Cauchy's integral theorem and formula: Visual walkthrough

Step 0 — What is a complex number, as a picture?

WHAT. A complex number $z$ is just a point on a flat sheet of paper. We give the sheet two axes: the horizontal one measures a number we call $x$ , the vertical one measures a number we call $y$ . The point sitting at horizontal $x$ and height $y$ is written

$z = how far right x + "turn 90°" marker i how far up y$

The symbol $i$ is not a mystery quantity — think of it as a label that says "this part points upward". So $z$ is an arrow from the origin to the dot.

WHY. Everything below is about walking a pen around a loop drawn on this sheet. To talk about a loop we first need to agree that each place the pen can be is one number $z$ .

PICTURE. The red dot is $z$ ; its shadow on the floor is $x$ , its shadow on the wall is $y$ .

Figure — Cauchy's integral theorem and formula

Step 1 — A function, a contour, and the meaning of $\oint f d z$

WHAT. A function $f$ takes each input point $z$ and hands back another point $f (z)$ — another arrow. A contour $γ$ (Greek letter "gamma") is a path we draw on the plane; a closed contour is a path that returns to where it started, like a loop of string.

The symbol $\oint_{γ} f (z) d z$ means: creep along the loop $γ$ in tiny steps; at each spot multiply the function's value $f (z)$ by the tiny step $d z$ you just took; add all those little products up.

$\oint$ (circle-integral) $=$ "sum along a closed loop".
$d z$ $=$ one tiny step arrow along the path (a small change in position, itself a complex number).
$f (z) d z$ $=$ the product of two arrows (complex multiplication rotates and scales).

WHY define it this way. Real integrals add up "height times tiny width". A complex integral adds up "arrow times tiny step". Because $d z$ carries a direction, walking the loop backwards flips every $d z$ — so the orientation of the loop matters. We always walk counter-clockwise (called positive orientation) unless told otherwise.

PICTURE. The loop is chopped into tiny step arrows $d z$ ; one of them is highlighted red.

Step 2 — Split the loop integral into two real loop integrals

WHAT. Write the function in parts, $f = u + i v$ , where $u (x, y)$ and $v (x, y)$ are ordinary real-valued functions (the "floor value" and the "wall value" of $f$ ). Write the step as $d z = d x + i d y$ . Multiply out:

$\oint_{γ} f d z = \oint_{γ} (u + i v) (d x + i d y) = real part \oint_{γ} (u d x - v d y) + i imaginary part \oint_{γ} (v d x + u d y)$

Term by term: $u, v$ are real functions of the point; $d x$ is a tiny sideways move, $d y$ a tiny up move; the $i^{2} = - 1$ is what turned $i v \cdot i d y$ into the negative $- v d y$ .

WHY. We do not yet know how to attack a complex loop integral, but we already own a tool for real loop integrals over the plane: Green's theorem. So we translate the one hard complex loop into two familiar real loops.

PICTURE. The same loop, now imagined as two ordinary "circulation" loops — one tracking $(u, - v)$ , one tracking $(v, u)$ .

Step 3 — Feed each real loop through Green's theorem

WHAT. Green's theorem converts circulation around a loop into a sum over the filled-in area $R$ :

$\oint_{γ} (P d x + Q d y) = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A$

Here $P, Q$ are the two real functions riding on $d x, d y$ ; $\frac{\partial Q}{\partial x}$ means "how fast $Q$ changes as you nudge right"; $\iint_{R} \dots d A$ means "add over every tiny patch of area inside the loop".

Apply it twice:

Real part ( $P = u, Q = - v$ ): $\iint_{R} (- v_{x} - u_{y}) d A$ .
Imag part ( $P = v, Q = u$ ): $\iint_{R} (u_{x} - v_{y}) d A$ .

(Subscripts are shorthand: $v_{x} = \partial v / \partial x$ , etc.)

WHY. Green's theorem is exactly the bridge from "what happens on the boundary" to "what happens inside" — the very spirit of Cauchy. It only works if $u, v$ are smooth on the whole filled region $R$ : no punctures allowed. Hold that thought for Step 6.

PICTURE. The boundary loop collapses inward into a mesh of little area patches, each carrying a "curl" value.

Step 4 — The collapse: Cauchy–Riemann kills both integrands

WHAT. A complex-differentiable function must give the same derivative no matter which direction you approach from. Forcing the real-axis approach to equal the imaginary-axis approach yields the Cauchy-Riemann equations:

$u_{x} = v_{y} and u_{y} = - v_{x}$

Plug these straight into Step 3's two area integrands:

$- v_{x} - u_{y} = - v_{x} - (- v_{x}) = 0, u_{x} - v_{y} = v_{y} - v_{y} = 0$

Both integrands are exactly zero at every point — not "small", zero. So both area sums are $0$ , and therefore

$\oint_{γ} f (z) d z = 0 (Cauchy’s Integral Theorem)$

WHY it works. Complex differentiability is a much stronger demand than real differentiability, and the Cauchy–Riemann equations are precisely that extra strength written out. That strength is what makes the curl vanish everywhere.

PICTURE. Every little patch's "curl" arrow shrinks to nothing — the whole mesh goes blank.

Recall Why

simply connected had to be there Green's theorem needed $f$ smooth on the entire filled region. A domain with no holes (simply connected) guarantees the loop can be filled with good patches. One bad point inside and this step is illegal — which is the whole story of the next steps.

Step 5 — The next miracle: the boundary determines the inside

WHAT. Pick a point $a$ inside the loop. Look at a new function built from the old one: $g (z) = \frac{f ( z )}{z - a}$ The denominator $z - a$ is the arrow from $a$ to the current point $z$ . It is fine everywhere except at $z = a$ , where it is zero and $g$ blows up. So $g$ has exactly one bad point — a pole — sitting at $a$ .

WHY this specific $g$ . We want a formula that reads off $f (a)$ from boundary data. Dividing by $z - a$ manufactures a controlled blow-up exactly at $a$ ; that blow-up is what will "catch" the value $f (a)$ for us.

PICTURE. The big loop $γ$ with the puncture $a$ marked; the arrow $z - a$ drawn in red for one boundary point $z$ .

Step 6 — Shrink the loop onto a tiny circle around $a$

WHAT. Between the big loop $γ$ and a tiny circle $C_{ε}$ of radius $ε$ centred at $a$ , the function $g$ has no bad points (the only pole, $a$ , is walled off inside the little circle). By Step 4, the integral of $g$ over any loop in a hole-free region is zero — so the outer and inner loops must give the same integral:

$\oint_{γ} \frac{f ( z )}{z - a} d z = \oint_{C_{ε}} \frac{f ( z )}{z - a} d z$

Now parametrise the tiny circle. Every point on it is $z = a + ε e^{i θ}$ :

$ε$ is the little radius, $θ$ the angle swept from $0$ to $2 π$ ,
$e^{i θ}$ is the unit arrow pointing at angle $θ$ (Euler's spin marker),
the tiny step is $d z = i ε e^{i θ} d θ$ .

Substitute: $\oint_{C_{ε}} \frac{f ( z )}{z - a} d z = \int_{0}^{2 π} \frac{f ( a + ε e ^{i θ} )}{ε e ^{i θ}} i ε e^{i θ} d θ = i \int_{0}^{2 π} f (a + ε e^{i θ}) d θ$

The $ε e^{i θ}$ on top and bottom cancel exactly — the blow-up and the tiny step were built to annihilate each other. That is the whole trick.

WHY shrink at all. On a tiny circle, $f$ barely changes: it is nearly constant equal to $f (a)$ . Shrinking replaces a hard integral with an almost-constant one.

PICTURE. The big loop deforms inward (dashed) until it hugs the small red circle $C_{ε}$ around $a$ ; the shaded ring between them is pole-free.

Step 7 — Let $ε \to 0$ and collect the answer

WHAT. As the radius $ε$ shrinks to $0$ , every point $a + ε e^{i θ}$ slides into $a$ , so by continuity $f (a + ε e^{i θ}) \to f (a)$ , a constant. A constant pulls out of the sum:

$i \int_{0}^{2 π} f (a + ε e^{i θ}) d θ ε \to 0 i f (a) \int_{0}^{2 π} d θ = i f (a) \cdot 2 π = 2 π i f (a)$

The lone $2 π$ is simply the total angle of one trip around a circle. Divide both sides by $2 π i$ :

$f (a) = \frac{1}{2 π i} \oint_{γ} \frac{f ( z )}{z - a} d z (Cauchy’s Integral Formula)$

WHY the shrink is legal. The value did not depend on $ε$ (Step 6 fixed it once and for all), so taking the limit only reveals the answer that was already true for every radius. The boundary $γ$ genuinely encodes $f (a)$ .

PICTURE. The circle shrinks to a dot; $f$ over it flattens to the single height $f (a)$ , and the $2 π$ of angle stamps out the answer.

Step 8 — The degenerate cases you must have seen

WHAT. Three edge scenarios, each its own picture-fact:

No pole inside ( $a$ outside $γ$ ): then $f / (z - a)$ is smooth on the whole filled loop, so Step 4 applies and the integral is $0$ . The formula quietly becomes the theorem.
The pure hole $\oint_{∣ z ∣ = 1} \frac{1}{z} d z$ : here $f = 1$ , $a = 0$ is inside, so the formula gives $2 π i \cdot f (0) = 2 π i$ . Not zero — because the region has a hole at $0$ , so the theorem was never allowed.
Higher-order pole (denominator $(z - a)^{n + 1}$ ): differentiating the formula $n$ times in $a$ (each derivative pulls one more factor out of $(z - a)^{- 1}$ , producing $n!$ ) gives $f^{(n)} (a) = \frac{n !}{2 π i} \oint_{γ} \frac{f ( z )}{( z - a ) ^{n + 1}} d z$ This is why one complex derivative secretly grants infinitely many.

WHY show all three. The reader must never meet a loop and freeze: pole outside → $0$ ; simple pole inside → $2 π i f (a)$ ; repeated pole → the $n!$ version.

PICTURE. Three mini-panels: pole outside (integral $0$ ), pole inside (integral $2 π i f (a)$ ), double pole (the $n!$ formula).

The one-picture summary

Everything above in a single diagram: the big loop $γ$ , the smooth (hole-free) interior where Cauchy–Riemann force $\oint f d z = 0$ , and the punctured case where shrinking onto $a$ reads off $f (a)$ from the boundary.

Recall Feynman retelling of the whole walkthrough

Draw a loop of string on a smooth sheet. Walking around it and adding up "value times step" gives zero — because the smoothness rule (Cauchy–Riemann) makes every tiny patch inside perfectly balanced, so Green's theorem sums nothing (Steps 1–4). Now poke one special point $a$ and divide your function by "the arrow from $a$ ", creating a single blow-up there (Step 5). Slide the outer loop inward until it hugs a tiny ring around $a$ — legal, since the doughnut between them is smooth (Step 6). On that tiny ring the blow-up and the tiny step cancel exactly, the function is basically constant $f (a)$ , and one full trip contributes an angle of $2 π$ ; the arithmetic spits out $2 π i f (a)$ (Step 7). Divide by $2 π i$ : the edge told you the inside. Finally, if the special point is outside you just get zero again, and if the blow-up is stacked several times you differentiate to get the $n!$ version (Step 8).

Recall Quick self-test

Why does $\oint f d z = 0$ hold only with no enclosed singularity? ::: Green's theorem needs $f$ smooth on the whole filled region; a pole breaks that. Where does the $2 π$ in the formula come from? ::: The total angle $\int_{0}^{2 π} d θ$ of one trip around the shrinking circle. What cancels the blow-up in Step 6? ::: The tiny step $d z = i ε e^{i θ} d θ$ kills the $ε e^{i θ}$ in the denominator.

Related: Cauchy-Riemann equations · Green's theorem · Residue theorem · Laurent series · Liouville's theorem · Maximum modulus principle