Complex analysis — analytic functions, Cauchy-Riemann equations: Visual walkthrough

Step 0 — The picture of a complex function (setup, no formula yet)

Look at the figure: the left plane holds the input point $z$ ; an arrow labelled $f$ carries it to the right plane, where it lands as the output with coordinates $(u, v)$ . Everything we do is about how that landing spot shifts when we nudge the input point.

Step 1 — What "derivative" even means here

The figure shows the input point with several coloured arrows $Δ z$ poking out in different directions. Complex differentiability demands that all of them, after being pushed through $f$ and divided out, report the identical $f^{'} (z)$ .

Step 2 — March in along the real axis (horizontal approach)

Set $Δ z = Δ x$ (a real number, purely horizontal). Then: $f^{'} (z) = lim_{Δ x \to 0} \frac{u ( x + Δ x , y ) + i v ( x + Δ x , y ) - u - i v}{Δ x} .$

Split the fraction into its right-part and up-part — each becomes an ordinary rate of change: $f^{'} (z) = u_{x} Δ x \to 0 lim \frac{u ( x + Δ x , y ) - u}{Δ x} + i v_{x} Δ x \to 0 lim \frac{v ( x + Δ x , y ) - v}{Δ x} .$

In the figure the nudge (mint arrow) points purely right. The two side-graphs show $u$ and $v$ as roads you walk along in $x$ ; their slopes are $u_{x}$ and $v_{x}$ — the real and imaginary halves of the derivative measured this way.

Step 3 — March in along the imaginary axis (vertical approach)

Now set $Δ z = i Δ y$ (purely vertical). The key algebra fact: $\frac{1}{i} = - i because i \cdot (- i) = - i^{2} = - (- 1) = 1.$

Compute: $f^{'} (z) = lim_{Δ y \to 0} \frac{u ( x , y + Δ y ) + i v ( x , y + Δ y ) - u - i v}{i Δ y} = \frac{1}{i} (u_{y} + i v_{y}) .$

Here $u_{y}, v_{y}$ are the slopes as you walk upward only ( $x$ held fixed). Multiply out using $\frac{1}{i} = - i$ : $\frac{1}{i} (u_{y} + i v_{y}) = - i u_{y} - i^{2} v_{y} = real v_{y} - i imag u_{y} .$

The figure repeats Step 2's setup but with a coral arrow pointing straight up, and highlights how the $1/ i$ turn rotates the little output arrow a quarter-turn clockwise — that rotation is exactly what rearranges $u_{y}, v_{y}$ into $v_{y} - i u_{y}$ .

Step 4 — Force the two answers to agree

$real, from A u_{x} + i imag, from A v_{x} = real, from B v_{y} - i imag, from B u_{y} .$

Match real parts, then imaginary parts: $real : u_{x} = v_{y}, imag : v_{x} = - u_{y} .$

The figure overlays the two derivative arrows (mint = horizontal result, coral = vertical result) and shows them collapsing onto one arrow when CR holds. The two component-matchings are labelled right where the arrows align.

Step 5 — What the equations mean geometrically (the special Jacobian)

Stack the four partials: $J = (u_{x} v_{x} u_{y} v_{y}) u_{x} = v_{y} u_{y} = - v_{x} (a b - b a), a := u_{x}, b := v_{x} .$

The figure shows a tiny square grid pushed through this matrix: it comes out rotated and uniformly resized — corners still square, orientation preserved. That "angles survive" property is why analytic maps are conformal (see Conformal mappings).

Step 6 — The degenerate case: CR at a lone point is not enough

Take $f (z) = ∣ z ∣^{2} = x^{2} + y^{2}$ , so $u = x^{2} + y^{2}$ , $v = 0$ . $u_{x} = 2 x, v_{y} = 0 \Rightarrow u_{x} = v_{y} needs x = 0; u_{y} = 2 y, v_{x} = 0 \Rightarrow u_{y} = - v_{x} needs y = 0.$

Both hold only at the single point $z = 0$ .

The figure marks $z = 0$ as the only place the two component conditions ( $x = 0$ line and $y = 0$ line) intersect — a lone point, not an open patch. Analyticity needs a full shaded disc; here we get a single pixel.

Step 7 — The degenerate case: a reflection breaks the handshake everywhere

Take $f (z) = \overset{z}{ˉ} = x - i y$ , so $u = x$ , $v = - y$ . $u_{x} = 1, v_{y} = - 1 \Rightarrow u_{x} \neq = v_{y} (fails, every point) .$

The figure pushes a grid through $\overset{z}{ˉ}$ : it comes out mirror-flipped (a letter "R" becomes its mirror image). No rotate-and-scale can do that, so the special Jacobian is impossible, and CR is broken at every point — nowhere analytic.

The one-picture summary

Read the summary left to right: nudge in from the horizontal ( $\to u_{x} + i v_{x}$ ) and from the vertical ( $\to v_{y} - i u_{y}$ ); the $1/ i$ quarter-turn (top) is what rearranges the second one. Forcing them equal splits into the two boxed CR equations, which reshape the Jacobian into the rotate-and-scale form — a map that spins and zooms but never flips.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a stretchy sheet with a point on it. I poke the point a tiny bit to the right and watch where the map sends it — that gives me one "rate," made of two numbers ( $u_{x}$ and $v_{x}$ ). Then I poke the same point a tiny bit upward and watch again — that gives another rate. But poking upward sneaks in a quarter-turn (that's the $1/ i$ ), which shuffles the two numbers around and flips one sign, so this rate reads as $v_{y}$ and $- u_{y}$ .

Now here's the rule of the game: a "nice" map has to zoom-and-spin the same way no matter which way I poked. So the rightward answer and the upward answer must be the identical complex number. Setting the right-coordinates equal gives $u_{x} = v_{y}$ ; setting the up-coordinates equal gives $u_{y} = - v_{x}$ . Those two lines are the Cauchy–Riemann equations, and the minus sign is just the leftover fingerprint of that quarter-turn.

Two warnings from the pictures: matching only right-and-up isn't always enough at a lone dot (like $∣ z ∣^{2}$ at the origin) — you need it on a whole patch. And a mirror map (like $\overset{z}{ˉ}$ ) flips the sheet over, which no spin-and-zoom can do, so it flunks everywhere.

Recall

Along the real axis, $f^{'} (z)$ equals what? ::: $u_{x} + i v_{x}$ Along the imaginary axis, $f^{'} (z)$ equals what? ::: $v_{y} - i u_{y}$ Where does the minus sign in $u_{y} = - v_{x}$ come from? ::: From $1/ i = - i$ in the vertical approach — the quarter-turn. Why isn't $f (z) = ∣ z ∣^{2}$ analytic even though CR holds at $0$ ? ::: A single point isn't an open region; analyticity needs CR on a whole neighbourhood. What shape does the Jacobian take when CR holds? ::: $(a b - b a)$ — a rotation-and-scaling (multiply by $a + ib$ ).

Active recall

What are the two directions we compare to derive CR, and the derivative each gives?

Horizontal (

Δ z = Δ x

) gives

u_{x} + i v_{x}

; vertical (

Δ z = i Δ y

) gives

v_{y} - i u_{y}

State the Cauchy–Riemann equations.

u_{x} = v_{y}

and

u_{y} = - v_{x}

Why can one complex equation split into two real ones?

Two complex numbers are equal only if both real and imaginary parts match separately.

What geometric transformation does the CR-Jacobian represent?

A pure rotation-and-scaling (multiplication by a single complex number

a + ib

) — orientation-preserving.