Cauchy's Integral Theorem

Integration along curves

Parameterized Curve

A complex function \(z(t)\) is said to be parametrized curve, if it maps a closed interval \([a,b]\subset \mathbb{R}\) to \(D\subset \mathbb{C}\).

A parametrized curve is said to be smooth, if \(z'(t)\) exists, \(z'(t)\neq 0, \forall t\in [a,b]\) and is continuous on \([a,b]\). At point \(t=a,b\), we intepret continuity and differentiability as one-side limits. (here readers could just view \(z\) as a function vector with respect to variable \(t\)). Similar situation for piecewise-smooth.

Equivalence of parametrized curve

Two parametrized curves \(z:[a,b]\mapsto \mathbb{C}\) and \(\tilde{z}:[c,d]\mapsto \mathbb{C}\) are said to ba equivalent, if there exsits a continuously differentiable bijection \(t\) \((t'(s)>0)\) which maps \([c,d]\) to \([a,b]\), such that

\[ \tilde{z}(s)=z(t(s)). \]

The above equivalence defines an equivalent family of \(z(t)\), the difference of which are the definition domains, varying among any closed interval in \(\mathbb{R}\). They are all the same in the range, so we denote the whole of them as \(\gamma\subset \mathbb{C}\), called smooth curve. Thus, a smooth curve could determine incountable parametrized curves.

The condition \((t'(s)>0)\) guarantee the direction of the curve. Later we would see that it determines the sign of integral.

Reversing the direction

For a given smooth curve \(\gamma\), and its one parametrized curve \(z:[a,b]\mapsto \mathbb{C}\), we could obtain one of its reverse curve \(\gamma^-\), by

\[ z^-=z(b+a-t). \]

so \(z^-:[a,b]\mapsto \mathbb{R}\). In fact, we could formulate infinite number of these equivalent curve, while the above one guarantees the definition domain is kept same as \(z\).

Simple curve, closed curve

A smooth curve is said to be simple, if it is not self-intersecting. Or in mathematical language, \(z(t)\neq z(s)\) unless \(s=t\).

A smooth curve is said to be closed, if \(z(a)=z(b)\) for any of its parametrizations.

For the following parts, we call the above piecewise-smooth curve a curve for brevity.

Example. Some examples of curves.

(i) Circle. Set like \(C_r(z_0)=\{z\in\mathbb{C}:|z-z_0|=r\}.\) are called circle, with its usual parametrization (positive and negative orientation respectively)

\[ z(t)=z_0+re^{i\theta},\quad t\in [0,2\pi], \]

\[ z(t)=z_0+re^{-i\theta},\quad t\in [0,2\pi]. \]

In the following chapters, we denote \(C\) as a general positive oriented circle.

Riemann integral of complex function f

Assume complex function \(f\) is defined on \(\gamma\), where \(\gamma\) is not necessarily smooth, but must be length-calculatable. For one parametrization \(z:[a,b]\mapsto \mathbb{C}\), we have partition

\[ \Delta_n: z(a)=z_0<z_1<\cdots<z_n=z(b) \]

Choose an arbitrary point \(\xi_k\) on arc from \(z_{k-1}\) to \(z_k\), \((k=1,\cdots,n)\) we have Riemann summation

\[ \sum_{k=1}^n f(\xi_k)(z_k-z_{k-1}). \]

Denote \(s_k\) to be the length of arc from \(z_{k-1}\) to \(z_k\). If \(\lambda=\max\{s_k: 1\leq k\leq n\}\rightarrow 0\), the above summation has a unique limit, then we call complex function \(f\) Riemann-integrable. And we denote the integral of \(f\) along \(\gamma\) as

\[ \int_\gamma f(z)dz=\lim_{\lambda\rightarrow 0}\sum_{k=1}^n f(\xi_k)(z_k-z_{k-1}). \]

Sufficient condition for Riemann integrability

If complex function \(f=u+iv\) is continuous on \(\gamma\) which is length-integrable, then it is Riemann-integrable on \(\gamma\), and

\[ \int_\gamma f(z)dz=\int_\gamma (u dx-vdy)+i\int_\gamma (vdx+udy). \]

Proof

Just compose complex number into two real parts. Denote \(z_k=x_k+iy_k\), \(\xi_k=\eta_k+i\zeta_k\), \(\Delta x_k=x_k-x_{k-1}, \Delta y_k=y_k-y_{k-1}\) then \(f(\xi_k)=u(\eta_k,\zeta_k)+iv(\eta_k,\zeta_k)\), so Riemann summation

\[ \begin{align*} \sum_{k=1}^n f(\xi_k)(z_k-z_{k-1})=&\sum_{k-1}^n [u(\eta_k,\zeta_k)+iv(\eta_k,\zeta_k)]\cdot [(x_k-x_{k-1})+i(y_k-y_{k-1})]\\ =&\sum_{k=1}^n\{ [u(\eta_k,\zeta_k)\Delta x_k-v(\eta_k,\zeta_k)\Delta y_k]+\\ &i [u(\eta_k,\zeta_k)\Delta y_k+v(\eta_k,\zeta_k)\Delta x_k]\} \end{align*} \]

when \(u, v\) is continuous on \(\gamma\), the above summation converges to the needed formula.

The above formula is easy to know by

\[ f(z)dz=(u+iv)(dx+idy)=udx-vdy+i(udy+vdx). \]

Integral along a smooth curve

Assume smooth curve \(\gamma\subset \mathbb{C}\) parametrized by \(z: [a,b]\mapsto \mathbb{C}\), and \(f\) is a continuous function on \(\gamma\). Then the integral of \(f\) along \(\gamma\)

\[ \int_\gamma f(z)dz=\int_a^b f(z(t))z'(t)dt. \]

Proof

The logic is simple, even a little rude: Parametrize the four real integral along \(\gamma\) and then sum them up. That is, \(f=u+iv\), \(z=x+iy\), then \(z'(t)=x'(t)+iv'(t)\).

Since the real and imaginary parts of integral are

\[ \int_\gamma udx-vdy=\int_a^b [u(x(t),y(t))x'(t)-v(x(t),y(t))y'(t)]dt \]

\[ \int_\gamma vdx+udy=\int_a^b [v(x(t),y(t))x'(t)+u(x(t),y(t))y'(t)]dt \]

we have

\[ \int_\gamma f(z)dz=\int_a^b [u(x(t),y(t))+iv(x(t),y(t))][x'(t)+iy(t)]dt=\int_a^b f(z(t))z'(t)dt. \]

Example. Calculate

(i)

\[ \int_\gamma dz=z(b)-z(a)\neq \int_a^b |z'(t)|dt=\text{length}(\gamma) \]

(ii)

\[ \int_\gamma \frac{dz}{(z-z_0)^n}, \]

where \(n\in \mathbb{Z}\), \(\gamma=C_r(z_0)\).

Answer for (ii)

(i) \(n=1\).

(ii) \(n\neq 1\).

Properties for Integral of complex function

If complex function \(f\) and \(g\) is continuous on \(\gamma\) which is length-calculatable, then

(i) \(\gamma^-\) is the reverse of \(\gamma\),

\[ \int_{\gamma^-}f(z)dz=-\int_{\gamma}f(z)dz. \]

(ii) For all \(\alpha, \beta\in \mathbb{C}\),

\[ \int_{\gamma} (\alpha f(z)+\beta g(z))dz=\alpha\int_{\gamma}f(z)dz+\beta \int_{\gamma}g(z)dz. \]

(iii) Inequation for absolute estimation

\[ \left|\int_\gamma f(z)dz\right|\leq \sup_{z\in \gamma}|f(z)|\cdot \text{length}(\gamma) \]

Cauchy's Theorem

Definition of Primitive function

Assume \(f\) is a function on the open set \(\Omega\). A Primitive for \(f\) on \(\Omega\) is a function \(F\) that is holomorphic on \(\Omega\), such that \(F'(z)=f(z),\forall z\in \Omega\).

Cauchy's Theorem

If a continuous function \(f\) has a primitive \(F\) on open set \(\Omega\), and \(\gamma\) is a piecewise-smooth curve begins at \(\omega_1\) and ends at \(\omega_2\), then

\[ \int_\gamma f(z)dz=F(\omega_2)-F(\omega_1). \]

Proof

(i) For \(\gamma\) is smooth.

Choose a parametrization \(z: [a,b]\mapsto \mathbb{C}\) of \(\gamma\), such that \(z(a)=\omega_1, z(b)=\omega_2\).

\[ \begin{align*} \int_\gamma f(z)dz&=\int_a^b f(z(t))z'(t)dt\\ &=\int_a^b F'(z(t))z'(t)dt\\ &=\int_a^b \frac{d}{dt}F(z(t))dt\\ &=F(z(b))-F(z(a))=F(\omega_2)-F(\omega_1). \end{align*} \]

where the third "\(=\)" is by chain rule, and the fouth "\(=\)" is by fundamental theorem of calculas.

(ii) For \(\gamma\) is piecewise-smooth, just add these smooth part up.

The condition and result of the above theorem could be replaced by the following statement.

Another language

\(F\) is holomorphic and \(F'\) is continuous on \(\Omega\), then

\[ \int_\gamma F'(z)dz=F(\omega_2)-F(\omega_1). \]

Corollary: Cauchy's closed curve theorem

If \(\gamma\) is a closed smooth curve in an open set \(\Omega\), and \(f\) is continuous and has a primitive in \(\Omega\), then

\[ \int_\gamma f(z)dz=0. \]

The condition and result of the above theorem could be replaced by the following statement.

Another language

\(F\) is holomorphic and \(F'\) is continuous on \(\Omega\), then

\[ \int_\gamma F'(z)dz=0. \]

Corollary 2: zero-derivatives means constant function

If \(f\) is holomorphic in a region \(\Omega\), and \(f'=0\), then \(f\) is a constant.

Proof

\(\forall \omega_0\in \Omega\), \(\forall \omega\),

\[ f(\omega)-f(\omega_0)=\int_\gamma f'(z)dz=0. \]

so \(f(\omega)=f(\omega_0)\).

Actually the above theorems also hold for general curves.

Goursat Theorem

Goursat's Theorem

Assume \(\Omega\) is an open set in \(\mathbb{C}\). If \(f\in H(\Omega)\), then for any a triangle \(T\subset \Omega\) whose interior is also contained in \(\Omega\),

\[ \int_T f(z)dz=0. \]

Proof

Bisect \(\mathcal{T}^0\), we have a sequence of triangles

\[ \mathcal{T}^0\supset \mathcal{T}^1\supset\cdots \]

use nested compact set theorem to get a point \(z_0\) in these subtriangles.

Use properties of holomorphic function,

\[ f(z)=f(z_0)+f'(z_0)(z-z_0)+\psi(z)(z-z_0) \]

where \(\lim\limits_{z\rightarrow z_0}\psi(z)=0\). The first and second items of the rightside equation equals zero after integrating over a closed curve, while last item is bounded by a small \(\varepsilon_n=\sup_{z\in T^{n}}|\psi(z)|\). That is,

\[ \left|\int_{T^n}fdz\right|=\left|\int_{T^n}\psi (z-z_0) dz\right|\leq \varepsilon_n d^{(n)}p^{(n)}. \]

where \(d^{(n)}, p^{(n)}\) denote the diameter and perimeter of the \(n\)th triangles. Each time after bisection, we have

\[ d^{(n+1)}\leq \frac{1}{2}d^{(n)},\quad p^{(n+1)}\leq \frac{1}{2}p^{(n)} \]

and the integral along \(T^{(n)}\), must hold for some \(j\in \{0,1,2,3\}\)

\[ \left|\int_{T^{(n)}} f(z)dz\right|\leq 4\left|\int_{T^{(n+1)}_j} f(z)dz\right| \]

So

\[ \begin{align*} \left|\int_{T^{(0)}} f(z)dz\right|&\leq 4^n \left|\int_{T^{(n)}_j} f(z)dz\right|\\ &\leq 4^n \varepsilon_n d^{(n)}p^{(n)}\\ &\leq 4^n \varepsilon_n 4^{-n} d^{(0)}p^{(0)}=\varepsilon_n\rightarrow 0(n\rightarrow \infty) \end{align*} \]

Corollary: integral in Rectangle curve

Assume \(f\in H(\Omega)\). For any rectangle \(R\subset \Omega\) whose interior is also contained in \(\Omega\),

\[ \int_T f(z)dz=0. \]

Local existence of Primitive

By constructing method.

A holomorphic function has a primitive

Assume \(f\) is a holomorphic function in an open disc, then \(f\) has a primitive in the disc.

Actually we use Goursat's Theorem and continuity of \(f\), without utilizing Holomorphism of \(f\). That is, a continuous function on an open set must have a primitive, if it satisfies that its integral on every triangle on \(\Omega\) equals zero. In the following parts, we will show that this equals to holomorphic.

Proof

Define

\[ F(z)=\int_{\gamma_z} f(\omega)d\omega, \]

where \(\gamma_z\) is a polygomal line from \(0\) to \(z\).

by Goursat's Theorem, we can show that

\[ \begin{align} F(z+h)-F(z)=\int_\eta f(\omega)d\omega,\label{primitive} \end{align} \]

where \(\eta\) is a straight line from \(z\) to \(z+h\).

Then use the continuity of \(f\) in \(\Omega\), i.e at point \(z\),

\[ f(\omega)=f(z)+\psi(\omega) \]

where \(\lim\limits_{\omega\rightarrow z}\psi(\omega)=0\). Substitute in equation \(\ref{primitive}\), we have

\[ \begin{align*} F(z+h)-F(z)&=f(z)\int_\eta d\omega + \int_\eta \psi(\omega)d\omega.\\ \frac{F(z+h)-F(z)}{h}&=f(z)+\frac{\int_\eta \psi(\omega)d\omega}{h} \end{align*} \]

the latter integral goes to \(0\) as \(h\) tend to \(0\) because

\[ \left|\frac{\int_\eta \psi(\omega)d\omega}{h} \right|\leq \frac{\sup\limits_{\omega\in \eta}|\psi(\omega)|\cdot |h|}{|h|} \]

So

\[ \lim_{h\rightarrow 0}\frac{F(z+h)-F(z)}{h}=f(z). \]

Corollary: Cauchy's Theorem for a disc

Assume \(D\) is an open disc, and \(f\in H(D)\). Then for closed curve \(\gamma\),

\[ \int_\gamma f(z)dz=0 \]

Proof

By using Corollary: Cauchy's closed curve theorem.

Actually, Cauchy's Theorem could apply to any closed curve with unambiguious interior, not just a disc, whose interior is clear.

Cauchy's Integral Formula

Using the following integral formula, we could show that holomorphic function has derivatives of arbitrary order \(n\), and could be expanded by power series.

Cauchy's integral formula

Assume \(f\) is holomorphic in an open set that contains the closure of a disc \(D\). If \(C\) denotes the boundary circle of this disc with the positive orientation, then for any point \(z\in D\),

\[ f(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta-z}d\zeta. \]

Proof

We could use a "key hole" to prove. Look the following image.

Let \(F(z)=\frac{f(\zeta)}{\zeta-z}\), then it is holomorphic away from point \(z\), so

\[ \int_{\gamma} F(\zeta)d\zeta=0=\int_{C_\varepsilon^+} F(\zeta)d\zeta + \int_{\Gamma^+}F(\zeta)d\zeta +\int_{\Gamma^-}F(\zeta)d\zeta +\int_{C_r^-}F(\zeta)d\zeta. \]

Let \(\delta\rightarrow 0\), we have \(\int_{\Gamma^+}F(\zeta)d\zeta +\int_{\Gamma^-}F(\zeta)d\zeta=0\). If we rewrite

\[ F(z)=\frac{f(\zeta)-f(z)}{\zeta-z}+\frac{f(z)}{\zeta-z}, \]

and by the example commen integral, we know

\[ \int_{C_\varepsilon^+}\frac{f(z)}{\zeta-z}d\zeta=2\pi i f(z), \]

and \(\frac{f(\zeta)-f(z)}{\zeta-z}\) is bounded, whose integral on \(C_\varepsilon^+\) converges to \(0\). So

\[ \int_{C_\varepsilon^+}F(\zeta)d\zeta = 2\pi i f(z). \]

and we are done.