Holomorphic Functions and Cauchy's Theorem

Complex differentiability replaces a real limit by a complex one, a small change in the definition that is enormously stronger. The difference quotient must converge as the increment approaches zero from every direction in the plane, and this forces a rigidity with no real analogue, that a function differentiable once is differentiable infinitely often and equals its own power series. The source of the rigidity is a closed contour integral that always vanishes, and from it the value of a function inside a region is determined by its values on the boundary. This post proves that chain of results, the Cauchy-Riemann equations, Goursat's theorem, and the Cauchy integral formula, building on the real differentiation and integration of the analysis track [1], [2].

#Holomorphic functions

Definition1

A function $f$ on an open set $\Omega\subseteq\C$ is holomorphic at $z_0$ when

f'(z_0)=\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} \tag{1}

exists as a complex limit, $h\in\C$ . It is holomorphic on $\Omega$ when holomorphic at every point, and entire when $\Omega=\C$ .

Writing $f=u+iv$ in real and imaginary parts and $z=x+iy$ , the limit must agree whether $h$ is real or imaginary, and that single demand couples the partials.

Proposition2

If $f=u+iv$ is holomorphic, then the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$ hold, and $f'=u_x+iv_x$ .

Proof

Take $h$ real in Equation (1). The limit is the partial in $x$ , $f'(z_0)=u_x+iv_x$ . Take $h=ik$ with $k$ real. Then $\frac{f(z_0+ik)-f(z_0)}{ik}=\frac1i(u_y+iv_y)=v_y-iu_y$ , the partial in $y$ divided by $i$ . The two expressions for $f'(z_0)$ must agree, so $u_x+iv_x=v_y-iu_y$ , and matching real and imaginary parts gives $u_x=v_y$ and $u_y=-v_x$ .

The Cauchy-Riemann equations say the gradient of $v$ is the gradient of $u$ rotated by a right angle, the geometric fact behind conformality where the derivative is nonzero. Their deeper consequence is the vanishing of contour integrals.

#Goursat's theorem

A contour integral $\int_\gamma f(z)\,dz$ of a continuous $f$ along a piecewise smooth path $\gamma$ is the Riemann integral $\int f(\gamma(t))\gamma'(t)\,dt$ , and it is bounded by the length of $\gamma$ times the maximum of $\abs f$ . The key fact is that it vanishes around the boundary of a triangle.

Theorem3

If $f$ is holomorphic on an open set containing a solid triangle $\Delta$ , then $\int_{\partial\Delta}f(z)\, dz=0$ .

Proof

Let $I=\int_{\partial\Delta}f$ . Join the midpoints of the sides to split $\Delta$ into four congruent sub-triangles. Summing their boundary integrals, the interior edges are traversed twice in opposite directions and cancel, so $I$ is the sum of the four, and at least one sub-triangle $\Delta_1$ has $\abs{\int_{\partial\Delta_1}f}\ge\abs I/4$ . Repeating produces nested triangles $\Delta\supseteq\Delta_1 \supseteq\cdots$ with $\abs{\int_{\partial\Delta_n}f}\ge\abs I/4^n$ , while the diameter and perimeter halve, $\operatorname{diam}\Delta_n=2^{-n}\operatorname{diam}\Delta$ and $\operatorname{len}\partial\Delta_n=2^{-n} \operatorname{len}\partial\Delta$ . The nested compact triangles with shrinking diameter intersect in a single point $z^\ast$ by completeness. Holomorphy at $z^\ast$ writes

f(z)=f(z^\ast)+f'(z^\ast)(z-z^\ast)+\psi(z)(z-z^\ast),\qquad\psi(z)\to 0\ \text{as }z\to z^\ast. \tag{2}

Here $\psi(z):=[f(z)-f(z^\ast)]/(z-z^\ast)-f'(z^\ast)$ for $z\ne z^\ast$ and $\psi(z^\ast):=0$ , continuous throughout (at $z^\ast$ by the definition of $f'(z^\ast)$ ), so $\psi(z)(z-z^\ast)$ is a continuous integrand on $\partial\Delta_n$ and the only property used downstream is $\sup_{\partial\Delta_n}\abs\psi\to 0$ . The first two terms have the primitive $f(z^\ast)z+\tfrac12 f'(z^\ast)(z-z^\ast)^2$ , so they integrate to $0$ around the closed $\partial\Delta_n$ by the fundamental theorem for contour integrals. Hence $\int_{\partial\Delta_n}f=\int_{\partial\Delta_n}\psi(z)(z-z^\ast)\,dz$ . Since $z^\ast$ lies in every closed $\Delta_n$ , any $z\in\partial\Delta_n$ has $\abs{z-z^\ast}\le\operatorname{diam}\Delta_n$ , so the integrand is bounded by $\varepsilon_n\operatorname{diam}\Delta_n$ and the integral in modulus by $\varepsilon_n \operatorname{diam}\Delta_n\cdot\operatorname{len}\partial\Delta_n$ , where $\varepsilon_n=\sup_{\partial\Delta _n}\abs\psi$ . Given $\delta>0$ , holomorphy gives $r>0$ with $\abs\psi<\delta$ for $0<\abs{z-z^\ast}<r$ , and $\operatorname{diam}\Delta_n=2^{-n}\operatorname{diam}\Delta\to 0$ forces $\partial\Delta_n$ inside that ball for large $n$ , so $\varepsilon_n\to 0$ . Combining with the lower bound, $\abs I/4^n\le\varepsilon_n\,4^{-n}\operatorname{diam} \Delta\cdot\operatorname{len}\partial\Delta$ , so $\abs I\le\varepsilon_n\operatorname{diam}\Delta\cdot \operatorname{len}\partial\Delta\to 0$ , forcing $I=0$ .

On a convex open set Goursat gives more. Fixing a base point $z^\ast$ and setting $F(z)=\int_{[z^\ast,z]}f$ along the straight segment, convexity puts the solid triangle $(z^\ast,z,z+h)$ inside the set, so Goursat applies and gives $F'(z)=f(z)$ . Thus $f$ has a primitive there and $\int_\gamma f=0$ for every closed $\gamma$ by the fundamental theorem for contour integrals. This is the form of Cauchy's theorem used below. The vanishing of the integral is what makes a holomorphic function recoverable from a boundary.

#The Cauchy integral formula

Theorem4

If $f$ is holomorphic on an open set containing the closed disk of radius $r$ about $z_0$ , then for the positively oriented circle $C_r=\{\abs{z-z_0}=r\}$ ,

f(z_0)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-z_0}\,dz. \tag{3}

Proof

The integrand is holomorphic on the disk except at $z_0$ . Cut the annulus between $C_r$ and a small circle $C_\rho$ of radius $\rho$ about $z_0$ along a thin keyhole slit; the resulting closed slit annulus is a region on which the integrand is holomorphic, and it triangulates into finitely many triangles. Goursat makes the integral around each triangle vanish, and summing them the interior edges cancel in opposite pairs, leaving the integral around the slit annulus boundary. As the slit width tends to $0$ the two slit traversals cancel, so the integral over $C_r$ equals the integral over $C_\rho$ . Split it,

\int_{C_\rho}\frac{f(z)}{z-z_0}\,dz=f(z_0)\int_{C_\rho}\frac{dz}{z-z_0}+\int_{C_\rho}\frac{f(z)-f(z_0)}{z-z_0 }\,dz. \tag{4}

Parametrising $z=z_0+\rho e^{i\theta}$ gives $\int_{C_\rho}\frac{dz}{z-z_0}=\int_0^{2\pi}\frac{i\rho e^{i \theta}}{\rho e^{i\theta}}\,d\theta=2\pi i$ . The second integral has integrand bounded by $\sup_{\abs{z-z_0} =\rho}\abs{f(z)-f(z_0)}/\rho$ over a circle of length $2\pi\rho$ , hence by $2\pi\sup\abs{f(z)-f(z_0)}$ , which tends to $0$ as $\rho\to 0$ by continuity of $f$ . So the left side is $2\pi i\,f(z_0)$ , independent of $\rho$ , and equals the integral over $C_r$ , giving Equation (3).

The formula yields at once infinite differentiability, analyticity, and the Cauchy estimates. The same keyhole argument applied to $f(z)/(z-\zeta)$ , holomorphic off $\zeta$ , gives the off-center form $f(\zeta)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-\zeta}\,dz$ for every $\zeta$ with $\abs{\zeta-z_0}<r$ , and the consequences follow from it. Differentiate in $\zeta$ through the difference quotient,

\frac{f(\zeta+h)-f(\zeta)}{h}=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{(z-\zeta-h)(z-\zeta)}\,dz. \tag{5}

For $\abs h<\tfrac12\operatorname{dist}(\zeta,C_r)=:d/2$ the kernel is bounded on $C_r$ by $1/(\tfrac{d}{2}d)$ and converges uniformly there to $(z-\zeta)^{-2}$ as $h\to 0$ , so the limit passes inside the integral, giving $f'(\zeta)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{(z-\zeta)^2}\,dz$ . Differentiating the order- $n$ formula by the same difference quotient, with the kernel now converging uniformly to $(z-\zeta)^{-(n+1)}$ , an identical induction on $n$ yields the derivatives of every order, $f^{(n)}(\zeta)=\frac{n!}{2\pi i}\int_{C_r}\frac{f(z)}{(z-\zeta)^{n+1}}\,dz$ . For $z$ on $C_r$ and $\abs{\zeta-z_0}<r$ expand the kernel, $\frac{1}{z-\zeta}=\frac{1}{(z-z_0)-(\zeta-z_0)}=\sum_{n\ge 0}\frac{(\zeta-z_0)^n}{(z-z_0)^{n+1}}$ , where $\abs{(\zeta-z_0)/(z-z_0)}=\abs{\zeta-z_0}/r<1$ is constant on $C_r$ , so the series converges uniformly in $z$ and may be integrated term by term, giving $f(\zeta)=\sum_{n\ge 0}\frac{f^{(n)}(z_0)}{n!}(\zeta-z_0)^n$ . Thus $f$ equals its Taylor series on every disk in $\Omega$ , so holomorphic functions are analytic. The integral bounds the derivatives by the Cauchy estimate $\abs{f^{(n)}(z_0)}\le n!\,M/r^n$ with $M=\max_{C_r}\abs f$ , and from it the rigidity of entire functions follows.

Theorem5

A bounded entire function is constant.

Proof

If $\abs f\le M$ everywhere, the Cauchy estimate for the first derivative on a circle of radius $r$ about any $z_0$ gives $\abs{f'(z_0)}\le M/r$ . Since $f$ is entire the estimate holds for every $r$ , and letting $r\to\infty$ gives $f'(z_0)=0$ . As $z_0$ was arbitrary, $f'\equiv 0$ , so $f$ is constant.

Liouville's theorem is the one-line proof of the fundamental theorem of algebra, since a nonvanishing polynomial would make $1/p$ a bounded entire function and hence constant. These rigidity phenomena, the maximum modulus principle, determination by boundary values, and the identity theorem, all flow from the vanishing contour integral of Goursat's theorem. The residue calculus of the next post turns that same vanishing into a method for evaluating integrals that real methods cannot reach, including the inversion integral of a characteristic function.

[1]

L. V. Ahlfors, Complex Analysis, 3rd ed. McGraw-Hill, 1979.

[2]

E. M. Stein and R. Shakarchi, Complex Analysis. in Princeton Lectures in Analysis. Princeton University Press, 2003.

Explore connections

see in the atlas

referenced by (1)

Residues and Contour Integration

cite

@misc{holomorphic-functions-and-cauchy,
  author = {Zac Kienzle},
  title  = {Holomorphic Functions and Cauchy's Theorem},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/holomorphic-functions-and-cauchy}
}

#Holomorphic functions

Definition1

A function $f$ on an open set $\Omega\subseteq\C$ is holomorphic at $z_0$ when

f'(z_0)=\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} \tag{1}

exists as a complex limit, $h\in\C$ . It is holomorphic on $\Omega$ when holomorphic at every point, and entire when $\Omega=\C$ .

Writing $f=u+iv$ in real and imaginary parts and $z=x+iy$ , the limit must agree whether $h$ is real or imaginary, and that single demand couples the partials.

Proposition2

If $f=u+iv$ is holomorphic, then the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$ hold, and $f'=u_x+iv_x$ .

Proof

#Goursat's theorem

Theorem3

If $f$ is holomorphic on an open set containing a solid triangle $\Delta$ , then $\int_{\partial\Delta}f(z)\, dz=0$ .

Proof

f(z)=f(z^\ast)+f'(z^\ast)(z-z^\ast)+\psi(z)(z-z^\ast),\qquad\psi(z)\to 0\ \text{as }z\to z^\ast. \tag{2}

#The Cauchy integral formula

Theorem4

If $f$ is holomorphic on an open set containing the closed disk of radius $r$ about $z_0$ , then for the positively oriented circle $C_r=\{\abs{z-z_0}=r\}$ ,

f(z_0)=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{z-z_0}\,dz. \tag{3}

Proof

\int_{C_\rho}\frac{f(z)}{z-z_0}\,dz=f(z_0)\int_{C_\rho}\frac{dz}{z-z_0}+\int_{C_\rho}\frac{f(z)-f(z_0)}{z-z_0 }\,dz. \tag{4}

\frac{f(\zeta+h)-f(\zeta)}{h}=\frac{1}{2\pi i}\int_{C_r}\frac{f(z)}{(z-\zeta-h)(z-\zeta)}\,dz. \tag{5}

Theorem5

A bounded entire function is constant.

Proof

[1]

L. V. Ahlfors, Complex Analysis, 3rd ed. McGraw-Hill, 1979.

[2]

E. M. Stein and R. Shakarchi, Complex Analysis. in Princeton Lectures in Analysis. Princeton University Press, 2003.

Explore connections

see in the atlas

referenced by (1)

Residues and Contour Integration

cite

@misc{holomorphic-functions-and-cauchy,
  author = {Zac Kienzle},
  title  = {Holomorphic Functions and Cauchy's Theorem},
  year   = {2026},
  month  = {06},
  url    = {https://zackienzle.com/blog/holomorphic-functions-and-cauchy}
}