Functions of a complex variable

Writing $f=u+iv$ and $dz=dx+i\,dy$, the contour integral splits into two real line integrals, each of which can be rewritten as a double integral over the region enclosed by $\gamma$ using Green's theorem. The two double integrals that appear are exactly $\iint (v_x - u_y)\,dA$ and $\iint(u_x - v_y)\,dA$ up to sign and a factor of $i$, and both vanish because of the Cauchy–Riemann equations $u_x=v_y$, $u_y=-v_x$.

Write $f(z) = u(x,y) + i v(x,y)$ and $dz = dx + i\,dy$ along the closed contour $\gamma = \partial D$ enclosing the region $D \subseteq U$. Expanding the product separates the contour integral into real and imaginary line integrals: $\oint_\gamma f(z)\,dz = \oint_\gamma (u\,dx - v\,dy) + i \oint_\gamma (v\,dx + u\,dy)$.

Since $u$ and $v$ have continuous partial derivatives on $U$, apply Green's theorem $\oint_{\partial D} (P\,dx + Q\,dy) = \iint_D (Q_x - P_y)\,dx\,dy$ to the real and imaginary parts separately to obtain $\oint_\gamma f(z)\,dz = \iint_D (-v_x - u_y)\,dx\,dy + i \iint_D (u_x - v_y)\,dx\,dy$.

Because $f$ is holomorphic, the Cauchy–Riemann equations $u_x = v_y$ and $u_y = -v_x$ hold at every point of $D$. Thus $-v_x - u_y = 0$ and $u_x - v_y = 0$ identically on the region, which gives $\oint_\gamma f(z)\,dz = 0 + i\cdot 0 = 0$.

Apply Cauchy's integral theorem to $g(z) = \dfrac{f(z)}{z-a}$ on the region between $\gamma$ and a tiny circle of radius $\varepsilon$ around $a$ (where $g$ is holomorphic, since $z=a$ is excluded): the two contour integrals of $g$ must agree. As $\varepsilon \to 0$, the integral over the tiny circle tends to $f(a)\cdot 2\pi i$, since $f$ is continuous and nearly constant, equal to $f(a)$, on a shrinking circle around $a$. This single formula shows that the values of a holomorphic function inside a region are completely determined by its values on the boundary.

Fix $a$ strictly inside $\gamma$, and choose $\varepsilon > 0$ small enough that the circle $C_\varepsilon = \{z : |z - a| = \varepsilon\}$ lies entirely inside the contour. Since $g(z) = \dfrac{f(z)}{z - a}$ is holomorphic on the region between the outer contour and the small circle, Cauchy's integral theorem gives $\oint_\gamma \dfrac{f(z)}{z - a}\,dz = \oint_{C_\varepsilon} \dfrac{f(z)}{z - a}\,dz$.

Parameterize the small circle by $z = a + \varepsilon e^{i\theta}$ for $\theta \in [0, 2\pi]$, so $dz = i\varepsilon e^{i\theta}\,d\theta$ and $\oint_{C_\varepsilon} \dfrac{1}{z - a}\,dz = \int_0^{2\pi} i\,d\theta = 2\pi i$. Splitting the numerator as $f(z) = f(a) + (f(z) - f(a))$ yields $\oint_{C_\varepsilon} \dfrac{f(z)}{z - a}\,dz = 2\pi i\,f(a) + \oint_{C_\varepsilon} \dfrac{f(z) - f(a)}{z - a}\,dz$.

Bounding the remainder integral by arc length gives $\left|\oint_{C_\varepsilon} \dfrac{f(z) - f(a)}{z - a}\,dz\right| \le \dfrac{\max_{C_\varepsilon}|f(z) - f(a)|}{\varepsilon}\cdot 2\pi\varepsilon = 2\pi \max_{C_\varepsilon}|f(z) - f(a)|$. Because the function is continuous at the center, as $\varepsilon \to 0$ we have $\max_{C_\varepsilon}|f(z) - f(a)| \to 0$, so the remainder vanishes and $f(a) = \dfrac{1}{2\pi i}\oint_\gamma \dfrac{f(z)}{z - a}\,dz$.

Cauchy's integral formula gives an estimate for the derivatives of a holomorphic function in terms of its size on a large circle: $|f'(z_0)| \le \dfrac{M}{R}$, where $M$ bounds $|f|$ on a circle of radius $R$ around $z_0$. If $f$ is entire and bounded by $M$ everywhere, this estimate holds for every $R$, however large; letting $R \to \infty$ forces $f'(z_0)=0$ at every point $z_0$, so $f$ is constant.

Suppose $f : \mathbb{C} \to \mathbb{C}$ is entire and bounded, so there exists a constant $M \ge 0$ such that $|f(z)| \le M$ for all $z \in \mathbb{C}$. Fix an arbitrary point $z_0 \in \mathbb{C}$ and any radius $R > 0$, and consider the circle $C_R = \{z : |z - z_0| = R\}$.

By Cauchy's integral formula for the first derivative, $f'(z_0) = \dfrac{1}{2\pi i}\oint_{C_R} \dfrac{f(z)}{(z - z_0)^2}\,dz$. Since the circle has arc length $2\pi R$ and satisfies $|z - z_0|^2 = R^2$ along the contour, the standard integral estimate yields $|f'(z_0)| \le \dfrac{1}{2\pi}\cdot \dfrac{M}{R^2}\cdot 2\pi R = \dfrac{M}{R}$.

Because the function is entire, this bound holds for every radius; letting $R \to \infty$ forces $f'(z_0) = 0$ at every point, so the function is constant.

Deform $\gamma$ into $k$ tiny circles, one around each singularity $z_j$, connected by thin corridors that cancel in pairs; by Cauchy's integral theorem applied to the (holomorphic) region between $\gamma$ and these circles, the integral over $\gamma$ equals the sum of the integrals over the small circles. Each small circle integral picks out exactly $2\pi i$ times the coefficient $c_{-1}$ of the Laurent series at $z_j$ — the residue — because $\oint (z-z_j)^n\,dz = 0$ for every power except $n=-1$, where it equals $2\pi i$.

Surround the isolated singularities with small disjoint positively oriented circles $C_1, \dots, C_k$ centered at $z_1, \dots, z_k$ of radii $\varepsilon_j > 0$ inside $\gamma$. Because the function is holomorphic on the multiply connected region between the outer contour and the small circles, Cauchy's integral theorem gives $\oint_\gamma f(z)\,dz = \sum_{j=1}^k \oint_{C_j} f(z)\,dz$.

On the punctured neighborhood of $z_j$, expand the function into its Laurent series $f(z) = \sum_{n=-\infty}^{\infty} c_{j,n}(z - z_j)^n$, which converges uniformly on $C_j$. Integrating term by term around the small circle yields $\oint_{C_j} f(z)\,dz = \sum_{n=-\infty}^{\infty} c_{j,n} \oint_{C_j} (z - z_j)^n\,dz$.

Using the parameterization $z = z_j + \varepsilon_j e^{i\theta}$, direct calculation gives $\oint_{C_j} (z - z_j)^n\,dz = 0$ for every integer $n \ne -1$ and $\oint_{C_j} (z - z_j)^{-1}\,dz = 2\pi i$. Hence only the residue term survives on each circle, giving $\oint_{C_j} f(z)\,dz = 2\pi i\,c_{j,-1} = 2\pi i \operatorname{Res}_{z=z_j} f(z)$, and summing over all circles yields $\oint_\gamma f(z)\,dz = 2\pi i \sum_{j=1}^k \operatorname{Res}_{z=z_j} f(z)$.