4.10.3 · D5Advanced Topics (Elite Level)
Question bank — Cauchy's integral theorem and formula
Reminders of the notation used below, so nothing is unexplained:
- means "add up as walks once around the closed loop ." A loop is closed when it returns to its start.
- Holomorphic = complex-differentiable (same limit from every direction) — built in the parent from the Cauchy-Riemann equations.
- Simply connected = "no holes": every loop shrinks to a point inside the region.
- A pole at is a point where blows up like .
- Positively oriented means the loop is traversed counter-clockwise, so the enclosed region stays on your left.
True or false — justify
Every line is statement ::: verdict + the real reason.
for every holomorphic and every closed loop .
False — only if encloses no singularity and is holomorphic on the whole enclosed region. around gives , not .
If is holomorphic everywhere in (entire), then for any closed .
True — an entire function has no singularities anywhere, so every enclosed region is singularity-free and Cauchy's theorem applies.
and give the same value.
True — both loops enclose the single pole at ; deforming out to crosses no new singularity, so the integral is unchanged.
Reversing the orientation of (clockwise instead of counter-clockwise) leaves unchanged.
False — reversing direction flips the sign: . Cauchy's formula would give .
The value depends on the exact shape of .
False — it depends only on whether is inside; any two loops enclosing (and no other singularity) give the same .
If for one particular loop, then must be holomorphic inside.
False — the integral can vanish by accident (e.g. symmetric cancellation) even with a singularity present; a single zero doesn't certify holomorphy.
A function that is complex-differentiable once on an open set is automatically infinitely differentiable there.
True — the generalized formula writes every as a contour integral of , which exists for all . This has no real-variable analogue.
Cauchy's theorem holds on an annulus (a disk with a hole) for any holomorphic .
False — an annulus is not simply connected; a loop around the hole need not shrink to a point, so can be nonzero even where is holomorphic on the annulus.
Spot the error
Each line quotes a flawed argument; the reveal names the flaw.
" is a ratio of polynomials, so it's holomorphic, so ."
The flaw: is not holomorphic at , which lies inside . Holomorphy must hold on the entire enclosed region, not just most of it.
" by the integral formula."
The flaw: is outside . With no pole enclosed the integrand is holomorphic inside, so the answer is , not .
" with ."
The flaw: the pole is order 3, so you need the generalized formula form with : answer , not .
" has poles at and inside , so use with ."
The flaw: the formula requires (the numerator) to be holomorphic on the whole interior; a second pole at violates this. Split by partial fractions or use the Residue theorem.
", so is constant."
The flaw: a vanishing loop integral only signals path-independence/holomorphy, not constancy. Constancy needs a global extra hypothesis — that is Liouville's bounded-entire argument.
"Since the interior of contains a pole, Green's theorem still lets me collapse the integral to ."
The flaw: Green's theorem demands the integrand be smooth on the entire region . A pole makes the double integral improper/undefined, so the Green's theorem collapse is illegal exactly there.
" because it's a nice rational function."
The flaw: has poles at , both inside . The integral is a sum of two residue contributions, not automatically .
Why questions
Why does the same-limit-from-every-direction demand of holomorphy give Cauchy's theorem, when real differentiability doesn't?
The directional agreement forces the Cauchy-Riemann equations, which make both Green's-theorem integrands vanish. Real differentiability only checks two directions, giving no such curl-free structure.
Why is "simply connected" essential rather than just cosmetic?
Green's theorem integrates over the whole enclosed region; a hole (enclosed singularity) is a point where the argument fails, and that failure is precisely what makes .
Why does the appear in the integral formula?
On the shrunken circle the factors cancel, leaving . The is literally the angle swept once around.
Why can we shrink down to a tiny circle around without changing the integral?
Between and the small circle the integrand is holomorphic (no pole there), so Cauchy's theorem makes the difference of the two loop integrals zero.
Why does the show up in the derivative formula?
Differentiating exactly times gives ; that factorial rides along when you differentiate under the integral sign times.
Why does knowing only on the boundary determine everywhere inside?
The formula uses only boundary values of . Holomorphy is rigid enough that the interior is forced — this rigidity also underlies the Maximum modulus principle.
Why is a holomorphic function's power series (its Laurent series with no negative terms) guaranteed to converge?
The generalized formula bounds every via the boundary integral, giving Cauchy estimates that force the Taylor coefficients to shrink fast enough to converge.
Edge cases
when lies exactly on (not inside, not outside)?
Undefined by the standard formula — the integrand blows up on the contour itself. One must take a principal value or indent the contour; the clean no longer holds.
when is a loop that winds around twice?
You get : the value is multiplied by the winding number, which counts how many net counter-clockwise turns makes around .
when is outside ?
Zero — no pole is enclosed, the integrand is holomorphic on the whole interior, so Cauchy's theorem applies.
The generalized formula with ?
It collapses to the ordinary integral formula, since and . The two formulas are one family.
over a loop of zero enclosed area (a contour that doubles back on itself, enclosing nothing)?
Zero — a degenerate loop enclosing no region contributes no net area to Green's theorem, so the integral vanishes trivially.
as a boundary point from inside?
The value still holds while is strictly inside, but it becomes singular in the limit — the formula degrades exactly as reaches the contour.
Two poles inside : does Cauchy's formula apply directly?
No — the formula picks off exactly one pole. With several enclosed poles you split by partial fractions or jump straight to the Residue theorem, which sums all enclosed residues.
holomorphic and bounded on all of : what does the machinery force?
The Cauchy estimates from the derivative formula drive , so is constant — this is Liouville's theorem, a direct child of the integral formula.
Recall One-line self-test before you leave
Theorem says loop = ? ::: , but only with no enclosed singularity. Formula says loop = ? ::: , harvesting the enclosed value. The single thing that breaks both when violated ::: an enclosed singularity / loss of simple connectivity.