\in\mathbb{C}$, not just two. Functions where this exists (on an open set) are holomorphic.

Deriving the Cauchy-Riemann equations

Two special directions pin down what "same limit everywhere" forces. Writing $f=u+iv$:

• $h=t$ real: the limit gives $u_x+iv_x$.
• $h=it$, $t$ real: the limit gives $v_y-iu_y$.

Setting these equal and matching real and imaginary parts gives the Cauchy-Riemann equations: $$u_x = v_y \qquad u_y = -v_x$$

(A common slip worth flagging explicitly: the imaginary-axis direction divides by $i$ along the way, and $1/i=-i$; it's easy to drop that sign and land on $+iu_y$ instead of $-iu_y$.)

CR is necessary for complex differentiability, though not quite sufficient on its own: sufficiency also needs the partial derivatives to be continuous.

Key takeaway: $u_x=v_y$, $u_y=-v_x$: necessary for complex differentiability.

Worked examples

$f(z)=z^2$: $u=x^2-y^2$, $v=2xy$ (from Lesson 0). $$u_x=2x=v_y \quad\checkmark \qquad u_y=-2y=-v_x \quad\checkmark$$ CR holds everywhere, so $z^2$ is holomorphic on all of $\mathbb{C}$.

$f(z)=\bar z$: $u=x$, $v=-y$. $$u_x=1, \qquad v_y=-1$$ CR fails everywhere: $\bar z$ is nowhere holomorphic. About as stark a failure as is possible.

$f(z)=|z|^2$: $u=x^2+y^2$, $v=0$. $$u_x=2x=v_y=0 \implies x=0, \qquad u_y=2y=-v_x=0 \implies y=0$$ CR holds only at $z=0$, a single isolated point, not an open set. This is worth pausing on: differentiable at an isolated point is not the same thing as holomorphic, which requires an open set of differentiability.

Key takeaway: $z^2$ is holomorphic on $\mathbb{C}$. $\bar z$ is nowhere holomorphic. $|z|^2$ is complex-differentiable only at the single point $z=0$, not holomorphic there.

Conformality: closing the Lesson 0 loop

Viewing $f=u+iv$ as a real map $\mathbb{R}^2\to\mathbb{R}^2$, its Jacobian is $$J = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}.$$ Substituting the Cauchy-Riemann relations $u_x=v_y=a$, $v_x=-u_y=b$ collapses this to $$J = \begin{pmatrix} a & -b \\ b & a \end{pmatrix},$$ which is exactly the form of a rotation matrix $\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}$ scaled by a constant factor: $J$ is a rotation by $\arg(a+ib)$ composed with a scaling by $\sqrt{a^2+b^2}=|f'(z_0)|$ (note $\det J = a^2+b^2$, the scaling factor squared).

That matrix computation just made four claims: the same rotation hits every tangent vector, the map collapses where $f'=0$, CR failing shows up as reflection, and $e^z$ has no bad points at all. Those are exactly the four things to watch below. As you scroll, the demo moves to each situation; it stays fully draggable the whole way.

A rotation-plus-scaling map preserves angles between curves by construction, so wherever $f'(z_0)\ne0$, $f$ is conformal at $z_0$. Exactly where $f'(z_0)=0$, the Jacobian collapses (no well-defined rotation angle), and conformality can break down. That's precisely what happened at $z=0$ for $f(z)=z^2$ in Lesson 0.

It's worth checking that CR really does force agreement along every direction, not just the two used to derive it. Writing $h=te^{i\theta}$ and substituting CR into the linear approximation, the $\theta$ dependence cancels completely: the limit comes out to $u_x-iu_y$ regardless of direction.

Key takeaway: $f$ holomorphic with $f'(z_0)\neq 0$ $\implies$ the Jacobian is a rotation-scaling $\implies$ $f$ is conformal at $z_0$. Conformality fails exactly where $f'=0$.

Harmonic functions (a natural closer)

Laplace's equation: $u_{xx}+u_{yy}=0$. Functions satisfying it are called harmonic, and they govern heat flow, electrostatics, and fluid flow.

It falls straight out of CR: differentiate $u_x=v_y$ with respect to $x$ and $u_y=-v_x$ with respect to $y$, then add: $$u_{xx}+u_{yy} = v_{yx} - v_{xy} = 0$$ (assuming mixed partials commute). The same argument applies to $v$. So the real and imaginary parts of any holomorphic function are automatically harmonic. Given holomorphic $f=u+iv$, $v$ is called a harmonic conjugate of $u$ — a one-way relation, not a mutual one. Swapping them generally breaks it: $v+iu$ holomorphic would require $u_y=v_x$, while CR for $u+iv$ gives $u_y=-v_x$. What does hold is that $-u$ is a harmonic conjugate of $v$, since $v-iu=-i(u+iv)$.

This is the seed of a genuinely useful bridge: complex analysis becomes a tool for solving Laplace's equation, one that pays off directly when conformal mapping enters later in the syllabus.