Cauchy's integral theorem and formula

1. Setup: what objects are we even talking about?

WHY this matters: real differentiability only checks left/right. Complex differentiability demands the same limit from every direction — a far stronger constraint. That extra rigidity is the engine behind every theorem below.

2. The Cauchy–Riemann equations (where rigidity is born)

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

Approach the limit along the real axis ($h=\Delta x$ real): $f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$

Approach along the imaginary axis ($h=i\,\Delta y$): $f'(z)=\frac{1}{i}\Big(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\Big)=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}$

These must be equal, so matching real and imaginary parts:

WHY this is the key lemma: these equations are exactly the condition that makes the contour integral curl-free, which is what kills the loop integral.

3. Deriving Cauchy's Theorem from scratch

HOW we derive it (via Green's theorem — the honest route):

Write $f=u+iv$, $dz=dx+i\,dy$. Then $\oint_\gamma f\,dz=\oint_\gamma (u+iv)(dx+i\,dy)=\underbrace{\oint (u\,dx-v\,dy)}_{\text{real part}}+i\underbrace{\oint (v\,dx+u\,dy)}_{\text{imag part}}$

Apply Green's theorem $\oint (P\,dx+Q\,dy)=\iint_R\big(\partial_x Q-\partial_y P\big)\,dA$ to each:

• Real part: $P=u,\ Q=-v$ gives $\iint(-v_x-u_y)\,dA$.
• Imag part: $P=v,\ Q=u$ gives $\iint(u_x-v_y)\,dA$.

WHY "simply connected" is non-negotiable: Green's theorem needs $f$ holomorphic on the whole region $R$ enclosed by $\gamma$. One hole (a point where $f$ blows up) and the argument breaks — which is precisely why the next formula is nonzero.

4. Cauchy's Integral Formula

HOW we derive it: The integrand $\dfrac{f(z)}{z-a}$ is holomorphic everywhere inside $\gamma$ except at $z=a$. Deform $\gamma$ to a tiny circle $C_\varepsilon$ of radius $\varepsilon$ centred at $a$ (legal, because in the region between them the integrand is holomorphic → Cauchy's theorem makes the difference vanish): $\oint_\gamma \frac{f(z)}{z-a}\,dz=\oint_{C_\varepsilon}\frac{f(z)}{z-a}\,dz.$

On $C_\varepsilon$ put $z=a+\varepsilon e^{i\theta}$, $dz=i\varepsilon e^{i\theta}\,d\theta$: $\oint_{C_\varepsilon}\frac{f(z)}{z-a}\,dz=\int_0^{2\pi}\frac{f(a+\varepsilon e^{i\theta})}{\varepsilon e^{i\theta}}\,i\varepsilon e^{i\theta}\,d\theta = i\int_0^{2\pi} f(a+\varepsilon e^{i\theta})\,d\theta.$

Let $\varepsilon\to 0$; by continuity $f(a+\varepsilon e^{i\theta})\to f(a)$: $= i\cdot 2\pi\, f(a)=2\pi i\,f(a).$

Divide by $2\pi i$. Done.

5. Worked Examples

6. Common Mistakes (Steel-manned)

7. Flashcards

Connections

• Cauchy-Riemann equations — the lemma that powers everything.
• Green's theorem — the bridge from contour to area integral.
• Residue theorem — the grand generalization for many poles.
• Laurent series — how the $1/(z-a)^{n+1}$ terms arise.
• Liouville's theorem — bounded entire functions are constant, a direct corollary.
• Maximum modulus principle — another rigidity consequence.

Concept Map

Hinglish (regional understanding)